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Article

Expansions for the Conditional Density and Distribution of a Standard Estimate

by
Christopher S. Withers
Formerly Industrial Research Ltd., Lower Hutt 6007, New Zealand
Stats 2025, 8(4), 98; https://doi.org/10.3390/stats8040098
Submission received: 22 August 2025 / Revised: 7 October 2025 / Accepted: 11 October 2025 / Published: 14 October 2025

Abstract

Conditioning is a very useful way of using correlated information to reduce the variability of an estimate. Conditioning an estimate on a correlated estimate, reduces its covariance, and so provides more precise inference than using an unconditioned estimate. Here we give expansions in powers of n 1 / 2 for the conditional density and distribution of any multivariate standard estimate based on a sample of size n. Standard estimates include most estimates of interest, including smooth functions of sample means and other empirical estimates. We also show that a conditional estimate is not a standard estimate, so that Edgeworth-Cornish-Fisher expansions cannot be applied directly.

1. Introduction and Summary

Given correlated estimates w ^ = ( w ^ 1 , w ^ 2 ) of unknown parameters w = ( w 1 , w 2 ) , inference on w 1 can be made more precise by conditioning on w ^ 2 . To see this, suppose that X is a bivariate normal with correlation ρ 0 . Then
v a r ( X 1 | X 2 ) = ( 1 ρ 2 ) v a r ( X 1 ) < v a r ( X 1 ) .
So if w ^ is an estimate of w based on a sample of size n, and it satisfies the Central Limit Theorem, (CLT),
X n = n 1 / 2 ( w ^ w ) L X N 2 ( 0 , V ) , as n ,
then
v a r ( w ^ 1 | w ^ 2 ) ( 1 ρ 2 ) v a r ( w ^ 1 ) < v a r ( w ^ 1 )
if ρ = correlation ( w ^ 1 , w ^ 2 ) is non-zero. A similar result holds when for i = 1 , 2 , w i has dimension q i , which we now assume. To apply this result to obtain inference on w 1 given w ^ 1 | w ^ 2 , we need to approximate its distribution, ideally beyond its first approximation given by the CLT. This paper uses expansions in powers of n 1 / 2 for the density and distribution of X n = n 1 / 2 ( w ^ w ) , for a wide class of estimates, called standard estimates.
Suppose that w ^ is a standard estimate of an unknown parameter w R q of a statistical model, based on a sample of size n. That is, w ^ is a consistent estimate, and for r 1 , its rth order cumulants have magnitude n ( r 1 ) / 2 and can be expanded in powers of n 1 . The coefficients in these expansions are called the cumulant coefficients. This is a very large class of estimates, with potential application to a range of practical problems. For example, w ^ may be a smooth function of one or more sample means, or a smooth functional of one or more empirical distributions. A smooth function of a standard estimate is also a standard estimate: see ref. [1]. Ref. [2] gave the multivariate Edgeworth expansions for the distribution and density of
X n = n 1 / 2 ( w ^ w ) ,
in powers of n 1 / 2 about the multivariate normal in terms of the Edgeworth coefficients of (3). (For typos, see p25 of ref. [1]. Also replace θ ^ by θ ^ / θ on 4th to last line p1121 and in (23). To line 3 p1138, add P 12 = B 23 / 2 ). Ref. [3] gave the Edgeworth coefficients explicitly for the Edgeworth expansions to O ( n 2 ) .
Choosing an estimate can be a tradeoff between simplicity and efficiency. Conventional point estimation emphasises efficiency as measured by mean square error. The maximum liklihood estimate is attractive as it is asymptotically efficient in this sense. However their cumulant coefficients generally take much more work to obtain than those of a moment estimate. See refs. [4,5]. But whether one chooses a simple estimate or a more complicated one, it will generally be a standard estimate.
Turning to conditioning, as noted, this is a very useful way of using correlated information to reduce the variability of estimates, and to make inference on unknown parameters more precise. This is the motivation for this paper. To emphasise when w i , w ^ i are vectors, we bold them. In Section 4 we take q 2 , and write w , w ^ and X n as w 1 w 2 , w ^ 1 w ^ 2 and X n 1 X n 2 of dimensions q 1 q 2 . Just as the distribution of X n allows inference on w, the conditional distribution of X n 1 given X n 2 , allows inference on w 1 for a given w 2 . The covariance of w ^ 1 | w ^ 2 can be substantially less than that of w ^ 1 . Only when w ^ 1 and w ^ 2 are uncorrelated, is there no advantage in conditioning. Given a statistical model, its unknown parameters w, will consist of one or more parameters of primary interest, w 1 , and the others. (For example, for an autoregressive time series with mean μ , autocorrelation ρ , and variance of residuals σ 2 , the parameter of primary interest is μ .) When conditioning one can choose w 2 to be all of the other parameters, or more simply, the single parameter w 2 which maximises the estimated correlation of w ^ 2 with w ^ 1 . This is another trade-off between efficiency and simplicity, as increasing q 2 will reduce the conditional variance.
We shall see that for V the asymptotic covariance of X n ,
as n , X n = n 1 / 2 ( w ^ w ) L X N q ( 0 , V ) ,
the multivariate normal on R q , with density and distribution
ϕ V ( x ) = ( 2 π ) q / 2 ( d e t V ) 1 / 2 exp ( x V 1 x / 2 ) , Φ V ( x ) = x ϕ V ( x ) d x . So , X n 1 | ( X n 2 = x 2 ) L X 1 | ( X 2 = x 2 ) N q 1 ( μ 1 2 , V 1 2 ) ,
where V 1 2 is a function of V, and μ 1 2 is a function of V and is also linear in x 2 . If q 1 = 1 , this leads in Section 4 to 1- or 2-sided confidence intervals for w 1 | ( X n 2 = x 2 ) , of error O ( n 1 / 2 ) or O ( n 1 ) . So unlike traditional confidence regions (including confidence intervals), the conditional versions depend on the value of the unknown w 2 . This gives a new level of sophistication to them over traditional confidence regions. While this paper does not deal with Studentized estimates, that next step can be done using ref. [6] or ref. [1]. The x 2 of most interest are small.
Theorems 1 and 2 give our main results: explicit expansions to O ( n 2 ) for the conditional density and distribution of X n 1 given X n 2 , that is, for the conditional density and distribution of w ^ 1 w 1 given w ^ 2 w 2 . In other words, it gives the likely position of w 1 for any given w 2 . The main difficulty is integrating the density. Theorem 2 does this in terms of I ¯ 1 k of (42), the integral of the multivariate Hermite polynomial, with respect to the conditional normal density. Note 1 gives I ¯ 1 k in terms of derivatives of the multivariate normal distribution. Theorem 3 gives I ¯ 1 k in terms of the partial moments of the conditional distribution. If q 1 = 1 , then Theorem 4 gives I ¯ 1 k in terms of the unit normal distribution and density.
Section 4 specialises to the case q 1 = q 2 = 1 . Examples are the condtional distribution and density of a bivariate sample mean, of entangled gamma random variables, and of a sample mean given the sample variance. Section 5 and Section 6 give conclusions, discussion, and suggestions for future research. Appendix A gives expansions for the conditional moments of X n 1 | ( X n 2 = x 2 ) . It shows that w ^ 1 given X n 2 , is neither a standard estimate, nor a Type B estimate, so that Edgeworth-Cornish-Fisher expansions do not apply to it.
Ref. [7] (pp. 34–36) argue that an ideal choice of conditioning variable w ^ 2 would be one whose distribution does not depend on w 1 . But this is generally not possible except for some exponential families. An example when it is true, is when w 1 and w 2 are location and scale parameters: on p54 they essentially suggest choosing w 2 = n v a r w ^ 1 . This is our motivation for Example 4. For some examples, see ref. [8]. Their (7.5) gave a form for the 3rd order expansion for the conditional density of a sample mean to O ( n 3 / 2 ) , but they did not attempt to integrate it.
Conditional expansions for the sample mean were given in Chapter 12 of [9], and used in Sections 2.3 and 2.5 of [10] to show bootstrap consistency. For some other results on conditional distributions, see refs. [11,12,13,14,15,16].

2. Multivariate Edgeworth Expansions

Suppose that w ^ is a standard estimate of w R q with respect to n. (n is typically the sample size.) That is, E w ^ w as n , where we use E for expected value, and for r 1 and 1 i 1 , , i r q , the rth order cumulants of w ^ = ( w ^ 1 , , w ^ q ) can be expanded as
k ¯ 1 r = k i 1 i r = κ ( w ^ i 1 , , w ^ i r ) d = r 1 n d k ¯ d 1 r , where k ¯ d 1 r = k d i 1 i r ,
where ≈ indicates an asymptotic expansion, and the cumulant coefficients  k ¯ d 1 r may depend on n but are bounded as n . So the bar replaces each i k by k. For example k ¯ 0 1 = w i 1 and k ¯ 1 12 = k 1 i 1 i 2 . We reserve i k for this bar notation to avoid double subscripts. (1) holds with V = ( k ¯ 1 12 ) , q × q . V may depend on n, but we assume that d e t V is bounded away from 0.
Let P ¯ r 1 k = P r i 1 i k be the rth Edgeworth coefficient of w ^ ,
for q 1 r 3 . These are Bell polynomials in the cumulant coefficients of (2), as defined and given in [3]. Their importance lies in their central role in the Edgeworth expansions of X n of (1). (When q = 1 and w ^ is a sample mean, the Edgeworth coefficients were given for all r in [17]. For typos, see pp. 24–25 of [1].)
Set P ( A ) = Probability A is true. By [2], or [1], for w ^ non-lattice, the distribution and density of X n can be expanded as
P ( X n x ) r = 0 n r / 2 P r ( x ) , p X n ( x ) r = 0 n r / 2 p r ( x ) , x R q ,
where P 0 ( x ) = Φ V ( x ) , p 0 ( x ) = ϕ V ( x ) , and for r 1 , P r ( x ) = k = 1 3 r [ P r k ( x ) : k r even ] ,
p r ( x ) / ϕ V ( x ) = k = 1 3 r [ p ˜ r k : k r even ] = p ˜ r ( x ) say ,
P r k ( x ) = P ¯ r 1 k H ¯ 1 k , p ˜ r k = P ¯ r 1 k H ¯ 1 k ,
H ¯ 1 k = H ¯ 1 k ( x , V ) = O ¯ 1 k Φ V ( x ) = x H ¯ 1 k ϕ V ( x ) d x ,
O ¯ 1 k = ( ¯ 1 ) ( ¯ k ) , ¯ k = i k , i = / x i ,
H ¯ 1 k = H i 1 i k = ϕ V ( x ) 1 O ¯ 1 k ϕ V ( x ) = E ( y ¯ 1 + I Y ¯ 1 ) ( y ¯ k + I Y ¯ k )
and I = 1 , y = V 1 x , Y = V 1 X N q ( 0 , V 1 ) .
H ¯ 1 k ( x , V ) = H ¯ 1 k is the multivariate Hermite polynomial. We use the tensor summation convention, repetition of i 1 , , i k in (6) implies their implicit summation over their range, 1 , , q . Ref. [3] gave H ¯ 1 k explicitly for k 6 and for k 9 when q = 2 .
Set μ ¯ 1 2 k = E Y ¯ 1 Y ¯ 2 k = 1.3 ( 2 k 1 ) V ¯ 12 V ¯ 2 k 1 , 2 k ,
where N f ¯ 1 2 k sums f ¯ 1 2 k over all N permutations of i 1 , , i 2 k giving distinct values. For example,
H ¯ 1 = y ¯ 1 , H ¯ 12 = y ¯ 1 y ¯ 2 V ¯ 12 , H ¯ 1 3 = y ¯ 1 y ¯ 2 y ¯ 3 3 y ¯ 1 V ¯ 23 = y ¯ 1 y ¯ 2 y ¯ 3 y ¯ 1 V ¯ 23 y ¯ 2 V ¯ 13 y ¯ 3 V ¯ 12 ,
H ¯ 1 = J ¯ 1 , H ¯ 12 = J ¯ 12 V ¯ 12 Φ V ( x ) , H ¯ 1 3 = J ¯ 123 3 J ¯ 1 V ¯ 23 , where
J ¯ 1 k = J ¯ 1 k ( x , V ) = E Y ¯ 1 Y ¯ k I ( X x ) = V ¯ 1 , k + 1 V ¯ k , 2 k M ¯ V k + 1 2 k ,
and M ¯ V a b = M ¯ a b ( x , V ) = x x ¯ a x ¯ b ϕ V ( x ) d x , for x ¯ a = x i a .
So , P 1 ( x ) = k = 1 , 3 P 1 k ( x ) , P 11 ( x ) = k ¯ 1 1 H ¯ 1 , P 13 ( x ) = k ¯ 2 1 3 H ¯ 1 3 / 6 ,
p ˜ 1 ( x ) = p 1 ( x ) / ϕ V ( x ) = k = 1 , 3 p ˜ 1 k , p ˜ 11 = k ¯ 1 1 H ¯ 1 , p ˜ 13 = k ¯ 2 1 3 H ¯ 1 3 / 6 .
(So the repeated i k + 1 , , i 2 k in (11) implies their repeated summatioin over 1 , , q .) P 2 ( x ) , P 3 ( x ) are given explicitly in [3]. So (4) with the P ¯ r 1 k in [3] give the Edgeworth expansions for the distribution and density of X n of (1) to O ( n 2 ) . p ˜ r k and P r k each have q k terms, but many are duplicates as P ¯ r 1 k is symmetric in i 1 , , i k . This is exploited by the notation of Section 4 of [3] to greatly reduce the number of terms in (6).
By (5), the density of X n relative to its asymptotic value is
p X n ( x ) / ϕ V ( x ) 1 + r = 1 n r / 2 p ˜ r ( x ) = 1 + n 1 / 2 p ˜ 1 ( x ) + O ( n 1 ) , for x R q ,
and for measurable C R q ,
P ( X n C ) Φ V ( C ) + r = 1 n r / 2 p r C , where for r 1 , p r C = E p r ( X ) I ( X C ) = C p r ( x ) ϕ V ( x ) d x = k = 1 3 r [ p ˜ r k ( C ) : k r even ] , p ˜ r k ( C ) = E p ˜ r k ( X ) I ( X C ) = C p ˜ r k ( x ) ϕ V ( x ) d x = P ¯ r 1 k H ¯ 1 k ( C ) , and H ¯ 1 k ( C ) = E H ¯ 1 k ( X , V ) I ( X C ) = C H ¯ 1 k ϕ V ( x ) d x .
If C = C , then for r odd, Q ¯ 1 r = p ˜ r k ( C ) = p r C = 0 , so that
P ( X n C ) Φ V ( C ) + r = 1 n r p 2 r C = Φ V ( C ) + n 1 p 2 C + O ( n 2 ) .
Examples 3 and 4 of [3] gave p 2 C for
C = { x : x V 1 x u } , and C = { x : | ( V 1 / 2 x ) j | u j , j = 1 , , q } .

3. The Conditional Density and Distribution

For q = q 1 + q 2 , q 1 1 , and q 2 1 , partition w , w ^ , X N q ( 0 , V ) , X n = n 1 / 2 ( w ^ w ) , x and y = V 1 x as w 1 w 2 , w ^ 1 w ^ 2 , X 1 X 2 , X n 1 X n 2 , x 1 x 2 and y 1 y 2 , where w i , w ^ i , X i , X n i , x i , y i are vectors of length q i . Partition V , V 1 as ( V i j ) , ( V i j ) , 2 × 2 , where V i j , V i j are q i × q j .
Set X 1 2 = X 1 | ( X 2 = x 2 ) , X n 1 2 = X n 1 | ( X n 2 = x 2 ) ,
w ^ 1 2 = w ^ 1 | ( X n 2 = x 2 ) = w 1 + n 1 / 2 X n 1 2 .
Now we come to the main purpose of this paper. Theorem 1 expands the conditional density of X n 1 2 about the conditional density of X 1 2 . Its derivation is straightforward, the only novel feature being the use of Lemma 2 to find the reciprocal of a series, using Bell polynomials. Theorem 2 integrates the conditional density to obtain the expansion for the conditional distribution of X n 1 2 about the conditional distribution of X 1 2 in terms of I ¯ 1 k of (42) below, the integral of the Hermite polynomial H ¯ 1 k of (8), with respect to the conditional normal density. Note 1 gives I ¯ 1 k in terms of derivatives of the multivariate normal distribution. Theorem 3 gives I ¯ 1 k in terms of the partial moments of the conditional normal distribution. For X 1 2 of (15), set
μ 1 2 = E X 1 2 = V 12 V 22 1 x 2 R q 1 ,
V 1 2 = c o v a r X 1 2 = V 11 V 12 V 22 1 V 21 = V 0 say , R q 1 × q 1 .
So , X 1 2 N q 1 ( μ 1 2 , V 1 2 ) .
Lemma 1.
The elements of ( V i j ) = V 1 are
V 11 = V 1 2 1 , V 12 = V 11 V 12 V 22 1 = V 11 1 V 12 V 22 , V 21 = V 22 V 21 V 11 1 = V 22 1 V 21 V 11 , V 22 = V 2 1 1 , where V 2 1 = V 22 V 21 V 11 1 V 12 R q 2 × q 2 .
For i = 1 , 2 , set A i = V i 1 V 12 V 22 1 + V i 2 . Then A 1 = 0 q 1 × q 2 , A 2 = V 22 1 .
Proof. 
V V 1 = V 1 V = I q gives 8 equations relating { V i j } and { V i j } . Now solve for { V i j } .
So A 1 = 0 q 1 × q 2 , A 2 = V 22 B V 22 1 for B = V 22 V 21 V 11 1 V 12 = ( V 22 ) 1 .
Since Q = V 11 V 1 2 0 q 1 × q 1 in the sense that x Q x 0 for x R q 1 , X 1 2 is less variable than X 1 , and X n 1 2 is less variable than X n 1 , unless X 1 and X 2 are uncorrelated, that is, V 12 is a matrix of zeros.
The conditional density of X n 1 2 is
p n 1 2 ( x 1 ) = p X n ( x ) / p X n 2 ( x 2 ) = ϕ 1 2 ( x 1 ) ( 1 + S ) / ( 1 + S 2 ) ,
where S = p X n ( x ) / ϕ V ( x ) 1 r = 1 n r / 2 p ˜ r ( x ) of ( 6 ) ,
S 2 = p X n 2 ( x 2 ) / ϕ V 22 ( x 2 ) 1 r = 1 n r / 2 f r , for f r = p r ( x 2 ) ,
where p r ( x 2 ) is p ˜ r ( x ) of (6) for X n 2 , and ϕ 1 2 ( x 1 ) is the density of X 1 2 of (15). By (4)–(6), Section 2.5 of [18], for V 0 of (18),
ϕ 1 2 ( x 1 ) = ϕ V ( x ) / ϕ V 22 ( x 2 ) = ϕ V 0 ( u ) , where u = x 1 μ 1 2 R q 1 .
So the distribution of X 1 | ( X 2 = x 2 ) is
Φ 1 2 ( x 1 ) = Φ V 0 ( u ) , for V 0 of ( 18 ) .
For μ 1 2 of (17), V 1 2 of (18), and v R q 1 , set
x 1 ( x 2 , v ) = μ 1 2 + V 1 2 1 / 2 v = V 12 V 22 1 x 2 + V 1 2 1 / 2 v .
Corollary 1.
Suppose that q 1 = 1 . Then for v = V 0 1 / 2 u of (23),
ϕ 1 2 ( x 1 ) = V 0 1 / 2 ϕ ( v ) , P ( X 1 2 < x 1 ( x 2 , v ) ) = Φ ( v ) , P ( w ^ 1 2 < w 1 + n 1 / 2 x 1 ( x 2 , v ) ) = P ( X n 1 2 < x 1 ( x 2 , v ) ) = Φ ( v ) + O ( n 1 / 2 ) , P ( | w 1 w ^ 1 2 | < n 1 / 2 x 1 ( x 2 , v ) )   = P ( | X n 1 2 | < x 1 ( x 2 , v ) ) = 2 Φ ( v ) 1 + O ( n 1 ) if v > 0 , P ( x 1 ( x 2 , v ) < X n 1 2 < x 1 ( x 2 , v ) ) = 2 Φ ( v ) 1 + O ( n 1 ) , if v > 0 .
Replacing V by an estimate will usually give 1- and 2-sided conditional confidence intervals of error O ( n 1 / 2 ) and O ( n 1 ) for w ^ 1 w 1 given w ^ 2 w 2 .
Set H ¯ q 1 k = H ¯ 1 k = H ¯ 1 k ( x , V ) , and H ¯ q 2 1 k = H ¯ 1 k ( x 2 , V 22 ) .
So H ¯ q 2 1 k is given by replacing y = V 1 x and ( V i j ) = V 1 in H ¯ q 1 k by
z = V 22 1 x 2 and ( U i j ) = V 22 1 . For example H ¯ q 2 12 = z ¯ 1 z ¯ 2 U ¯ 12 .
By (5) and (6), for r 1 , p r ( x 2 ) of (22) is given by
p r ( x 2 ) = k = 1 3 r [ p r k : k r even ] , where p r k = P ¯ r 1 k H ¯ q 2 1 k ,
and implicit summation in (28) for i 1 , , i k is now over  q 1 + 1 , , q . So,
p 1 ( x 2 ) = k = 1 , 3 p 1 k , p 11 = i 1 = q 1 + 1 q k ¯ 1 1 H ¯ q 2 1 , p 13 = i 1 , i 2 , i 3 = q 1 + 1 q k ¯ 2 1 3 H ¯ q 2 1 3 / 6 , where H ¯ q 2 1 = z ¯ 1 , H ¯ q 2 1 3 = z ¯ 1 z ¯ 2 z ¯ 3 3 U ¯ 12 z ¯ 3 , p 2 ( x 2 ) = k = 2 , 4 , 6 p 2 k , p 3 ( x 2 ) = k = 1 , 3 , 5 , 7 , 9 p 3 k , for p r k of ( 28 ) .
Ordinary Bell polynomials. For a sequence e = ( e 1 , e 2 , ) from R, the partial ordinary Bell polynomial  B ˜ r s = B ˜ r s ( e ) , is defined by the identity
for s = 0 , 1 , 2 , and z R , S s = r = s z r B ˜ r s ( e ) , where S = r = 1 z r e r .
So , B ˜ r 0 = δ r 0 , B ˜ r 1 = e r , B ˜ r r = e 1 r , B ˜ 32 = 2 e 1 e 2 ,
where δ 00 = 1 , δ r 0 = 0 for r 0 . They are tabled on p309 of [19]. To obtain (21), we use
Lemma 2.
Take B ˜ r s ( e ) of (29). Set S 2 = r = 1 z r f r for f r R . Then
( 1 + S 2 ) 1 = r = 0 z r C r , where C r = B r ( f ) , B r ( e ) = s = 0 r B ˜ r s ( e ) .
So , B 0 ( e ) = 1 , B 1 ( e ) = e 1 , B 2 ( e ) = e 2 + e 1 2 , B 3 ( e ) = e 3 + 2 e 1 e 2 + e 1 3 ,
C 0 = 1 , C 1 = f 1 , C 2 = f 1 2 f 2 , C 3 = f 1 3 + 2 f 1 f 2 f 3 .
Proof. 
( 1 + S 2 ) 1 = s = 0 ( S 2 ) s , and ( S 2 ) s = r = s z r B ˜ r s ( f ) .
Now swap summations. □
Theorem 1.
Take p ˜ r ( x ) of (6) and C r = B r ( f ) of (30) with
f r = p r ( x 2 ) of ( 28 ) .
The conditional density p n 1 2 ( x 1 ) of (21), relative to ϕ 1 2 ( x 1 ) of (23), is
p n 1 2 ( x 1 ) / ϕ 1 2 ( x 1 ) r = 0 n r / 2 D r , where D r = C r p ˜ r ( x ) ,
and for sequences ( a 0 , a 1 , ) and ( b 0 , b 1 , ) , a r b r = i = 0 r a i b r i . So,
D 0 = p ˜ 0 ( x ) = 1 , D 1 = C 1 + p ˜ 1 ( x ) , D 2 = C 2 + C 1 p ˜ 1 ( x ) + p ˜ 2 ( x ) ,
D 3 = C 3 + C 2 p ˜ 1 ( x ) + C 1 p ˜ 2 ( x ) + p ˜ 3 ( x ) .
Proof. 
This follows from (21) and Lemma 2. □
So D 0 , , D 3 of (34) and (35) give the conditional density to O ( n 2 ) . We call (33) the relative conditional density. We now give our main result, an expansion for the conditional distribution of X n 1 | ( X n 2 = x 2 ) . As noted, Theorem 2 gives this in terms of I ¯ 1 k of (42) below, an integral of the Hermite polynomial H ¯ 1 k of (8), and Note 1 gives I ¯ 1 k in terms of derivatives of the multivariate normal distribution. Theorem 3 gives I ¯ 1 k in terms of the partial moments of the conditional distribution Φ 1 2 ( x 1 ) of (24).
When q 1 = 1 , Theorem 4 gives I ¯ 1 k in terms of Φ ( v ) and ϕ ( v ) for
v = V 1 2 1 / 2 u = V 1 2 1 / 2 ( x 1 μ 1 2 ) R q 1 .
Theorem 2.
Take C r , D r of Theorem 1. Set p ˜ 0 ( x ) = 1 . The conditional distribution of X n 1 given X n 2 , about Φ 1 2 ( x 1 ) of (24), has the expansion
P n 1 2 ( x 1 ) = P ( X n 1 2 ) x 1 r = 0 n r / 2 G r ,
where G r = x 1 D r d Φ 1 2 ( x 1 ) = C r g r , and
g r = x 1 p ˜ r ( x ) d Φ 1 2 ( x 1 ) . So , G 0 = g 0 = Φ 1 2 ( x 1 ) = Φ V 0 ( u ) of ( 24 ) ,
for r 1 , g r = k = 1 3 r [ g r k : k r even ] , where for p ˜ r k of ( 6 ) ,
g r k = x 1 p ˜ r k d Φ 1 2 ( x 1 ) = P ¯ r 1 k I ¯ 1 k , and for 1 i 1 , , i k q ,
I ¯ 1 k = I i 1 i k = x 1 H ¯ q 1 k ϕ 1 2 ( x 1 ) d x 1 , for ϕ 1 2 ( x 1 ) of ( 23 ) .
Proof. 
(40) holds by (6). (41) holds by (6). Now use (23). □
So P n 1 2 ( x 1 ) = Φ V 0 ( u ) + r = 1 3 n r / 2 G r + O ( n 2 ) , where G 1 = g 1 f 1 g 0 ,
G 2 = g 2 f 1 g 1 + C 2 g 0 , G 3 = g 3 f 1 g 2 + C 2 g 1 + C 3 g 0 ,
for C r of (31). g 1 = g 11 + g 13 is given by I ¯ 1 , I ¯ 1 3 , g 2 = g 22 + g 24 + g 26 is given by I ¯ 12 , I ¯ 1 4 , I ¯ 1 6 , and g 3 = ( g 3 k : k = 1 , 3 , 5 , 7 , 9 ) is given by I ¯ 1 , I ¯ 1 3 , , I ¯ 1 9 .
Note 1.
Set  = Π i = q 1 + 1 q i .  By (23),
I ¯ 1 k = ϕ V 22 ( x 2 ) 1 L ¯ 1 k , where
L ¯ 1 k = x 1 H ¯ q 1 k ϕ V ( x ) d x 1 = ( ¯ 1 ) ( ¯ k ) Φ V ( x ) .
Comparing  L ¯ 1 k  with the Hermite function H ¯ 1 k of (7), we can call  L ¯ 1 k the partial Hermite function. When  q = 2 , see (53).
By (39), G r in (37) is given by C r of (32) and g r of (40). Viewing H ¯ q 1 k as a polynomial in x 1 = μ 1 2 + u for u of (23), I ¯ 1 k is linear in
x 1 x i 1 x i s ϕ 1 2 ( x 1 ) d x 1 = u ( μ 1 2 + u ) i 1 ( μ 1 2 + u ) i s ϕ V 0 ( u ) d u
for 0 s k , 1 i 1 , , i s q 1 . So I ¯ 1 k can be expanded in terms of the partial moments of
M = Φ 1 2 ( x 1 ) = Φ V 0 ( u ) ,
M ¯ a b = M ¯ a b ( u , V 0 ) = M i a i b ( u , V 0 ) = u u i a u i b ϕ V 0 ( u ) d u .
This has only q 1 integrals, while (12) has q integrals.
Lemma 3.
For u = x 1 μ 1 2 , y = V 1 x = α + Λ u , where
Λ = V 11 V 21 R q × q 1 , α = α 1 α 2 , α 1 = 0 q 1 , α 2 = V 22 1 x 2 .
Proof. 
y = y 1 y 2 , where y i = V i 1 x 1 + V i 2 x 2 = α i + V i 1 u , and α i = A i x 2 for A i of (20). □
Our main result, Theorem 2, gave the conditional distribution expansion in terms of I ¯ 1 k of (42). Note 4.1 gave these in terms of the derivatives of Φ V ( x ) . We now give I ¯ 1 k in terms of J ¯ 1 k , the partial moments of the conditional distribution Φ 1 2 ( x 1 ) of (24). As in (10), for any π = ( m , , n ) , set N c π = c π summed over all, N say, permutations of π giving distinct c π . For example, 2 c 23 = c 23 + c 32 .
Theorem 3.
Take J ¯ 1 k ( x , V ) of (11), u of (24), M of (45), M ¯ a b of (46), Λ , α of (47), and 1 i 1 , , i k q . Set
K ¯ 1 k = K i 1 i k = u ( Λ u ) i 1 ( Λ u ) i k ϕ V 0 ( u ) d u = Λ i 1 j 1 Λ i k j k M j 1 , j k ,
where j 1 , j k sum over their range 1 , , q 1 . So
K ¯ 1 k = Λ ¯ 1 , k + 1 Λ ¯ k , 2 k M ¯ k + 1 2 k . For example , K ¯ 1 = Λ ¯ 12 M ¯ 2 , K ¯ 12 = Λ ¯ 13 Λ ¯ 24 M ¯ 34 , K ¯ 123 = Λ ¯ 14 Λ ¯ 25 Λ ¯ 36 M ¯ 456 .
Set J ¯ 0 1 k = J ¯ 1 k ( u , V 0 ) = u y ¯ 1 y ¯ k ϕ V 0 ( u ) d u : J ¯ 0 1 k = α ¯ 1 α ¯ k M + k 1 α ¯ 1 α ¯ k 1 K ¯ k + k 2 α ¯ 1 α ¯ k 2 K ¯ k 1 , k + + K ¯ 1 k .
For example , J ¯ 0 1 = α ¯ 1 M + K ¯ 1 , J ¯ 0 12 = α ¯ 1 α ¯ 2 M + 2 α ¯ 1 K ¯ 2 + K ¯ 12 , J ¯ 0 1 3 = α ¯ 1 α ¯ 2 α ¯ 3 M + 3 ( α ¯ 1 α ¯ 2 K ¯ 3 + α ¯ 1 K ¯ 23 ) + K ¯ 1 3 , J ¯ 0 1 4 = α ¯ 1 α ¯ 4 M + 4 ( α ¯ 1 α ¯ 2 α ¯ 3 K ¯ 4 + α ¯ 1 K ¯ 234 ) + 6 α ¯ 1 α ¯ 2 K ¯ 34 + K ¯ 1 4 , J ¯ 0 1 5 = α ¯ 1 α ¯ 5 M + 5 ( α ¯ 1 α ¯ 4 K ¯ 5 + α ¯ 1 K ¯ 2 5 ) + 10 ( α ¯ 1 α ¯ 2 α ¯ 3 K ¯ 45 + α ¯ 1 α ¯ 2 K ¯ 3 5 ) + K ¯ 1 5 , J ¯ 0 1 6 = α ¯ 1 α ¯ 6 M + 6 ( α ¯ 1 α ¯ 5 K ¯ 6 + α ¯ 1 K ¯ 2 6 ) + 15 ( α ¯ 1 α ¯ 4 K ¯ 56 + α ¯ 1 α ¯ 2 K ¯ 3 6 ) + 20 α ¯ 1 α ¯ 2 α ¯ 3 K ¯ 4 6 + K ¯ 1 6 , For μ ¯ 1 2 k of ( 10 ) , I ¯ 1 = J ¯ 0 1 , I ¯ 12 = J ¯ 0 12 M V ¯ 12 , I ¯ 1 3 = J ¯ 0 1 3 3 J ¯ 0 1 V ¯ 23 , I ¯ 1 4 = J ¯ 0 1 4 6 J ¯ 0 12 V ¯ 34 + M μ ¯ 1 4 , I ¯ 1 5 = J ¯ 0 1 5 10 J ¯ 0 1 3 V ¯ 45 + 5 J ¯ 0 1 μ ¯ 2 5 , I ¯ 1 6 = J ¯ 0 1 6 15 J ¯ 0 1 4 V ¯ 56 + 15 J ¯ 0 12 μ ¯ 3 6 M μ ¯ 1 6 .
Proof. 
Since x 1 = μ + u , y = Λ x 1 + V 12 V 22 x 2 = α + Λ u R q . Substitute y = α + Λ u into the expressions for H ¯ 1 k . Now multiply by ϕ V 0 ( u ) and integrate from to u. □
This gives the I ¯ 1 k needed for g 1 , g 2 , G 1 , G 2 . The I ¯ 1 k , k = 7 , 9 needed for g 3 , G 3 can be written down similarly in terms of the partial moments using H ¯ q 1 k for k = 7 , 9 . We now show that if q 1 = 1 , we only need the partial moments of Φ ( v ) at v of (36), and that these are easily written in terms of Φ ( v ) and ϕ ( v ) × a polynomial in v of (36).
The case  q 1 = 1 . So w 1 = w 1 , w ^ 1 = w ^ 1 , X 1 = X 1 , X n 1 = X n 1 , V 11 = V 11 .
Theorem 4.
For q 1 = 1 , 1 k 6 , I ¯ 1 k is given by Theorem 3 with
α 1 = 0 , K ¯ 1 k = Λ ¯ 1 Λ ¯ k σ k γ k , where Λ = V 11 V 21 R q ,
σ = V 1 2 1 / 2 of ( 18 ) , γ k = v v k ϕ ( v ) d v , for v of ( 36 ) .
So γ 0 = Φ ( v ) , γ 1 = ϕ ( v ) , γ k = v k 1 γ 1 + ( k 1 ) γ k 2 , for k 2 :
γ 2 = γ 0 + v γ 1 , γ 3 = ( v 2 + 2 ) γ 1 , γ 4 = 3 γ 0 + ( v 3 + 3 v ) γ 1 , γ 5 = ( v 4 + 4 v 2 + 4.2 ) γ 1 , γ 6 = 5.3 γ 0 + ( v 5 + 5 v 3 + 5.3 v ) γ 1 , γ 7 = ( v 6 + 6 v 4 + 6.4 v 2 + 6 . 4.2 ) γ 1 , γ 8 = 7 . 5.3 γ 0 + ( v 7 + 7 v 5 + 7.5 v 3 + 7 . 5.3 v ) γ 1 , γ 9 = ( v 8 + 8 v 6 + 8.6 v 4 + 8 . 6.4 v 2 + 8.6 . 4.2 ) γ 1 ,
where dot denotes multiplication. Also, G 0 = Φ ( v ) .
Proof. 
For v of (36), by (23), ϕ 1 2 ( x 1 ) = σ 1 ϕ ( v ) . (51) follows from integration by parts. By (48), K 1 k = Λ ¯ 1 Λ ¯ k M 1 k where M 1 k = u u k d Φ ( u / σ ) = σ k γ k . That G 0 = Φ ( v ) , follows from (39). □
By (37), for C r of (32) and v of (36), the conditional distribution of X n 1 2 is
P ( X n 1 2 / σ v ) Φ ( v ) + r = 1 n r / 2 G r , where G r = C r g r ,
as in (43), and g r is given by (40) in terms of the integrated Hermite polynomial, I ¯ 1 k of (42) given by Theorems 3 and 4.

4. The Case q 1 = q 2 = 1

Theorem 3.2 gave the conditional Edgeworth expansion in terms of I ¯ 1 k of (42). Theorem 3.3 gave I ¯ 1 k needed for g r k of (41) and G 1 , G 2 of (37), in terms of the partial moments M ¯ a b of (46). When q 1 = 1 , Theorem 4 gave I ¯ 1 k in terms of Φ ( v ) and its partial moments for v of (36). But now q = 2 so that i 1 , , i k = 1 or 2. So for ( I , y , Y ) of (9), we switch notation to
H a b = ( 1 ) a ( 2 ) b ϕ V ( x ) = E ( y 1 + I Y 1 ) a ( y 2 + I Y 2 ) b , H a b = ( 1 ) a ( 2 ) b Φ V ( x ) = x H a b ϕ V ( x ) d x . So H a b = H a 1 , b 1 ϕ V ( x ) if a 2 , b 1 , H 10 = x 2 ϕ V ( x ) d x 2 = 1 Φ V ( x ) , H a 0 = ( 1 ) a 1 H 10 if a 2 , H 01 = x 1 ϕ V ( x ) d x 1 = 2 Φ V ( x ) , H 0 b = ( 2 ) b 1 H 01 if b 1 ,
L 1 a 2 b = ( 1 ) a ( 2 ) b + 1 Φ V ( x ) = H a , b + 1 ,
for L ¯ 1 k of (44). Similarly, write (2) as
κ a b ( w ^ 1 , w ^ 2 ) d = a + b 1 n d k a b d , for a + b 1 , where k a b d = k d 1 a 2 b , and
set K a b = K 1 a 2 b , J a b = J 0 1 a 2 b , I a b = I 1 a 2 b = x 1 H a b ( x ) ϕ 1 2 ( x 1 ) d x 1 .
Also, we switch from P ¯ r 1 k to
P r ( a b ) = a + b a P r 1 a 2 b .
given for r 3 in Section 4 of [3]. So,
p ˜ r k = b = 0 k P r ( k b , b ) H k b , b , P r k ( x ) = b = 0 k P r ( k b , b ) H k b , b .
So , p ˜ r 1 = P r ( 10 ) H 10 + P r ( 01 ) H 01 , p ˜ 11 = k 101 y 1 + k 011 y 2 , p ˜ 13 = P 1 ( 30 ) H 30 + P 1 ( 21 ) H 21 + P 1 ( 12 ) H 12 + P 1 ( 03 ) H 03 .
P r ( b a ) is just P r ( a b ) with 1 and 2 reversed. For the other p ˜ r k and P r k ( x ) needed for r 3 , see Section 4 of [3]. Our main result for this section, Theorem 7, gives simple formulas for I a b and for g r of (40), the main ingredient needed in Theorem 2 for the expansion of the conditional distribution.
Theorem 5.
The conditional density of X n 1 2 of (15), is given by Theorem 1 where f r = p r ( x 2 ) is given by (28) in terms of
p r k = P r ( 0 k ) H k , where H k = H k ( z , V 22 ) and z = V 22 1 x 2 .
For example , H 1 = z , H 2 = z 2 V 22 1 , H 3 = z 3 3 z V 22 1 ,
H 4 = z 4 15 z 2 V 22 1 + 3 V 22 2 , H 5 = z 5 10 z 3 V 22 1 + 15 z V 22 2 ,
H 6 = z 6 15 z 4 V 22 1 + 45 z 2 V 22 2 15 V 22 3 .
Proof. 
This follows from Theorem 1. □
Theorem 6 gives a laborious expression for the conditional distribution.
However Theorem 7 gives a huge simplification.
Theorem 6.
The conditional distribution of X n 1 2 of (15), is given by Theorem 2 with Λ , σ , γ s of Theorem 4 as follows. For k r even, g r k of (41) is given by
g r k = b = 0 k P r ( k b , b ) I k b , b ,
where I a b of (54) is given for a + b = k , as follows in terms of Λ i = V i 1 .
K a b = Λ 1 a Λ 2 b σ a + b γ a + b , and J k 0 = s = 0 k α 1 k s K s 0 , for K s 0 = ( Λ 1 σ ) s γ s .
For k = 1 : I 10 = J 10 = α 1 γ 0 + Λ 1 σ γ 1 . For k = 2 : I 20 = J 20 γ 0 V 11 , I 11 = J 11 γ 0 V 12 ,
J 11 = α 1 α 2 γ 0 + σ γ 1 2 α 1 Λ 2 + Λ 1 Λ 2 σ 2 γ 2 . For k = 3 : I 30 = J 30 3 J 10 V 11 , I 21 = J 21 ( 2 J 10 V 12 + J 01 V 11 ) , J 21 = α 1 2 α 2 γ 0 + X 21 + X 12 + K 21 , where X 21 = α 1 2 K 01 + 2 α 1 α 2 K 10 , X 12 = 2 α 1 K 11 + α 2 K 20 . For k = 4 : I 40 = J 40 6 J 20 V 11 + 3 γ 0 ( V 11 ) 2 , I 31 = J 31 S 6 + γ 0 S 3 , where S 6 = 3 J 20 V 22 + 3 J 11 V 12 , S 3 = 3 V 11 V 12 , J 31 = α 1 3 α 2 γ 0 + X 31 + X 22 + X 13 + K 31 , where X 31 = α 1 3 K 01 + 3 α 1 2 α 2 K 10 , X 22 = 4 α 1 2 K 11 + 2 α 1 α 2 K 20 , X 13 = 3 α 1 K 21 + 6 α 2 K 30 , I 22 = J 22 S 6 + γ 0 S 3 , where S 6 = J 20 V 22 + 4 J 11 V 12 + J 02 V 11 , S 3 = μ 1122 = V 11 V 22 + 2 ( V 12 ) 2 , J 22 = α 1 2 α 2 2 γ 0 + X 31 + X 22 + X 13 + K 22 , where X 31 = 2 α 1 2 α 2 K 01 + 2 α 1 α 2 2 K 02 , X 22 = α 1 2 K 02 + 4 α 1 α 2 K 11 + α 2 2 K 20 , X 13 = 2 α 1 K 12 + 2 α 2 K 21 . For k = 5 : I 50 = J 50 10 J 30 V 11 + 15 J 10 ( V 11 ) 2 , I 41 = J 41 S 10 + S 15 , where S 10 = 6 J 21 V 11 + 4 J 30 V 12 , S 15 = 12 J 10 V 11 V 12 + 3 J 01 ( V 11 ) 2 , J 41 = α 1 4 α 2 γ 0 + X 41 + X 32 + X 23 + X 14 + K 41 , where , X 41 = 4 α 1 3 α 2 K 10 + α 1 4 K 01 , X 32 = 5 α 1 2 α 2 K 20 + 5 α 1 3 K 11 , X 23 = 6 α 1 2 K 21 + 4 α 1 α 2 K 30 , X 14 = 4 α 1 K 31 + α 2 K 40 , I 32 = J 32 S 10 + S 15 , where S 10 = 3 J 12 V 11 + 6 J 21 V 12 + J 30 V 22 , S 15 = 3 J 10 μ 1122 + 6 J 01 V 11 V 12 , J 32 = α 1 3 α 2 2 γ 0 + X 41 + X 32 + X 23 + X 14 + K 32 , where , X 41 = 3 α 1 2 α 2 2 K 10 + 2 α 1 3 α 2 K 01 , X 32 = 3 α 1 α 2 2 K 20 + 6 α 1 2 α 2 K 11 + α 1 3 K 02 , X 23 = 3 α 1 2 K 12 + 6 α 1 α 2 K 21 + α 2 2 K 30 , X 14 = 3 α 1 K 22 + 2 α 2 K 31 . For k = 6 : I 60 = J 60 15 J 40 V 11 + 45 J 20 ( V 11 ) 2 15 γ 0 ( V 11 ) 3 , I 51 = J 51 S 15 + S 45 γ 0 S 15 , where S 15 = 5 V 12 J 40 + 10 V 11 J 31 , S 45 = 30 V 11 V 12 J 20 + 15 ( V 11 ) 2 J 11 , S 15 = 15 γ 0 ( V 11 ) 2 V 12 , J 51 = α 1 5 α 2 γ 0 + X 51 + X 42 + X 33 + X 24 + X 15 + K 51 , where X 51 = α 1 5 K 01 + 5 α 1 4 α 2 K 10 , X 42 = 5 α 1 4 K 11 + 10 α 1 3 α 2 K 20 , X 33 = 10 α 1 3 K 21 + 10 α 1 2 α 2 K 30 , X 24 = 10 α 1 2 K 31 + 5 α 1 α 2 K 40 , X 15 = α 2 K 60 + 5 α 1 K 51 , I 42 = J 42 S 15 + S 45 γ 0 S 15 , where S 15 = V 22 J 40 + 6 V 11 J 22 + 8 V 12 J 31 , S 45 = 3 ( V 11 ) 2 J 02 + 6 μ 1122 J 20 + 24 V 11 V 12 J 11 , S 15 = 3 ( V 11 ) 2 V 22 + 24 ( V 11 V 12 ) 2 + 6 V 11 μ 1122 , J 42 = α 1 4 α 2 2 γ 0 + X 51 + X 42 + X 33 + X 24 + X 15 + K 42 , where X 51 = 2 α 1 5 K 01 + 4 α 1 4 α 2 K 10 , X 42 = α 1 4 K 02 + 8 α 1 3 α 2 K 11 + 6 α 1 2 α 2 2 K 20 , X 33 = 10 α 1 2 α 2 K 12 + 10 α 1 α 2 2 K 21 , X 24 = α 2 2 K 40 + 8 α 1 α 2 K 31 + 6 α 1 2 K 22 , X 15 = 2 α 2 K 50 + 4 α 1 K 41 , I 33 = J 33 S 15 + S 45 γ 0 S 15 , where S 15 = 6 V 11 J 04 + 9 V 12 J 22 + 6 V 22 J 40 , S 45 = 9 V 12 V 22 J 20 + 9 μ 1122 J 11 + 9 V 12 V 11 J 02 , S 15 = 6 V 11 V 12 V 22 + 3 V 12 μ 1122 , J 33 = α 1 3 α 2 3 γ 0 + X 51 + X 42 + X 33 + X 24 + X 15 + K 33 , where X 51 = 2 α 1 2 α 2 3 K 10 + 3 α 1 3 α 2 2 K 01 , X 42 = 6 α 1 α 2 3 K 20 + 3 α 1 3 α 2 3 K 11 + 6 α 1 3 α 2 K 02 , X 33 = α 1 3 K 03 + 9 α 1 2 α 2 K 12 + 9 α 1 α 2 2 K 21 + α 2 3 K 30 , X 24 = 6 α 1 2 K 13 + 3 α 1 α 2 K 22 + 8 α 2 2 K 31 , X 15 = 3 α 1 K 23 + 3 α 2 K 32 .
Also J b a , I b a are J a b , I a b with α 1 , Λ 1 = V 11 and α 2 , Λ 2 = V 21 of (47) reversed, before setting α 1 = 0 and α 2 = z = V 22 1 x 2 by (27). For example, by (61), for Λ , σ , γ s of Theorem 3.4,
I 10 = α 1 γ 0 + Λ 1 σ γ 1 = V 11 σ γ 1 , I 01 = α 2 γ 0 + Λ 2 σ γ 1 = z γ 0 + V 11 σ γ 1 ,
J k 0 = K k 0 = ( V 11 σ ) k γ k , J 0 k = s = 0 k z k s K 0 s = z J 0 , k 1 + K 0 k
where K 0 s = ( V 21 σ ) s γ s .
Proof 
This follows from Theorems 3 and 4. □
This gives the I ¯ a b needed for g 1 , g 2 , G 1 , G 2 for the conditional distribution of (37)–(39) to O ( n 3 / 2 ) . The I ¯ a b needed for g 3 , G 3 can be written down similarly. We now give a much simpler method for obtaining g r k of (41), and so g r by (40), and G r needed for (37) by (38). Theorem 7 gives g r k and g r in terms of I 0 k of (54). Theorem 4.4 gives I 0 k in terms of J 0 k of (63), a function of ( Λ , σ , γ s ) of Theorem 3.4.
Theorem 7.
For v of (36), I a b of (54) is given by
I a b = H a 1 , b σ 1 ϕ ( v ) , for a 1 .
For k r 1 and k r even, g r k of (41) is given by
g r k = P r ( 0 k ) I 0 k b r k σ 1 ϕ ( v ) , for b r k = a = 1 k P r ( a , k a ) H a 1 , k a .
So by (40), for r 1 , g r of (39) is given by
g r = k = 1 3 r [ P r ( 0 k ) I 0 k b r k σ 1 ϕ ( v ) : k r even ] .
Proof. 
By (60), g r k = P r ( 0 k ) I 0 k + a = 1 k P r ( a , k a ) I a , k a .
By (23), ϕ 1 2 ( x 1 ) / ϕ V ( x ) = θ 1 where θ = ϕ V 22 ( x 2 ) and ϕ 1 2 ( x 1 ) = σ 1 ϕ ( v ) .
So for a 1 , H a b ϕ V ( x ) = ( 1 ) a ( 2 ) b = 1 [ H a 1 , b ϕ V ( x ) ] . So I a b = θ 1 x 1 H a b ϕ V ( x ) d x 1 = θ 1 H a 1 , b ϕ V ( x ) = H a 1 , b σ 1 ϕ ( v ) .
This proves (64). So,
g r k = P r ( 0 k ) I 0 k θ 1 ϕ V ( x ) [ P r ( a b ) H a 1 , b : a + b = k , a 1 ] .
(65) follows. (66) now follows from (28). □
Note 2.
b r k  is just  p ˜ r k  of (55) with  ( H 0 b , H a b )  replaced by  ( 0 , H a 1 , b )  for  a 1 .
So for r = 1 , 2 , 3 , b r k is given in terms of P r ( . ) of Section 4, by
b r 1 = P r ( 10 ) = k r 1 = k 10 r , b r 3 = P r ( 30 ) H 20 + P r ( 21 ) H 11 + P r ( 12 ) H 02 ,
b 22 = P 2 ( 20 ) H 10 + P 2 ( 11 ) H 01 ,
b 24 = P 2 ( 40 ) H 30 + P 2 ( 31 ) H 21 + P 2 ( 22 ) H 12 + P 2 ( 13 ) H 03 .
b 26 = P 2 ( 60 ) H 50 + P 2 ( 51 ) H 41 + P 2 ( 42 ) H 32 + P 2 ( 33 ) H 23 + P ( 24 ) H 14 + P 2 ( 15 ) H 05 , b 35 = P 3 ( 50 ) H 40 + P 3 ( 41 ) H 31 + P 3 ( 32 ) H 22 + P 3 ( 23 ) H 13 + P 3 ( 14 ) H 04 , b 37 = P 3 ( 70 ) H 60 + P 3 ( 61 ) H 51 + P 3 ( 52 ) H 42 + P 3 ( 43 ) H 33 + P 3 ( 34 ) H 24 + P 3 ( 25 ) H 15 + P 3 ( 16 ) H 06 ,
b 39 = P 3 ( 90 ) H 80 + P 3 ( 81 ) H 71 + P 3 ( 72 ) H 62 + P 3 ( 63 ) H 53 + P 3 ( 54 ) H 44 + P 3 ( 45 ) H 35 + P 3 ( 36 ) H 26 + P 3 ( 27 ) H 17 + P 3 ( 18 ) H 08 .
This gives g r k and g r of (40) for r 3 , and so the conditional distribution P 1 2 ( x 1 ) of (37), to O ( n 2 ) , in terms of I 0 k of (54) and the coefficients P r ( a b ) .
Theorem 8.
The I 0 k needed for g 1 , g 2 , g 3 of (66) and (38) are given in terms of γ 0 = Φ ( v ) , v of (36), and J 0 k of (63), by
I 01 = J 01 , I 02 = J 02 γ 0 V 22 , I 03 = J 03 3 J 01 V 22 , I 04 = J 04 6 J 02 V 22 + 3 γ 0 ( V 22 ) 2 , I 05 = J 05 10 J 03 V 22 + 15 J 01 ( V 22 ) 2 , I 06 = J 06 15 J 04 V 22 + 45 J 02 ( V 22 ) 2 15 γ 0 ( V 22 ) 3 , I 07 = J 07 21 J 05 V 22 + 105 J 03 ( V 22 ) 2 105 J 01 ( V 22 ) 3 , I 08 = J 08 28 J 06 V 22 + 210 J 04 ( V 22 ) 2 420 J 02 ( V 22 ) 3 + 105 γ 0 ( V 22 ) 4 , I 09 = J 09 36 J 07 V 22 + 378 J 05 ( V 22 ) 2 1260 J 03 ( V 22 ) 3 + 945 J 01 ( V 22 ) 4 .
Proof. 
For k 6 , I 0 k follow from Theorem 2.
By the proof of Theorem 3, I 0 k can be read off [3] and the univariate Hermite polynomials H k ( u ) given in terms of I = 1 by expanding
H k = H k ( u ) = ϕ ( u ) 1 ( d / d u ) k ϕ ( u ) = E ( u + I N ) k , for k 0 .
To summarise, the conditional density of X n 1 2 of (15), is given by Theorem 5, and the conditional distribution is given by (37), (41) in terms of g r of (66) and I 0 k of Theorem 8.
Example 1.
Conditioning when w ^ R 2 is the mean of a sample with cumulants κ a b . The non-zero P r ( a b ) were given in Example 6 of [3]. So b r k = 0 for ( r k ) = ( 11 ) , ( 22 ) , ( 31 ) , ( 33 ) , and for r = 1 , 2 , 3 , other b r k are given by (67)–(70) starting
6 b 13 = κ 30 H 20 + 3 κ 21 H 11 + κ 12 H 02 ,
24 b 24 = κ 40 H 30 + 4 κ 31 H 21 + 6 κ 22 H 12 + 4 κ 13 H 03 ,
72 b 26 = κ 30 2 H 50 + 6 κ 30 κ 21 H 41 + 3 ( 2 κ 30 κ 12 + 3 κ 21 2 ) H 32 + 12 ( κ 30 κ 03 + 9 κ 21 κ 12 ) H 23 + 3 ( 2 κ 03 κ 21 + 3 κ 12 2 ) H 14 + 6 κ 03 κ 12 H 05 .
The relative conditional density is given to O ( n 2 ) by (33) in terms of p ˜ r of (6), p ˜ r k of (55), f r = p r ( x 2 ) of (28) for r 3 , and H k of (56) for k 9 .
So , f 1 = p 13 = P 1 ( 03 ) H 3 , P 1 ( 03 ) = κ 03 / 6 , f 2 = p 24 + p 26 , p 24 = P 2 ( 04 ) H 4 , P 2 ( 04 ) = κ 04 / 24 , p 26 = P 2 ( 06 ) H 6 , P 2 ( 06 ) = κ 03 2 / 72 ,
f 3 = k = 5 , 7 , 9 p 3 k , p 3 k = P 3 ( 0 k ) H k , P 3 ( 05 ) = κ 05 / 120 , P 3 ( 07 ) = κ 04 κ 03 / 144 , P 3 ( 09 ) = ( κ 03 / 6 ) 3 .
The conditional distribution is given by (52) with g r of (66), starting
G 0 = g 0 = Φ ( v ) , g 1 = κ 03 I 03 / 6 b 13 σ 1 ϕ ( v ) ,
for v of ( 36 ) , with σ 2 = V 0 = κ 20 κ 11 2 / κ 02 , μ 1 2 = κ 11 κ 02 1 x 2 ,
I 03 of Theorem 8, and b 13 of (71). As noted this is a far simpler result than using Theorem 6.
Similarly , g 2 = κ 04 I 04 / 24 + κ 03 2 I 06 / 72 ( b 24 + b 26 ) σ 1 ϕ ( v ) , g 3 = k = 5 , 7 , 9 [ P 3 ( 0 k ) I 0 k b 3 k σ 1 ϕ ( v ) ] ,
for b 24 , b 26 of (72), (73) and b 3 k above.
Example 2.
We now build on  the entangled gamma  model of Example 7 of [3], which gave the P r ( a b ) needed. Let G 0 , G 1 , G 2 be independent gamma random variables with means γ = γ 0 , γ 1 , γ 2 . For i = 1 , 2 , set X i = G 0 + G i , w i = E X i = γ + γ i , and let w ^ be the mean of a random sample of size n distributed as ( X 1 , X 2 ) . So, E w ^ = w , and n w ^ = L ( G n 0 + G n 1 , G n 0 + G n 2 ) where G n 0 , G n 1 , G n 2 are independent gamma random variables with means n γ , n γ 1 , n γ 2 . The rth order cumulants of X = ( X 1 , X 2 ) are κ i r = ( r 1 ) ! w i , and otherwise ( r 1 ) ! γ . Now suppose that γ i 1 , the entangled exponential model. So q = 2 , X n 1 and X n 2 have correlation 1 / 2 ,
V = 2 1 1 2 , V 12 V 22 1 = 1 / 2 , V 1 2 = 3 / 2 , V 1 = 2 1 1 2 / 3 , P ( X n 1 | ( X n 2 = x 2 ) < x 1 ( x 2 , u ) ) = Φ ( u ) + O ( n 1 / 2 ) ,
for x 1 ( x 2 , v ) of (25), that is, x 1 ( x 2 , u ) = x 2 / 2 + ( 3 / 2 ) 1 / 2 u . Figure 1 plots the conditional asymptotic quantiles of X n 1 2 , that is, x 1 ( x 2 , u ) , for Φ ( u ) = 0.01 , 0.0025 , 0.1 , 0 , 0.9 , 0.975 , 0.99 . To O ( n 1 / 2 ) , given n and w ^ , this figure is equivalent to a figure of w 1 versus w 2 . That is, Figure 1 shows to O ( n 1 / 2 ) , the likely value of w 1 = w ^ 1 n 1 / 2 x 1 for a given value of w 2 = w ^ 2 n 1 / 2 x 2 . In fact by (26), X n 1 2 = n 1 / 2 ( w 1 w ^ 1 2 ) lies between the outer limits with probability 0.98+ O ( n 1 ) . So although labelled as x 1 versus x 2 , the figure can be viewed as showing the likely value of w 1 = w ^ 1 n 1 / 2 x 1 for a given value of w 2 = w ^ 2 n 1 / 2 x 2 .
We now give C r of (31), D r of (33), H k and p r k of (56), and g r for G r , the coefficients of the expansion for the conditional distribution of (37).
So , P 1 ( 03 ) = 2 / 3 , P 1 ( 21 ) = 1 , P 2 ( 04 ) = 1 / 2 , P 2 ( 31 ) = 1 , P 2 ( 22 ) = 3 / 2 , P 2 ( 06 ) = 2 / 9 , P 2 ( 51 ) = 2 / 3 , P 2 ( 42 ) = 7 / 6 , P 2 ( 33 ) = 26 / 3 P 3 ( 05 ) = 2 / 5 , P 3 ( 41 ) = 1 , P 3 ( 32 ) = 2 , P 3 ( 07 ) = 1 / 3 , P 3 ( 61 ) = P 3 ( 52 ) = 5 / 2 , P 3 ( 43 ) = 3 , P 3 ( 09 ) = 8 / 27 , P 3 ( 81 ) = 4 / 27 , P 3 ( 72 ) = 10 / 27 , P 3 ( 63 ) = 47 / 756 , P 3 ( 54 ) = 59 / 945 . By Theorem 8 , to 3 decimal places , I 03 = J 03 2 J 01 = 0.586 , I 04 = J 04 4 J 02 + 4 γ 0 / 3 = 0.871 , I 05 = J 05 20 J 03 / 3 + 20 J 01 / 9 = 0.709 , I 06 = J 06 10 J 04 + 20 J 02 40 γ 0 / 9 = 3.187 , I 07 = J 07 14 J 05 + 140 J 03 / 3 280 J 01 / 9 = 12.857 , I 08 = J 08 56 J 06 / 3 + 280 J 04 / 3 1120 J 02 / 3 + 560 γ 0 / 27 = 12.077 , I 09 = J 09 24 J 07 + 168 J 05 1120 J 03 + 560 J 01 / 3 = 28.278 .
By Note 2, p ˜ r k of Example 7 of [3], symmetry, and (66),
b 13 = 5 H 20 / 3 + H 11 , b 24 = 3 H 40 / 2 + 2 H 31 , b 26 = [ 7 H 50 + 12 H 41 + 19 H 32 ] / 9 , b 35 = [ 3 H 40 + 2 H 31 + H 22 ] / 5 , b 37 = [ 9 H 60 + 19 H 51 + 30 H 42 + 18 H 33 ] / 6 , b 39 = [ 44 H 80 + 83 H 71 + 206 H 62 + 159 H 53 ] / 27 , g 1 = 2 I 03 / 3 + b 13 σ 1 γ 1 , g 2 = I 04 / 2 + 2 I 06 / 9 b 24 b 26 , g 3 = 2 I 05 / 5 + I 07 / 3 + 8 I 09 / 27 b 35 b 37 b 39 .
Let us work through 2 numerical examples to get the conditional distribution to O ( n 2 ) . We build on Example 7 of [3]. By Theorem 5, if x 2 = 1 , then z = 1 / 2 ,
H 3 = 5 / 8 , H 4 = 17 / 16 , H 6 = 89 / 64 , H 5 = 41 / 32 , H 7 = 461 / 2 7 , H 9 = 6481 / 2 9 , C 1 = f 1 = 5 / 12 , p 24 = 17 / 32 , p 26 = 89 / 288 , f 2 = 121 / 144 , p 35 = 41 / 80 , p 37 = 461 / 384 , p 39 = 6481 / 1728 , f 3 = 52921 / 17280 , C 2 = 83 / 72 , C 3 = 39571 / 17280 .
We worked to 8 significant figures, but display less. If x = ( 1 , 1 ) , then
D 1 = 113 / 324 = 0.349 , D 2 = 120199 / 2 3 3 8 = 2.290 , D 3 = 8896102087 / 2 7 3 12 5 = 26.156 .
So to O ( n 2 ) the relative conditional density of (33) for n = 4 , 16 , 64 , is
( 1 , 1 , 1 ) ( 2 1 , 4 1 , 8 1 ) 0.349 + ( 4 1 , 16 1 , 64 1 ) 2.290 + ( 8 1 , 64 1 , 2 9 ) 26.156 = 1 0.174 + 0.573 + 3.269 1 0.087 + 0.143 + 0.409 1 0.044 + 0.036 + 0.051 ,
so that for n = 4 and 16 we can only include two terms, and for n = 64 , only three terms. We now give the 1st 3 g r , G r , needed by (37) for the conditional distribution to O ( n 2 ) . By (50), σ 2 = 3 / 2 , σ 2 = 1.225 . By (17), μ 1 2 = x 2 / 2 .
For x = ( 1 , 1 ) , μ 1 2 = 1 / 2 , and by ( 18 ) , v = 6 1 / 2 = 0.408 G 0 = g 0 = γ 0 = Φ ( v ) = 0.658 , γ 1 = ϕ ( v ) = 0.367 , γ 2 = 0.509 , γ 3 = 0.795 , γ 4 = 1.501 , γ 5 = 3.191 , γ 6 = 7.500 , γ 7 = 19.150 , γ 8 = 52.500 , γ 9 = 153.200 . K 0 s = ( 6 1 / 2 ) s γ s K 01 = 0.150 , K 02 = 0.0848 , K 03 = 0.0541 , K 04 = 0.0417 , K 05 = 0.0362 , K 06 = 0.03472 , K 07 = 0.0362 , K 08 = 0.0405 , K 09 = 0.0483 . So by ( 63 ) , J 01 = 0.479 , J 02 = 0.0203 , J 03 = 0.372 , J 04 = 0.0738 , J 05 = 0.0731 , J 06 = 0.0713 , J 07 = 0.0718 , J 08 = 0.441 , J 09 = 0.269 . So for x = ( 1 , 1 ) , b 13 = 13 / 27 = 0.481 , g 1 = 0.246 , b 24 = 47 / 54 , b 26 = 2726 / 2107 , g 2 = 0.696 , b 35 = 10 / 27 , b 37 = 9371 / 4374 , b 39 = 163806 / 59049 = 2.774 , g 3 = 3.375 . By ( 38 ) , G 1 = g 1 + C 1 g 0 = 0.0281 , G 2 = 0.040 , G 3 = 0.762 .
For example for n = 4 , 16 , 64 , to O ( n 2 ) , P ( X n 1 < 1 | X n 2 = 1 ) =
0.658 + 0.0141 0.01000 + 0.0952 , n = 4 , 0.658 + 0.00703 0.00250 + 0.0119 , n = 16 , 0.658 + 0.00351 0.000625 + 0.00149 , n = 64 ,
so that divergence begins with the 4th term.
If x 2 = 2 then z = 1 , H 3 = 1 / 2 , H 4 = 23 / 4 , H 6 = 23 / 8 , H 5 = 1 / 4 , H 7 = 29 / 8 , H 9 = 175 / 16 , C 1 = f 1 = 1 / 3 , p 24 = 23 / 4 , p 26 = 23 / 36 , f 2 = 161 / 72 , p 35 = 1 / 10 , p 37 = 29 / 24 , p 39 = 175 / 54 , f 3 = 2303 / 1080 = 2.132 , C 2 = 17 / 8 = 2.125 , C 3 = 733 / 1080 = 0.679 . If x = ( 2 , 2 ) , then D 1 = 37 / 81 = 0.457 , D 2 = 0.387 , D 3 = 13.313 .
So to O ( n 2 ) the relative conditional density of (33) for n = 4 , 16 , 64 , is
( 1 , 1 , 1 ) ( 2 1 , 4 1 , 8 1 ) 0.457 + ( 4 1 , 16 1 , 64 1 ) 0.387 + ( 8 1 , 64 1 , 2 9 ) 13.313 = 1 0.228 + 0.0969 + 1.664 1 0.114 + 0.0242 + 0.208 1 0.0571 + 0.00605 + 0.0260 , ,
so that we can only include three terms. Finally, we now give the 1st three g r , G r , needed by (37) for the conditional distribution to O ( n 2 ) .
For x = ( 2 , 2 ) , μ 1 2 = 1 , v = ( 2 / 3 ) 1 / 2 = 0.816 , G 0 = γ 0 = Φ ( v ) = 0.793 , γ 1 = ϕ ( v ) = 0.286 , γ 2 = 0.559 , γ 3 = 0.762 , γ 4 = 1.522 , γ 5 = 3.176 , γ 6 = 7.511 , γ 7 = 19.142 , γ 8 = 52.505 , γ 9 = 153.190 . K 0 s = ( 6 1 / 2 ) s γ s K 01 = 0.117 , K 02 = 0.0932 , K 03 = 0.0519 , K 04 = 0.0423 , K 05 = 0.0360 , K 06 = 0.0348 , K 07 = 0.0362 , K 08 = 0.0405 , K 09 = 0.0483 . So by ( 63 ) , J 01 = 0.910 , J 02 = 1.0028 , J 03 = 1.055 , J 04 = 1.097 , J 05 = 1.133 , J 06 = 1.168 , J 07 = 1.204 , J 08 = 1.249 , J 09 = 1.293 . I 03 = 0.764 , I 04 = 1.877 , I 05 = 3.877 , I 06 = 6.731 , I 07 = 6.263 , I 08 = 276.110 , I 09 = 848.735 , b 13 = 11 / 27 , g 1 = 0.605 , b 24 = 26 / 27 , b 26 = 1660 / 2187 , g 2 = 0.771 , b 35 = 138 / 405 , b 37 = 20 , 128 / 4374 , b 39 = 1 , 795 , 048 / 3 10 , g 3 = 224.802 . By ( 38 ) , G 1 = g 1 + C 1 g 0 = 0.0180 , G 2 = 2.463 , G 3 = 4.204 .
For example for n = 4 , 16 , 64 , to O ( n 2 ) , P ( X n 1 < 2 | X n 2 = 2 ) =
0.793 + 0.00902 0.616 + 0.525 , n = 4 , 0.793 + 0.00451 0.154 + 0.131 , n = 16 , 0.793 + 0.00226 0.0385 + 0.0164 , n = 64 ,
so that divergence begins with the 3rd term.
Example 3.
Conditioning when the distribution of w ^ is symmetric about w. Then for r odd, C r = D r = g r k = g r = 0 . By (33), the conditional density is
p n 1 2 ( x 1 ) = σ 1 ϕ ( v ) [ 1 + n 1 D 2 + O ( n 2 ) ] , where D 2 = p ˜ 2 ( x ) p 2 ( x 2 ) ,
for p ˜ 2 ( x ) of Example 1 of [3], H k of (56), and
p 2 ( x 2 ) = k 022 H 2 / 2 + k 043 H 4 / 24 .
By (52), the conditional distribution of X n 1 2 is
Φ ( v ) + n 1 G 2 + O ( n 2 ) , where G 2 = g 2 p 2 ( x 2 ) Φ ( v ) , g 2 = k = 2 , 4 [ P 2 ( 0 k ) I 2 ( 0 k ) b 2 k σ 1 ϕ ( v ) ] ,
for b 2 k of (68) and (69).
Example 4.
Discussions of pivotal statistics advocate using the distribution of a sample mean, given the sample variance. So q = 2 . Let w ^ 2 , w ^ 2 be the usual unbiased estimates of the mean and variance from a univariate random sample of size n from a distribution with rth cumulant κ r . So w 1 = κ 1 , w 2 = κ 2 . By the last 2 equations of Section 12.15 and (12.35)–(12.38) of [20], the cumulant coefficients needed for P ¯ r 1 k of (3) for r 3 , – the coefficients needed for the conditional density to O ( n 2 ) , in terms of ( i 1 j 1 i 2 j 2 ) = κ i 1 j 1 κ i 2 j 2 , are
k 201 = κ 2 , k 111 = κ 3 , k 021 = κ 4 + 2 κ 2 2 , V = κ 2 κ 3 κ 3 κ 4 + 2 κ 2 2 , k 101 = k 011 = 0 , k 302 = κ 3 , k 212 = k 122 = 0 , k 032 = ( 6 ) + 12 ( 24 ) + 4 ( 3 2 ) + 8 ( 2 3 ) , k 202 = k 112 = 0 , k 022 = 2 ( 2 2 ) , k 403 = ( 4 ) , k 313 = ( 5 ) , k 223 = k 133 = 0 , k 043 = ( 8 ) + 24 ( 26 ) + 32 ( 35 ) + 32 ( 4 2 ) + 144 ( 2 2 4 ) + 96 ( 23 2 ) + 48 ( 2 4 ) , k 102 = k 012 = k 303 = k 213 = k 123 = 0 , k 033 = 12 ( 24 ) + 16 ( 2 3 ) , k 504 = k 324 = k 234 = k 144 = 0 , k 414 = ( 6 ) , k 054 = ( 10 ) + 40 ( 28 ) + 80 ( 37 ) + 200 ( 46 ) + 96 ( 5 2 ) + 480 ( 2 2 6 ) + 1280 ( 235 ) + 1280 ( 24 2 ) + 960 ( 3 2 4 ) + 1920 ( 2 3 4 ) + 1920 ( 2 2 3 2 ) + 384 ( 2 5 ) .
(33) gives D r in terms of p ˜ r and p r , that is, in terms of p ˜ r k and p r k of (28) in terms of P r ( a b ) . In this example, many of these are 0. The non-zero P r ( a b ) are in order needed,
P 1 ( 30 ) = κ 3 / 6 , P 1 ( 03 ) = k 032 / 6 , P 2 ( 02 ) = κ 2 2 , P 2 ( 40 ) = κ 4 / 24 , P 2 ( 04 ) = k 043 / 24 , P 2 ( 32 ) = κ 5 / 6 , P 2 ( 60 ) = κ 3 2 / 72 , P 2 ( 06 ) = k 032 2 / 72 , P 2 ( 33 ) = κ 3 k 032 / 36 . P 3 ( 03 ) = k 033 / 6 . P 3 ( 05 ) = k 054 / 120 + k 022 k 032 / 12 . P 3 ( 70 ) = κ 3 κ 4 / 144 , P 3 ( 07 ) = k 032 k 042 / 144 , P 3 ( 62 ) = κ 3 k 313 / 36 , P 3 ( 43 ) = k 032 κ 4 / 144 , P 3 ( 34 ) = ( κ 3 k 043 + k 022 k 313 ) / 144 , P 3 ( 90 ) = κ 3 / 6 3 , P 3 ( 09 ) = ( k 032 / 6 ) 3 , P 3 ( 63 ) = 3 κ 3 2 k 032 / 6 3 , P 3 ( 36 ) = 3 κ 3 k 032 2 / 6 3 . So , p ˜ 11 = 0 , p ˜ 13 = P 1 ( 30 ) H 30 + P 1 ( 03 ) H 03 , p ˜ 22 = P 2 ( 02 ) H 02 , p ˜ 24 = P 2 ( 40 ) H 40 + P 2 ( 04 ) H 04 + P 2 ( 32 ) H 31 , p ˜ 26 = P 2 ( 60 ) H 60 + P 2 ( 06 ) H 06 + P 2 ( 33 ) H 33 , p ˜ 31 = 0 , p ˜ 33 = P 3 ( 30 ) H 30 + P 3 ( 03 ) H 03 , p ˜ 35 = P 3 ( 05 ) H 05 , p ˜ 37 = P 3 ( 70 ) H 70 + P 3 ( 2 7 ) H 07 + P 3 ( 62 ) H 61 + P 3 ( 43 ) H 43 + P 3 ( 34 ) H 34 + P 3 ( 25 ) H 25 + P 3 ( 16 ) H 16 + P 3 ( 07 ) H 07 , p ˜ 39 = P 3 ( 90 ) H 90 + P 3 ( 63 ) H 63 + P 3 ( 36 ) H 36 + P 3 ( 09 ) H 09 . Also , b 13 = P 1 ( 30 ) H 20 , b 22 = P 2 ( 20 ) H 10 + P 2 ( 11 ) H 01 , b 24 = P 2 ( 40 ) H 30 + P 2 ( 31 ) H 21 , b 26 = P 2 ( 60 ) H 50 + P 2 ( 33 ) H 23 , b 31 = 0 , b 33 = P 3 ( 30 ) H 20 , b 35 = 0 , b 37 = P 3 ( 70 ) H 60 + P 3 ( 61 ) H 51 + P 3 ( 43 ) H 33 + P 3 ( 34 ) H 24 , b 39 = P 3 ( 90 ) H 80 + P 3 ( 63 ) H 53 + P 3 ( 36 ) H 26 .
For r = 1 , 2 , 3 , p ˜ r ( x ) is now given by (13), p r ( x ) , and Section 2 of [3]. By (4) and (33), this gives the conditional density p n 1 2 ( x 1 ) to O ( n 2 ) . And (66) gives g r needed for the conditional distribution P n 1 2 ( x 1 ) to O ( n 2 ) .

5. Conclusions

Conditioning is a very useful and basic way to use correlated information to reduce the variability of an estimate. These results provide the means to obtain conditional regions and 1- or 2-sided conditional confidence intervals.
Section 4 gave the conditional density and distribution of X n 1 given X n 2 to O ( n 2 ) where X n 1 X n 2 is any partition of X n = n 1 / 2 ( w ^ w ) . The expansion (33) gave the conditional density of any multivariate standard estimate. Our main result, an explicit expansion for the conditional distribution (37) to O ( n 2 ) , is given in terms of the leading I ¯ 1 k of (42). These are given explicitly by Theorems 3.3 and 3.4.
When q 1 = q 2 = 1 , Theorem 4.1 simplified the conditional density expansion, and Theorem 7 gave a huge simplification, and the coefficients of the conditional distribution expansion in terms of I 0 k = I 2 k of Theorem 4.4.
Cumulant coefficients can also be used to obtain estimates of bias O ( n k ) for k 2 : see [21,22,23].

6. Discussion

Ref. [3] gave the density and distribution of X n = n 1 / 2 ( w ^ w ) to O ( n 2 ) , for w ^ R q any standard estimate, in terms of functions of the cumulant coefficients k ¯ j 1 r of (2), called the Edgeworth coefficients, P ¯ r 1 k .
Most estimates of interest are standard estimates, including smooth functions of sample moments, like the sample skewness, kurtosis, correlation, and any multivariate function of k-statistics. (These are unbiased estimates of cumulants and their products, the most common example being that for a variance.) Unbiased estimates are not needed for Edgeworth expansions, although this does simplify the Edgeworth coefficients, as seen in Examples 1, 2, and 4. However unbiased estimates are not available for most parameters or functions of them, such as the ratio of two means or variances, except for special cases of exponential families. Ref. [1] gave the cumulant coefficients for smooth functions of standard estimates.
A good approximation for the distribution of an estimate, is vital for accurate inference. It enables one to explore the distribution’s dependence on underlying parameters. Our analytic method avoids the need for simulation or jack-knife or bootstrap methods while providing greater accuracy than any of them. Ref. [10] used the Edgeworth expansion to show that the bootstrap gives accuracy to O ( n 1 ) . Ref. [24] said that “2nd order correctness usually cannot be bettered”. But this is not true using our analytic method. Simulation, while popular, can at best shine a light on behaviour, only when there is a small number of parameters, and only for limited values of their range.
Estimates based on a sample of independent, but not identically distributed random vectors, are also generally standard estimates. For example, for a univariate sample mean X ¯ = n 1 j = 1 n X j n where X j n has rth cumulant κ r j n , then κ r ( X ¯ ) = n 1 r κ r where κ r = n 1 j = 1 n κ r j n is the average rth cumulant. For some examples, see [2,25,26,27] The last is for a function of a weighted mean of complex random matrices.
For conditions for the validity of multivariate Edgeworth expansions, see [28] and its references, and Appendix C of [3].
While the use of Edgeworth-Cornish-Fisher expansions is widespread, few papers address how to deal with their divergence for small sample sizes. Refs. [29,30] avoided this question as it did not arise in their examples. In contrast we confronted this in Example 2, the examples of Withers (1984), and in Example 7 of [3].
We now turn to conditioning. Conditioning on w ^ 2 makes inference on w 1 more precise by reducing the covariance of the estimate. The covariance of w ^ 1 | w ^ 2 can be substantially less than that of w ^ 1 . See the references at the end of Section 1.
A conditional distribution by tilting, was first given by [31] up to O ( n 1 ) , for a bivariate sample mean. See Chapter 4 of [32], Compare [8]. Tilting (also known as small sample asympotics, or saddlepoint expansioins), was first used in statistics by [33]. He gave an approximation to the density of a sample mean, good for the whole line, not just in the region where the Central Limit Theorem approximation holds.
Future directions.
1. The results here give the first step for constructing confidence intervals and confidence regions of higher order accuracy. See [6,34]. What is needed next, is an application of [1] to obtain the cumulant coefficients of θ ^ i = V ^ i i 1 / 2 ( w ^ i w i ) , i = 1 , 2 , or those of θ ^ = V ^ 1 / 2 ( w ^ w ) . This should be straightforward.
2. When q 1 = 1 , our expansion for the conditional distribution of X n 1 2 of (15), can be inverted using the Lagrange Inversion Theorem, to give expansions for its percentiles. This should be straightforward. (The quantile expansions of [29] and Withers (1984) do not apply as Appendix A shows that conditional estimates of standard estimates are not standard estimates.)
3. Here we have only considered expansions about the normal. However expansions about other distributions can greatly reduce the number of terms by matching the leading bias coefficient. The framework for this is [2], building on [34]. For expansions about a matching gamma, see [35,36].
4. The results here can be extended to tilted (saddlepoint) expansions by applying the results of [2]. The tilted version of the multivariate distribution and density of a standard estimate are given by Corollaries 3, 4 there, and that of the conditional distribution and density follow from these. For the entangled gamma of Example 2, this requires solving a cubic. See also [37].
5. A possible alternative approach to finding the conditional distribution, is to use conditional cumulants, when these can be found. Section 6.2 of [38] uses conditional cumulants to give the conditional density of a sample mean to O ( n 3 / 2 ) . Section 5.6 of [39] gave formulas for the 1st 4 cumulants conditional on X 2 = x 2 only when X 1 and X 2 are uncorrelated. He says that this assumption can be removed, but gives no details how. That is unlikely to give an alternative to our approach, for as well as giving expansions for the first 3 conditional cumulants, Appendix A shows that the conditional estimate is not a standard estimate.
6. Lastly we discuss numerical computation. We have used [40] for our calculations. Its input is V 11 , V 12 , V 22 and y 1 , y 2 , - not V 11 , V 12 , V 22 and x 1 , x 2 . There is a function sub2(sb1,sb2) which takes as argument the two subscripts of mu, and returns the value. If global variables mu20, mu02, mu11 are symbolic variables (defined using sympy) then it returns the answer in terms of those, but if they are numeric then it returns a numeric answer. There is another function called biHermite(n, m, y1, y2) which takes the 2 subscripts of H. If y1 and y2 are symbolic, then it returns a symbolic answer, but if they are numeric it returns a numeric answer. A numerical example is given by Example 2, that is, for the case V 11 = 2 , V 12 = 1 , V 22 = 2 and x 1 = x 2 = 1 or x 1 = x 2 = 2 .
Similar software for numerical calculations for Theorems 5, 7 and 8 would be invaluable, as would software for applying the Lagrange Inversion Theorem. (We mention R-4.4.1 for Windows: dmvnorm for the density function of the multivariate normal, mvtnorm for the multivariate normal, qmvnorm for quantiles, and rmvnorm to generate multivariate normal variables.) On bivariate Hermite polynomials, see cran.r-project.org/web/packages/calculus/vignettes/hermite.html accessed on 20 December 2024.

Funding

This research received no external funding.

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The author declares no conflict of interest.

Appendix A. Conditional Moments

Here we give expansions for the conditional moments of X n 1 2 of (15), in terms of the conditional normal moments of X 1 2 , of (15). And we show that
w ^ 1 2 = w ^ 1 | ( X n 2 = x 2 )
is neither a standard estimate of w 1 , nor a Type B estimate, as defined below.
Consider the case q 1 = 1 . By (19),
X 1 2 = μ + σ N , where μ = μ 1 2 , σ 2 = V 1 2 of ( 18 ) , N N 1 ( 0 , 1 ) . Set M s = E X 1 2 s . So M 1 = μ , M 2 = μ 2 + σ 2 , M 3 = μ 3 + 3 μ σ 2 , M 4 = μ 4 + 6 μ 2 σ 4 + 3 σ 4 . For r = 0 , 1 , , set A r B r = a + b = r A a B b , A r B r C r = a + b + c = r A a B b C c .
Non-central moments.
Theorem A1.
Take C r , D r of Theorem 3.1. Set p ˜ 0 ( x ) = 1 . For s > 0 , the sth conditional moment of X n 1 2 of (15) about Φ 1 2 ( x 1 ) of (24), has the expansion
m n s = E X n 1 2 s = n s / 2 E ( w ^ 1 2 w 1 ) s r = 0 n r / 2 G r s
where G r s = C r g r s , g r s = E X 1 2 s P r , and P r = p ˜ r ( x ) at x 1 = X 1 2 . So , G 0 s = g 0 s = M s = E X 1 2 s .
g r s = k = 1 3 r [ g r k s : k r even ] , for r 1 , where for p ˜ r k of ( 6 ) ,
g r k s = E X 1 2 s P ˜ r k = P ¯ r 1 k I ¯ s 1 k , P ˜ r k = p ˜ r k at x 1 = X 1 2 ,
and for 1 i 1 , , i k q , I ¯ s 1 k = I s i 1 i k = E X 1 2 s H ¯ 1 k ( X 1 2 ) ,
for H ¯ 1 k ( X 1 2 ) = H ¯ 1 k at x 1 = X 1 2 .
Proof. 
This follows from Theorem 1. □
So by (A3), the sth conditional moment of X n 1 2 is
m n s = M s + n 1 / 2 G 1 s + n 1 G 2 s + O ( n 3 / 2 ) , where G 1 s = C 1 M s + g 1 s , G 2 s = C 2 M s + C 1 g 1 s + g 2 s , g 1 s = g 11 s + g 13 s , g 2 s = g 22 s + g 24 s + g 26 s ,
of (A5) and (A6). For example,
g 11 s = k ¯ 1 1 E X 1 2 s H ¯ 1 ( X 1 2 ) , and g 13 s = k ¯ 2 1 3 E X 1 2 s H ¯ 1 3 ( X 1 2 ) / 6 .
So w ^ 1 2 of (A1) is not a standard estimate, as by (A3), the expansion for its mean is a power series in n 1 / 2 , not n 1 . Is it a Type B estimate? These are defined as for a standard estimate, but with cumulant expansions being series in n 1 / 2 , not n 1 . We shall see. Take q 2 = q 1 = 1 . By Theorem 6, for P r ( a b ) of (3), g r k s of (A4) is given by
g r k s = b = 0 k P r ( k b , b ) I k b , b s , where I a b s = E X 1 2 s H a b ( X 1 2 ) , and H a b ( X 1 2 ) = H a b at x 1 = X 1 2 .
For example,
g r 1 s = P r ( 10 ) I 10 s + P r ( 01 ) I 01 s g r 3 s = P r ( 30 ) I 30 s + P r ( 21 ) I 21 s + P r ( 12 ) I 12 s + P r ( 03 ) I 03 s .
Finding the  I a b s .
The H a b needed are given in Appendix B of [3] in terms of
y = V 1 x : y i = V i 1 x 1 + V i 2 x 2 , y 1 = μ 20 x 1 + μ 11 x 2 , y 2 = μ 11 x 1 + μ 02 x 2 .
For example,
H 10 = y 1 = μ 20 x 1 + μ 11 x 2 , H 01 = y 2 = μ 11 x 1 + μ 02 x 2 , H 30 = y 1 3 3 y 1 μ 20 = ( μ 20 x 1 + μ 11 x 2 ) 3 3 ( μ 20 x 1 + μ 11 x 2 ) μ 20 , H 03 = y 2 3 3 y 2 μ 02 = ( μ 11 x 1 + μ 02 x 2 ) 3 3 ( μ 11 x 1 + μ 02 x 2 ) μ 20 , H 21 = y 2 ( y 1 2 μ 20 ) 2 y 1 μ 11 , H 12 = y 1 ( y 2 2 μ 02 ) 2 y 2 μ 11 .
Let us write H a b in terms of M s of (A2), as
H a b = k = 0 a + b H a b k x 1 k . Then , I a b s = k = 0 a + b [ H a b k M s + k : s + k even ] . So , I 10 s = H 10 0 M s + H 10 1 M s + 1 , I 01 s = H 01 0 M s + H 01 1 M s + 1 : for odd s , I 10 s = H 10 1 M s + 1 = μ 20 M s + 1 , I 01 s = H 01 1 M s + 1 = μ 02 x 2 M s + 1 , and for even s , I 10 s = H 10 0 M s = μ 11 x 2 M s , I 01 s = H 01 0 M s = μ 02 x 2 M s . So , H 10 0 = μ 11 x 2 , H 10 1 = μ 20 , H 01 0 = μ 02 x 2 , H 01 1 = μ 11 , H 30 0 = ( μ 11 x 2 ) 3 3 μ 11 x 2 μ 20 , H 30 1 = 3 μ 20 [ ( μ 11 x 2 ) 2 μ 20 ] , H 30 2 = 3 μ 20 2 μ 11 x 2 , H 30 3 = μ 20 3 , H 03 0 = ( μ 02 x 2 ) 3 3 μ 02 x 2 μ 20 , H 03 1 = 3 μ 11 [ ( μ 02 x 2 ) 2 μ 11 ] , H 03 2 = 3 μ 11 2 μ 02 x 2 , H 03 3 = μ 11 3 , H 21 0 = μ 02 x 2 [ ( μ 11 x 2 ) 2 μ 20 ] 2 μ 11 2 x 2 , H 21 1 = μ 11 [ ( μ 11 x 2 ) 2 μ 20 ] + 2 μ 20 μ 11 ( μ 02 x 2 2 1 ) , H 21 2 = μ 20 μ 22 x 2 since μ 22 = μ 20 μ 02 + 2 μ 11 2 , H 21 3 = μ 11 μ 20 2 , H 12 0 = μ 11 x 2 [ ( μ 02 x 2 ) 2 μ 02 ] 2 μ 02 μ 11 x 2 , H 12 1 = μ 20 [ ( μ 02 x 2 ) 2 μ 02 ] + 2 μ 11 2 ( μ 02 x 2 2 1 ) , H 12 2 = μ 11 μ 22 x 2 , H 12 3 = μ 20 μ 11 2 .
To get a general formula for H a b k , set
c 1 = V 11 x 1 , c 2 = I V 11 X 1 , c 3 = V 12 x 2 , c 4 = I V 12 X 2 . So , y 1 + I Y 1 = c 1 + c 2 , y 2 + I Y 2 = c 3 + c 4 , H a b = E ( c 1 + c 2 ) a ( c 3 + c 4 ) b = j = 0 a a j c 1 a j k = 0 b b k c 3 b k C j k where C j k = E c 2 j c 4 k = I j + k V 11 j V 12 k μ j k , μ j k = E X 1 j X 2 k . So , H a b a j = a j V 11 a j k = 0 b b k c 3 b k C j k ,
where C j k = 0 if j + k is odd. μ j k is just μ j k of Appendix B of [3] with V replaced by V 1 .
Central moments. Set m s ( X ) = E X s and μ s ( X ) = E ( X E X ) s .
For m s = m s 1 2 of (A3), set
μ s = μ s 1 2 = μ s ( X n 1 2 ) = n s / 2 μ s ( w ^ 1 2 ) . So by ( A 3 ) , μ 2 = m 2 m 1 2 r = 0 n r / 2 μ 2 r , where μ 2 r = G r 2 G r 1 G r 1 , and μ 3 = m 3 3 m 1 m 2 + 2 m 1 3 r = 0 n r / 2 μ 3 r where μ 3 r = G r 3 3 G r 1 G r 2 + 2 G r 1 G r 1 G r 1 .
Is the conditional estimate w ^ 1 2 a Type B estimate? This requires its rth cumulant to have magnitude O ( n 1 r ) for r 1 . This is true for r = 1 and 2 but not for r = 3 , as μ r ( w ^ 1 2 ) has magnitude O ( n r / 2 ) , since μ s 1 2 = O ( 1 ) .

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Figure 1. x 1 ( x 2 , v ) = x 2 / 2 + ( 3 / 2 ) 1 / 2 v of (25) for Φ ( v ) = 0.01 , 0.1 , 0 , 0.9 , 0.99 —courtesy of Dr Paul Teal.
Figure 1. x 1 ( x 2 , v ) = x 2 / 2 + ( 3 / 2 ) 1 / 2 v of (25) for Φ ( v ) = 0.01 , 0.1 , 0 , 0.9 , 0.99 —courtesy of Dr Paul Teal.
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