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What is the correlation of a bivariate normal distribution after truncating the tails of both variables at $\alpha$ standard deviations?

In symbols, what is $$E\left[XY\Big|X| \leq z_{\alpha/2}, |Y| \leq z_{\alpha/2}\right]\ ?$$

I’d like a formula for this; for now, here is a computation in R with $\alpha = .1$ and $\rho = .9$.

# generating the data set
set.seed(42)
n <- 5000
R <- matrix(c( 1,.9,
              .9, 1), 2, 2)
jd <- MASS::mvrnorm(empirical = TRUE,
                    mu = c(0,0),
                    Sigma = R,
                    n = n)

remove truncated values


alpha <- .10
crit <- qnorm(alpha/2, lower.tail = FALSE)
jd[abs(jd)>(crit)] <- NA
jd <- na.omit(jd)


Correlation


cor(jd)

The original correlation was .9 and the truncated correlation is .8382. This is not simply sampling variation; with a high sample size I also found find approximately the same difference (.837).

Can this correlation be derived analytically for any $\alpha$ and any correlation $\rho$?

POC
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    This answer maybe helpful: https://stats.stackexchange.com/questions/517347/ It provides links to discussion about truncated nultivariate normals. The answer contains a link to paywalled papers, but the last one is on arXiv and is very thorough: https://arxiv.org/abs/1206.5387. – Chad Sexington Jul 12 '23 at 16:53
  • Can you state the problem in math expression? Is $E[XY||X| \leq z_\alpha, |Y| \leq z_\alpha]$ what you want to compute? – Zhanxiong Jul 12 '23 at 17:16
  • Thank you @Zhanxiong. Yes, it is. – POC Jul 12 '23 at 17:18
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    The solution comes down to finding integrals of the type https://stats.stackexchange.com/questions/498851 and with the same integrand multiplied by $x^2,$ which can be found by differentiating the result twice with respect to $b.$ Yes, it's messy, but taking that derivative is elementary. – whuber Jul 18 '23 at 18:06
  • @whuber, an expression with Owen’s T would not be much simpler than the original integral. One could make the “analytical” requirement in the question precise by asking: Does the truncated correlation have an expression in terms of well-known functions of one argument, like $\text{erf}$ or $\Phi$? Then I think the answer would be no. – Matt F. Jul 20 '23 at 09:42
  • @MattF. The sense of "not simpler" depends on how well you understand Owen's T, your capabilities of computing with it and using for analysis, and on mathematical relationships between it and other transcendental functions. I accept that it's a fairly obscure function, but it has been analyzed; it does appear in other contexts; and its definition is tractable, leading to various potential methods of numerical evaluation. – whuber Jul 20 '23 at 13:57

1 Answers1

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Thank you for everyone in the comments, I finaly understood how to compute the answer. From Manjunath & Wilhelm and their R package R package, we can get the solution.

alpha <- .10
crit <- qnorm(alpha/2, lower.tail = FALSE)
a <- -c(crit, crit); b <- c(crit,crit)
mu <- c(0,0); sigma <- R
moments <- tmvtnorm::mtmvnorm(mean=mu, sigma=sigma,
                    lower=a, upper=b)

> moments
$tmean
[1] 0 0


$tvar
          [,1]      [,2]
[1,] 0.5747640 0.4811731
[2,] 0.4811731 0.5747640


> cov2cor(moments$tvar)
          [,1]      [,2]
[1,] 1.0000000 0.8371664
[2,] 0.8371664 1.0000000
POC
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    This is just a numerical calculation. Your question explicitly asks for an "analytical solution." – whuber Jul 20 '23 at 20:14