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I am trying to obtain the following conditional expectation

$$E[X|Z]$$

where $Z= \max(X,Y)$ and $X,Y$ are independent Gaussian random variables. Can you help me with this?

Thank you!

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1 Answer

This is an interesting question, more involved than it may look at first sight. Here are the explicit formulae which answer it.


Let us first assume that $X$ and $Y$ follow the same distribution with cumulative distribution function $F$, hence $F(x)=\mathrm P(X\leqslant x)=\mathrm P(Y\leqslant x)$ for every real number $x$. Then, $\mathrm E(X\mid Z)=g(Z)$ almost surely, with $$ g(z)=z-\frac1{2F(z)}\int_{-\infty}^zF(x)\mathrm dx. $$ When the common distribution of $X$ and $Y$ is standard Gaussian, $F=\Phi$ and an integration by parts yields $$ g(z)=\frac12\left(z-\frac{\mathrm e^{-z^2/2}}{\sqrt{2\pi}\Phi(z)}\right). $$


In somewhat more generality now, assume that the distribution of $X$ has density $f_X$ and cumulative distribution function $F_X$ and the distribution of $Y$ has density $f_Y$ and cumulative distribution function $F_Y$. Then $\color{red}{\mathrm E(X\mid Z)=g(Z)}$ almost surely, with $$ g(z)=\frac{zf_X(z)F_Y(z)+f_Y(z)\,\mathrm E(X;X\leqslant z)}{f_X(z)F_Y(z)+f_Y(z)F_X(z)}. $$ The only term in $g(z)$ which is not $z$, $f_X(z)$, $f_Y(z)$, $F_X(z)$ or $F_Y(z)$ is $$ \mathrm E(X;X\leqslant z)=\int_{-\infty}^zxf_X(x)\mathrm dx=zF_X(z)-\int_{-\infty}^zF_X(x)\mathrm dx, $$ which yields the definitive expression $$ \color{red}{g(z)=z-\frac{f_Y(z)}{f_X(z)F_Y(z)+f_Y(z)F_X(z)}\int_{-\infty}^zF_X(x)\mathrm dx}. $$ This holds for any independent random variables $X$ and $Y$ with continuous distributions. Two remarks. First, when $X$ and $Y$ are identically distributed, one recognizes the expression written in the first part of this answer. Second, I call this expression definitive because I do not think there exists a much simpler expression for general Gaussian random variables.

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Please let me know what you know, what you tried and why this failed, and I might add some details. –  Did Jan 27 '12 at 15:23
    
Thank you for your quick response. Could you please add some more details to your explanation? I don't fully understand it. I'm a researcher (not an student :) ). In my research, the above expected value appears as a consequence of the following relationship: $Z= \max(X,Y)$ where $X \sim N_X( \mu_X, \sigma_X^2)$ and $Y \sim N_Y(\mu_Y, \sigma_Y^2)$. I've already computed the pdf $P(Z)$: $P(Z=z)= P(X=z) \Phi_Y( Y \leq z) + P(Y=z) \Phi_X( X \leq z)$ Then, I also need to obtain $E[X|Z=z]$. Thank you for your time. –  Jose Jan 27 '12 at 15:54
    
@DidierPiau : The question didn't say there was a common distribution. –  Michael Hardy Jan 27 '12 at 16:55
    
Yes, the distributions of $X$ and $Y$ are different. I've been thinking in the following. If $\Phi_Y(Y\leq z)= 1$, then $P(X|Z)= 1$ and the expectation should be $E[X|Z=z]=z$. On the other hand, $P(X|Z)= P(X)$ if $\Phi_X(X \leq z)= 1$. –  Jose Jan 27 '12 at 17:17
    
Jose: The revised version deals with the general case. –  Did Jan 28 '12 at 7:25
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