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Let $X_1$ and $X_2$ be two Random variables with a standard normal distribution, and the two variables are independent.

Find $E[X_1|X_1>X_2]$

My answer is far.

If we knew $X_2$, then the answer would be: $\frac{\phi(X_2)}{1-\Phi(X_2)}$

But, since we don't know X_2 either I have

$\int_{-\infty}^{\infty} \frac{\phi^2(X_2)}{1-\Phi(X_2)}dX_2$

I cannot solve this integral, I've tried Integration by parts, but I get stuck.

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  • $\begingroup$ In what sense you 'do not know ' $X_2$? $X_2 \sim N(0,1)$ $\endgroup$ – Alex Jan 25 '13 at 22:53
  • $\begingroup$ Sorry, I'm not sure understand your comment. I started off assuming both $X_1$ and $X_2$ are $N(0,1)$ $\endgroup$ – Greg Jan 25 '13 at 23:50
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Let $\phi$ and $\Phi$ be the pdf and cdf of the standard normal distribution, and for a condition $A$, let $[A]$ be $1$ if $A$ is true and $0$ otherwise. Then

\begin{eqnarray*} {\Bbb E}[X_1\mid X_1>X_2] &=& \frac{{\Bbb E}[X_1 [X_1>X_2]]}{{\Bbb P}[X_1>X_2]}\\ &=& \frac{1}{{\Bbb P}[X_1>X_2]} \int_{x>y} x \phi(x) \phi(y) \, dy\, dx\\ &=& \frac{1}{{\Bbb P}[X_1>X_2]} \int_{\Bbb R} x \phi(x) \left( \int_{y<x} \phi(y) \, dy\right) \, dx\\ &=& \frac{1}{{\Bbb P}[X_1>X_2]} \int_{\Bbb R} x \phi(x) \Phi(x) \, dx. \end{eqnarray*} Since ${\Bbb P}[X_1>X_2]=\frac12$, this equals $$ 2 \int_{\Bbb R} x \phi(x) \Phi(x) \, dx. $$ You can integrate this by parts by setting $u=\Phi(x)$, $dv=x \phi(x) dx$.

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  • $\begingroup$ Thanks for this. Could you elaborate on that first step, I don't quite see how you got there? $\endgroup$ – Greg Jan 26 '13 at 9:15
  • $\begingroup$ I expanded the answer. $\endgroup$ – David Moews Jan 26 '13 at 9:52
  • $\begingroup$ so Integrating by parts, I get $-\Phi(x)*\phi(x)+\int \phi^2(x)dx$, which I get to be 0? $\endgroup$ – Greg Jan 26 '13 at 16:20
  • $\begingroup$ No, that's not right. After all, you expect to have ${\Bbb E}[X_1\mid X_1>X_2]>{\Bbb E}[X_1]=0$. $\endgroup$ – David Moews Jan 26 '13 at 18:52
  • $\begingroup$ Correct, sorry made a mistake with the integral $\int \phi^2(x)$. $\endgroup$ – Greg Jan 26 '13 at 20:14

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