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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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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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