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I have encountered the following integral in my research which does not give-in to my attempts: $$ \int_\mathbb{R} x \left( \frac{1}{\sigma_1} \phi\left(\frac{x}{\sigma_1}\right) \Phi\left(\frac{x-\mu}{\sigma_2}\right) + \frac{1}{\sigma_2} \phi\left(\frac{x-\mu}{\sigma_2}\right) \Phi\left(\frac{x}{\sigma_1}\right) \right) dx $$ where $\phi(x)$ denote probability density function and $\Phi(x)$ a cumulative density function of the standard normal distribution. $\sigma_1$ and $\sigma_2$ are positive, and $\mu$ is real.

I would appreciate hints on how to evaluate it. Thank you!

2 Answers 2

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Let's rewrite the integral a little. Denote $X$ a Gaussian random variable with zero mean and variance $\sigma_1^2$, and $Y$ an independent Gaussian random variable with mean $\mu$ and variance $\sigma_2^2$. Then your integral reads: $$ \begin{eqnarray} \int_\mathbb{R} \left( x f_X(x) F_Y(x) + x f_Y(x) F_X(x) \right) \mathbb{x} &=& \int_\mathbb{R} x f_X(x) F_Y(x) \mathbb{x} + \int_\mathbb{R} y f_Y(y) F_X(y) \mathbb{y} \\ &=& \mathbb{E}\left( X [ Y \leqslant X] + Y [X < Y] \right) \\ &=& \mathbb{E}\left(\max(X,Y) \right) = \mathbb{E}\left(\max(X-Y, 0)+Y \right) \\ &=& \mathbb{E}\left(\max(X-Y, 0) \right) + \mu \end{eqnarray} $$ where $[X

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Is that the expected value of the max of two independent normals ? If so, you have the following simple argument: with m = min(X,Y), M = Max(X,Y), m+M = X+ Y and M-m = |X - Y|. As you have the expectations of the rhs (this is special to normal) you can find expected value of m,M. The first is $\mu$ and the second is the expected value of the absolute value of a $N(\mu, \sigma_1^2 + \sigma_2^2)$, which is straightforward. If answer to first question is 'yes' one expects this comes to same expression as in previous.

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    Could you please point out a simple way of evaluating the rhs, if different from answers already provided. References will do as well. Thank you.2012-09-14
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    i added a little detail, the issue it comes down to, $\mathbb E |X|$ where X is normal , really is straightforward. I've no reason to think previous post (Sasha) is incorrect.2012-09-14