Can we evaluate the following integral ?

$$\int_0^\infty x e^{-x^2} \Phi(ax+b)\,\mathrm dx$$

Here $\Phi(\cdot)$ is the cumulative probability distribution function of a standard normal random variable, and the question is supposed to be finding $E[\frac 12\Phi(aX+b)]$ where $X$ is a Rayleigh random variable with density function $2xe^{-x^2}\mathbf 1_{x: x \geq 0}$.

  • $\begingroup$ given the answer by Chinny84, what are you after? the result will be in terms of cdf of standard normal. $\endgroup$ – Math-fun May 5 '15 at 10:34
  • $\begingroup$ Actually Chinny84's answer is satisfactory, writing the solution in terms of od CDF is better than other ugly expressions $\endgroup$ – Alireza May 5 '15 at 15:57

$$ \int_0^{\infty} \left(-\frac{1}{2}\dfrac{d}{dx}\mathrm{e}^{-x^2}\right)\phi(ax+b)dx $$ use integration by parts $$ \left[-\frac{1}{2}\mathrm{e}^{-x^2}\phi(ax+b)\right]_0^{\infty} +\int_0^{\infty} \frac{1}{2}\mathrm{e}^{-x^2}\phi'(ax+b) dx $$ where $$ \phi'(ax+b) = \dfrac{d}{dx}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{ax+b}\mathrm{e}^{-s^2}ds = \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-(ax+b)^2}a $$ thus you end up getting $$ \left[-\frac{1}{2}\mathrm{e}^{-x^2}\phi(ax+b)\right]_0^{\infty} +\frac{a}{2\sqrt{2\pi}}\int_0^{\infty} \mathrm{e}^{-x^2}\mathrm{e}^{-(ax+b)^2} dx $$ so you should be able to finish it off. (if I haven't made a mistake).


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