Discussion
If \(X\) is a random variable with the cumulative distribution function
\[ F(x) = \begin{cases} 0, & x \leq 0, \\ 1 - e^{-\frac{x^{2}}{2}}, & x \gt 0, \end{cases} \]
then \(E(X)\) equals
Answer: A
The mean is \(\sqrt{\frac{\pi}{2}}\).
For a non-negative random variable, \(E(X) = \int_{0}^{\infty} (1 - F(x))\,dx\). Here \(1 - F(x) = e^{-x^{2}/2}\), so
\[ \begin{aligned} E(X) &= \int_{0}^{\infty} e^{-\frac{x^{2}}{2}}\,dx \\ &= \frac{1}{2}\sqrt{2\pi} = \sqrt{\frac{\pi}{2}} \end{aligned} \]
using half of the Gaussian integral \(\int_{-\infty}^{\infty} e^{-x^{2}/2}dx = \sqrt{2\pi}\). This is the Rayleigh distribution with scale 1, whose mean is about 1.25.
\(\sqrt{2\pi}\) is the full Gaussian integral, forgetting that the range is only \((0, \infty)\). \(\sqrt{\frac{2}{\pi}}\) is the reciprocal. 1 is the mode of this distribution, where the density \(xe^{-x^{2}/2}\) peaks, not its mean.
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