Let $X$ be a normally distributed random variable with mean $0$ and variance $1$. Let $\Phi$ be the cumulative distribution function of the variable $X$. The find the expectation of $\Phi(X)$.

I have $$ E(\Phi(X))= \int\limits_{-\infty}^{\infty}\Phi(x)\frac{e^{-x^2/2}}{\sqrt{2\pi}}\;\mathrm{d}x, \quad\text{where}\quad \Phi(x)=\int\limits_{-\infty}^{x}\frac{e^{-t^2/2}}{\sqrt{2\pi}}\;\mathrm{d}t. $$ I am stuck here. How do I proceed?


There is an easier approach.

Note that $\Phi(x)$ is a continuous increasing function going from $0$ to $1$. Let $Y=\Phi(X)$, so $Y$ is in the interval $(0,1)$. Then $$F(y)=\Pr(Y \le y) =\Pr(\Phi(X) \le y) = y$$ so $f(y)=1$ when $y \in (0,1)$ and $E[Y]=\int_0^1 y \,f(y) \, dy =\frac12$.

This works for any continuous distribution.

  • $\begingroup$ I did not understand why $\Pr(\Phi(X) \le y) = y$. $\endgroup$ – saubhik Jun 19 '15 at 9:03
  • $\begingroup$ $\Phi(X)\le y$ when $X \le \Phi^{-1}(y)$, and this has probability $\Phi(\Phi^{-1}(y))$, which is $y$. $\endgroup$ – Henry Jun 19 '15 at 16:21

The answer above is called the probability integral transform. Aside from using that, we can directly compute the expectation with integration by parts. First, note that the pdf is the derivative of the cdf. $$ \begin{align*} \mathbb{E}\left[\Phi\left(X\right)\right] &= \int_{-\infty}^{\infty} \Phi\left(x\right)\phi\left(x\right)\, dx\\ &= \left.\Phi^{2}\left(x\right)\right|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \Phi\left(x\right)\phi\left(x\right) \, dx\\ &= 1 - \int_{-\infty}^{\infty} \Phi\left(x\right)\phi\left(x\right) \, dx \qquad \left(=1-\mathbb{E}\left[\Phi\left(X\right)\right] \right)\\ 2\mathbb{E}\left[\Phi\left(X\right)\right] &= 1 \end{align*} $$ Therefore, $$ \mathbb{E}\left[\Phi\left(X\right)\right] = 1/2 $$


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