What is $\mathbb{E}\left( \Phi^{-1}(U)\right)$, $U \sim \mathcal{U}(0,1)$? Let $\Phi(\cdot)$ denote the CDF of a standard normal random variable and let $U \sim \mathcal{U}(0,1)$. What can we say about
$$\mathbb{E}\left( \Phi^{-1}(U)\right)?$$

Let $X$ have PDF $f_X$ and CDF $F_X$ and define the transformation $U= F_X(X)$.  Then it may be shown that $U \sim Uniform(0,1)$. Inverting this transformation, we have $X = F^{-1}_X(U)$.  Using this formula one can express the moments of $X$ using the PDF of a standard uniform distribution.  For completeness, the expected value of $X$ is defined as
\begin{eqnarray*}
\mbox{E}[X]=\int_{-\infty}^{\infty} x f_X (x) \mbox{d} x.
\end{eqnarray*}
Define the transformation $u=F_X(x)$ with $\mbox{d}u = f_X(x) \mbox{d}x$.  Therefore, $x=F^{-1}_X(u)$ and $\mbox{d}u = f_X\left(F^{-1}_X(u)\right) \mbox{d}x.$  Plugging these values into the above integral, we have that
\begin{eqnarray*}
\mbox{E}[X]=\int_{0}^{1} F^{-1}_X(u) \mbox{d} u.
\end{eqnarray*}
