# 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 $$\mathbb{E}\left( \Phi^{-1}(U)\right)?$$

Inverse CDF is a method to create random samples from given distributions. So, $$\Phi^{-1}(U)$$ is standard normal, and its mean is $$0$$.
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 $$\begin{eqnarray*} \mbox{E}[X]=\int_{0}^{1} F^{-1}_X(u) \mbox{d} u. \end{eqnarray*}$$