1
$\begingroup$

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

$\endgroup$
2

2 Answers 2

3
$\begingroup$

Inverse CDF is a method to create random samples from given distributions. So, $\Phi^{-1}(U)$ is standard normal, and its mean is $0$.

$\endgroup$
2
$\begingroup$

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*}

$\endgroup$
1
  • $\begingroup$ @ user277126: Great answer! I really like all the different mathematical notations you used and with the cumulative probability distribution! $\endgroup$
    – stats_noob
    Aug 13, 2022 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.