Let X $\sim$ $\mathcal{N(\mu, \sigma^2})$. Find the Expectation of $\left(-log_{e} \left(\Phi\left(\frac{X - \mu}{\sigma}\right)\right)\right)^3$ , where $\Phi \left(. \right)$ denotes the cumulative distributive function of $\mathcal{N}\left(0, 1\right)$ random variable.
Please, provide some help and ideas on how to get this expectation. The answer is given as 6.
I have tried to find the $P \left(\left(-log_{e}\left(\Phi(Z)\right)\right)^3 \le z\right) $, where $Z = \frac{X - \mu}{\sigma}$ and then take the derivative to get the density function. finally calculated the expectation, but I didn't get the answer as 6.
I appreciate any help you can provide.
Note that, using properties of the Normal Distribution, $\dfrac{X-\mu}{\sigma}\sim N(0,1)$. Then, as the variable is continuous, a standard result gives us that $$\Phi\left(\dfrac{X-\mu}{\sigma}\right)\sim U(0,1)$$
Then, the variable which expectation we want to find, call It $Y$, verifies that $$Y\sim -log^3(T),$$ where $T \sim U(0,1)$
But $-log(T)\sim Exp(1)$ as $T \sim U(0,1)$.
Thus, we just need to find $E[Z^3]$, where $Z\sim Exp(1)$, that is,
$$\int_0^{\infty} x^3 e^{-x} dx$$
You may do this integral by parts, or just remember that, for a Exponential Variable, $$E[Z^k]=k!$$
In any case, the result is 6, as desired.