Expected value of a probability

Question

I know this might be a very basic question, but I think I might be missing something fundamental. What is the expected value of a probability of a random variable? For example, if $X$ is a random variable with some probability density, let's say Gaussian. There is an answer here that shows how to derive $$E_{X}[\Phi(X)] = \int_{R} \Phi(x)\phi(x)dx$$ where $\Phi(x) = P(T < x)$ is the Gaussian cumulative function, the answer shows that it is just another probability although I am not sure how they got to that answer in the first equality on the first equation line. But my question is a bit more general, for example, what if we have the cumulative function is not of a Gaussian. Or even if it is Gaussian, what if we have products of probabilities? i.e $E_X[\Phi(X)\dots\Phi(X)]$. Is there a simple way to explain this? I tried looking at the probability as just a fixed function, and this is how I understood the answer above by looking at it as an integration by parts, although I would really like to understand how to retrieve the general case or even in particular the first equality in the linked answer above.

Answer 1

For a continuous random variable $X\sim F_X$: $$ \mathsf E (F_X^n(X))=\int_{-\infty}^\infty F_X^n(x)f_X(x)\,\mathrm dx=\int_{-\infty}^\infty F_X^n(x)\,\mathrm dF_X(x)=\frac{F_X^{n+1}(x)}{n+1}\bigg|_{-\infty}^\infty=\frac{1}{n+1}. $$ Since we are integrating w.r.t. $F_X$ the antiderivative follows from the power rule $\int x^n\,\mathrm dx=x^{n+1}/(n+1)$. For example if $X\sim\mathcal N(0,1)$ then $$ \mathsf E(\Phi(X))=\frac{1}{1+1}=\frac{1}{2}. $$

Answer 2

If we look at $\Phi(X)$ with $X\sim N(0,1)$, we have a transformation of the random variable $X$. Using the probability integral transform, we have that $$ \Phi(X)\sim U[0,1], $$ where $U[0,1]$ is the uniform distribution on $[0,1]$. Hence, $$ \operatorname E\Phi(X)=\frac12. $$ This actually holds for any continuous random variable, not necessarily Gaussian (we have to replace $\Phi$ with another cumulative distribution function of course).

I hope this helps.