If M(x) is the cumulative Gaussian function and X is N(0,1) then what is E[M(X)]?
Thus if X ~ N(0,1):
$M(x) = P(X \leq x)=\Phi(x)$
The answer given is for x in (0,1):
\begin{align} P(M(X) \leq x) &= P(\Phi(X) \leq x) \\ &=P(X \leq \Phi^{-1}(x)) \\ &=x \end{align}
Hence M(X) has a uniform distribution on (0,1) and hence an expectation of 0.5.
My understanding:
$\Phi(x)$ is not a r.v. it returns for a given x the Gaussian cdf for that value (so it's range is between 0 and 1).
In the second line the inverse Gaussian cdf is applied, thus on the right side of the inequality you have the value of the RV distributed N(0,1) corresponding to each value of x, where x is restricted to be between 0 and 1. The whole of line 2 is the probability that X is less than or equal to this value.
Is my understanding of the first two lines correct?
What I don't understand how you end up with a uniformly distributed variable is the third line. Algebraically I can see that you are applying $\Phi(\Phi^{-1}(x))$ so you get x which s a value between zero and one but how do we know that these values are uniformly distributed?.
Thanks
If you know that $P(X \le t) = \Phi(t)$, then the last line is simply this equation applied with $t := \Phi^{-1}(x)$, i.e. $$P(X \le \Phi^{-1}(x)) = \Phi(\Phi^{-1}(x)) = x.$$ To conclude, you have shown the CDF of $M(X)$ is $P(M(X) \le x)=x$ for $x \in [0,1]$, which is the CDF of the uniform distribution.