Let $Y\vert X \sim N(X,X^2)$ We can write $$E[Y] = E[E[Y\vert X]] =E[X]$$ Where the first equality follows from the law of iterated expectations, and the second equality confuses me. Where does the second equality come from? $X$ is distribution uniformly over $[0,1]$ if that helps.
I thought maybe if I wrote out the integrals (using the definition of expected value) the result would follow, but I did not see an obvious way to get from the integrals to $E[x]$, so I don't think that is the correct approach.
Thanks.
If $$Y \mid X \sim \operatorname{Normal}(\mu = X, \sigma^2 = X^2),$$ then what is $$\operatorname{E}[Y \mid X]?$$ This is obviously simply $X$: Given the value of $X$, $Y$ is normal with mean $\mu = X$, thus the expected value of $Y$ given $X$ is $X$. So $$\operatorname{E}[Y \mid X] = X.$$
Next, just take the expectation with respect to $X$: we have $$\operatorname{E}[\operatorname{E}[Y \mid X]] = \operatorname{E}[X].$$ Note that the two expectations are not with respect to the same variable: the inner expectation is with respect to $Y$ for a given $X$; the outer expectation is with respect to $X$ alone.