Proving Two Standard Normal Variables Are Uncorrelated 
Let $X \sim N(0,1)$, a random variable $U$ does not depend on $X$, and $P(U = 1) = P(U = -1) = 1/2$. Finally $Y = UX$. Prove that $X$ and $Y$ are uncorrelated.

$Y$ should also be standard normal due to the symmetry of the normal distribution. I know that independent implies uncorrelated, but $X$ and $Y$ are clearly dependent variables.
 A: It may sound counterintuitive, but having them connected via an equation doesn't mean they're correlated. Clearly, $$\operatorname{cov}(X,Y)=E[XY]-E[X]E[Y]=E[UX^2]=E[U]E[X^2]=0$$
And, having two normal RVs uncorrelated doesn't mean they're independent. Independence is implied only when the two are jointly normal.
A: The idea behind this exercise is that when you randomly switch the sign of a variable (with equal probabilities of positive and negative), you eliminate any correlation it might have with any other variable, provided that correlation existed in the first place.
In this general setting, then, let $(X,Z)$ be any bivariate random variable for which both $X$ and $Z$ have finite nonzero variances (which, after choosing suitable units of measurement, can be taken to equal $1$), let $U$ be independent of $X$ as in the question, and define $Y=UZ.$ We will show $Y$ is uncorrelated with $X.$

Observe, from the definition of variance (first and fourth lines below) and the fact $U^2=1$ (third line) that
$$\eqalign{
\operatorname{Var}(Y) &= E[Y^2]-E[Y]^2 = E[(UZ)^2]-E[UZ]^2 \\&\le E[(UZ)^2] = E[U^2Z^2] \\&=E[Z^2] \\&= E[Z]^2 + \operatorname{Var}(Z) \\&= E[Z]^2+1.
}$$
Because this implies $$|\operatorname{Cov}(X,Y)|^2\le \operatorname{Var}(X)\operatorname{Var}(Y)=\operatorname{Var}(Y)\le E[Z]^2+1 \lt \infty,$$ the covariance of $X$ and $Y$ exists and is finite.
Since $U$ and $-U$ are identically distributed and independent of $X,$
$$\operatorname{Cov}(X,Y)=\operatorname{Cov}(X,UZ)=\operatorname{Cov}(X,-UZ)=-\operatorname{Cov}(X,UZ)=-\operatorname{Cov}(X,Y).$$
The only finite number equal to its negative is $0.$  The correlation of $X$ and $Y$ is a multiple of this covariance and therefore also is zero, QED.

The original problem concerns the special case where $Z=X$ and $X$ has a standard Normal distribution.
