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Let $X \sim \mathcal{N}(0, 1)$ and $Y$ be a random variable independent of $X$ such that \begin{align*} P(Y=y) = \begin{cases} \frac{1}{2} & y = -1\\ \frac{1}{2} & y = 1\\ 0 & otherwise \end{cases} \end{align*} If $Z = XY$, are $Z$ and $X$ independent?

I've found that the correlation between $X$ and $Z$ is 0; however, I know that this does not say anything about their independence.

From the problem, it is quite obvious that the two variables are not independent; however, I am having trouble showing this formally. I want to find some way to show $f_{X,Z}(x,z)\neq f_X(x) \cdot f_Z(z)$ for some $(x, z)$, but I'm not sure what the joint density function actually is.

Since both $X$ and $Z$ follow $\mathcal{N}(0,1)$, I assume that the density function for both variables is the normal density. What, then, would be the joint density?

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  • $\begingroup$ Independence gives you correlation $0$, not the other way around. $X$ and $Z$ will not be independent, in this case. If $Z=z$, then $X$ has a 50/50 shot of being $z$ or $-z$. For two variables to be independent, their distributions can't change depending on the other variable. Here, $X$ moves from a normal distribution to a 2-point distribution when we're given information about $Z$. This means the two are not independent. $\endgroup$ – Benjamin Roycraft Dec 11 '14 at 9:46
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Based on: $$P\left(X\leq x\wedge Z\leq x\right)=$$$$P\left(X\leq x\wedge Z\leq x\mid Y=-1\right)P\left(Y=-1\right)+P\left(X\leq x\wedge Z\leq x\mid Y=1\right)P\left(Y=1\right)$$

we find for $x<0$:

$$P\left(X\leq x\wedge Z\leq x\right)=\frac{1}{2}P\left(X\leq x\wedge-X\leq x\right)+\frac{1}{2}P\left(X\leq x\wedge X\leq x\right)=\frac{1}{2}P\left(X\leq x\right)$$

So independence demands that $P\left(Z\leq x\right)=\frac{1}{2}$ for each $x<0$, which cannot be true.

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  • $\begingroup$ Thanks for the edit, that's rightful and a bit embarassing $\endgroup$ – Ilya Dec 12 '14 at 7:31
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It's easy to specific the value of $x$ and $z$ to disprove the independence.

Let $x=1$ and $z=3$. Firstly we have $f_{XZ}(1,3)=0$ because that $z$ can only be equal to $x$ or $-x$.

While $f_X(1)*f_Z(3) = f_X(1)*(0.5f_X(3)+0.5f_X(-3))=f_X(1)*f_X(3) \ne0 $

So $X$ and $Z$ are not independent.

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A faster way to notice lack of independence is as follows. If $X$ and $Z$ are independent, then knowing the value of $X$ you still need $Z$ to have the same distribution. It is easy to notice that $Z$ has continuous distribution, but given $X = x$, you know for sure that $Z\in\{-x,x\}$ so that conditional distribution of $Z$ much be different from the unconditional one. If you feel uncomfortable with such statements, you may still want to verify it formally as in drhab's answer, but I think even before that it's a good mental check to know which result to expect from formal computations.

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