The product of a normal and Rademacher variables, independent from each other

Question

Let $X\sim N(0,1)$ and let $Z$ be a random variable independent of $X$ such that:

\begin{equation} \Pr(Z=z) = \begin{cases} \frac{1}{2} & \mbox{if $z = -1$ or $z=1$}, \\ \\ 0 & \mbox{otherwise}. \end{cases} \end{equation}

1. Define $Y = ZX$.
2. Show that $Y \sim N(0,1)$.
3. Determine $\operatorname{Cor}(X,Y)$.
4. Are $X$ and $Y$ independent?

Progress:

I'm basically stuck at the beginning. I got as far as $P(ZX < y) = P(Z < y/X)$, which I'm not even sure is correct. I'm also not clear on what it means by defining $Y = ZX$ - does it mean I need to state the probability density function or something? And how do I show that $Y \sim N(0,1)$?

Answer 1

Let $Q(x)$ be the Q-function, i.e., $Q(x)=P(N\geq x)$, where $N$ is a Gaussian random variable with mean $0$ and variance $1$. Then

$$ \begin{align*} P(Y \geq y) & =P(XZ \geq y) \\ &= P(Z=1)P(XZ \geq y|Z=1) + P(Z=-1)P(XZ \geq y|Z=-1) \\ &= \frac{1}{2}P(X \geq y|Z=1) + \frac{1}{2}P(-X \geq y|Z=-1) \\ &= \frac{1}{2}P(X \geq y) + \frac{1}{2}P(X \leq -y) \\ &= \frac{1}{2}Q(y) + \frac{1}{2}Q(y) \\ & = Q(y) \end{align*} $$

Hence $Y \sim \mathcal{N}(0,1)$. The fifth equality follows since Gaussian distribution with mean zero is symmetric around $y$-axis.

Next,

$$ \begin{align*} Cor(X,Y) &= \mathbb{E}\left[ XY\right] \\ &=P(Z=1)\mathbb{E}\left[ XY|Z=1\right] + P(Z=-1)\mathbb{E}\left[ XY|Z=-1\right] \\ &=\frac{1}{2} \mathbb{E}\left[ X^2\right] + \frac{1}{2} \mathbb{E}\left[ -X^2\right]\\ &= 0 \end{align*} $$

Finally, $X$ and $Y$ are not independent because conditioned on $X$, $Y$ can take only two different values, whereas its marginal distribution is $\mathcal{N}(0,1)$.

At first sight, the answers to parts $3$ and $4$ might seem contradictory, since uncorrelatedness implies independence for jointly Gaussian random variables (although this is not true for arbitrary random variables). However, $X$ and $Y$ are not jointly Gaussian distributed, hence this implication is not true in this case.

Answer 2

We have $Y\le y$ if and only if $X\le y$ and $Z=-1$ or if $X \gt -y$ and $Z=-1$.

If $f_X$ is the density function of the standard normal, then $$F_Y(y)=\frac{1}{2}\int_{-\infty}^y f_X(x)\,dx+\frac{1}{2}\int_{-y}^\infty f_X(x)\,dx.$$

Differentiate, using the Fundamental Theorem of Calculus. We get $f_X(y)$.