If $X$ standard normal and $Y$ symmetric Bernoulli are independent then $Y$ and $XY$ are independent?

Question

If $X$ standard normal and $Y$ symmetric Bernoulli are independent then $Y$ and $Z=XY$ are independent?

It is quite intuivively obvious that $Z\sim\mathcal N(0,1)$ and that would imply that $Y$ and $Z$ are independent, but how to prove it?

$E(Z)=E(XY)=E(X)E(Y)=0\cdot0=0$

$Var(Z)=E((E(XY)-XY)^2)=E(X^2Y^2)=E(X^2)=Var(X)+E(X)^2=1+0=1$

But that merely gives us the expected value and variance, this probably isn't enough to conclude that $Z\sim\mathcal N(0,1)$

Would $X$ and $Z$ also be independent?

Update:

Greedoid's answer led me to this:

$$P(Z\le z)= P(Z\le z|Y=1)P(Y=1)+P(Z\le z|Y=-1)P(Y=-1) \\ = P(X\le z)\cdot{1\over2}+P(-X\le z)\cdot{1\over2}\\ ={1\over2}\phi(z)+{1\over2}\phi(z)=\phi(z)$$

where $\phi$ is the cumulative distribution function of the $\mathcal N(0,1)$ law

And this is true because $P(-X\le z)=P(X\le z)$ because $-X\sim\mathcal N(0,1)$

Answer 1

Let $f$ and $g$ be bounded measurable functions.

Note that $$\begin{align}E(f(XY))E(g(Y))&=E(f(X)1_{Y=1}+f(-X)1_{Y=-1})E(g(Y))\\ &=[E(f(X))E(1_{Y=1})+E(f(-X))E(1_{Y=-1})]E(g(Y)) \tag1\\ &=[E(f(X))\frac 12+E(f(X))\frac 12]E(g(Y)) \tag2\\ &= E(f(X))E(g(Y)) \end{align} $$ $(1)$: $X$ and $Y$ are independent
$(2)$: $X$ is symmetric

and $$\begin{align} E(f(XY)g(Y))&= E(E(f(XY)g(Y)|Y)) = E(g(Y) E(f(XY)|Y))\\ &= E(g(Y)[1_{Y=1}E(f(X)) + 1_{Y=-1}E(f(-X))])\tag 1\\ &= E(f(X)) [E(g(Y)(1_{Y=1}+1_{Y=-1})])\tag2\\ &= E(f(X)) E(g(Y)) \end{align}$$

$(1)$: $E(f(XY)|Y=y)=E(f(yX)|Y=y)=\int f(yx)dP_{X|Y=y}(x)=\int f(yx)dP_{X}(x)=E(yX)$
$(2)$: $X$ is symmetric

Thus $E(f(XY)g(Y))= E(f(XY))E(g(Y))$, hence $XY$ and $Y$ are independent.

Note that the only property of $X$ I used is that its distribution is symmetric. Normality does not matter.