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Let $Z\sim\mathcal{N}(0,1)$ and $X$ be a discrete random variable with pmf
$\mathbb{P}(X = -1) = \mathbb{P}(X = 1) = \frac{1}{2}$ and is independent of $Z$. Define $Y = X|Z|$ and show that $Y$ has the same distribution as $Z$.

How would I start this?

Following TheoreticalEconomist's suggestion:

$$\mathbb{P}(Y\leq y) = \mathbb{P}(X|Z| \leq y) \\ = \mathbb{P}(-y \leq XZ \leq y) \\ = \mathbb{P}(-y \leq Z \leq y | X= 1) \mathbb{P}(X=1) + \mathbb{P}(-y\leq -Z \leq y|X=-1)\mathbb{P}(X=-1) \\ = \frac{1}{2}(2\mathbb{P}(-y\leq Z\leq y)) \\ = \mathbb{P}(-y \leq Z \leq y) $$ which doesn't seem to be the CDF of $Z$

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  • $\begingroup$ Have you tried computing $\Pr(Y \le y)$? You could also try calculating the characteristic function of $Y$, but I suspect that will be harder than my first suggestion. $\endgroup$ – Theoretical Economist Apr 13 '18 at 10:37
  • $\begingroup$ Yep, I edited it in $\endgroup$ – Mr. Bromwich I Apr 13 '18 at 10:44
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    $\begingroup$ Your second $=$ sign is wrong, because $X|Z|$ could be arbitrarily negative. $\endgroup$ – J.G. Apr 13 '18 at 10:47
  • $\begingroup$ How would I compute this correectly? $\endgroup$ – Mr. Bromwich I Apr 13 '18 at 10:53
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Suppose WLOG $y\geq 0,$ the other case is similar. You have: $$\mathbb{P}(Y\leq y) = \mathbb{P}(X|Z| \leq y)= \\ =\mathbb{P}(\{X=-1, -|Z|\leq y\} \cup \{X=1, |Z|\leq y\})=\\=\mathbb{P}(X=-1,-|Z|\leq y)+\mathbb{P}(X=1, |Z|\leq y)=\\=\mathbb{P}(X=-1)\mathbb{P}(-|Z|\leq y)+\mathbb{P}(X=1)\mathbb{P}(|Z|\leq y)=\\=\frac{1}{2}\mathbb{P}(-|Z|\leq y)+\frac{1}{2}\mathbb{P}(|Z| \leq y)= \frac{1}{2}\cdot 1+ \frac{1}{2}\mathbb{P}(-y \leq Z \leq y)=\\=\frac{1}{2}(\mathbb{P}(Z\leq y)+P(Z \geq y)+ \mathbb{P}(-y \leq Z \leq y))=\\=\frac{1}{2}(\mathbb{P}(Z\leq y)+P(Z \leq -y)+ \mathbb{P}(-y \leq Z \leq y))=\\=\frac{1}{2}(2\cdot \mathbb{P}(Z\leq y))=\mathbb{P}(Z\leq y) $$ where I used the independence of the variables to split the summands into 2 products, and the (centered) gaussianity to turn $\mathbb{P}(Z\geq -y)$ into $\mathbb{P}(Z\leq y).$

Notice that I've used the fact that $y\geq 0$ when I said that $\{-|Z|\leq y\}$ is an event of probability one, and I've split $1$ into $\mathbb{P}(Z\leq y)+P(Z \geq y).$

One last comment: notice that gaussianity hasn't really been used for this solution. What you really need is the independence of X and Z and the symmetry of the distribution of Z.

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    $\begingroup$ The answer is wrong, let me edit it. $\endgroup$ – Riccardo Ceccon Apr 13 '18 at 11:07
  • $\begingroup$ Thanks, how did you form your second equality? (with the union) How did you get rid of the absolute values? $\endgroup$ – Mr. Bromwich I Apr 13 '18 at 11:07
  • $\begingroup$ 2 minutes and I'll correct the answer :) $\endgroup$ – Riccardo Ceccon Apr 13 '18 at 11:09
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    $\begingroup$ No worries and no rush $\endgroup$ – Mr. Bromwich I Apr 13 '18 at 11:10
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    $\begingroup$ Shouldn't it be a conditional probability for the second equality? So it becomes: $$\mathbb{P}(-|Z| \leq y | X = -1)\mathbb{P}(X=-1) + \mathbb{P}(|Z| \leq y|X=1)\mathbb{P}(X=1) $$ (I think, though, that they equal because of independence but in other situations, should it be the above?) $\endgroup$ – Mr. Bromwich I Apr 13 '18 at 11:19

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