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 $Z$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ No worries and no rush $\endgroup$ – Mr. Bromwich I Apr 13 '18 at 11:10
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.