Determine, the distribution of $Z(w)=\left\{\begin{matrix} X(w)& \mbox{if}\ Y(w)\geq 0 \\ -X(w)&\mbox{if} \ Y(w)<0 \end{matrix}\right.$

Question

Hey I want to check my solutions for this problem:

Let $X$ and $Y$ be standard normally distributed random variables on a probability space $(\Omega, \mathcal F, \mathbb P)$ and define

$Z(w)=\left\{\begin{matrix} X(w)& \mbox{if}\ Y(w)\geq 0 \\ -X(w)&\mbox{if} \ Y(w)<0 \end{matrix}\right.$

Why is Z a random variable? Determine, as concretely as possible, the distribution of Z.

So a random variable is a measurable function on a probabilty space. Therefore a construct of measurable functions is in turn a measurable function. Therefore is the construct of $Z(w)$ a random variable.

For the second part I have not so many ideas: What I thougt is:

$F_Z(z)=P(Z\leq z)=\left\{\begin{matrix} P_X(-X\leq z)=P_X(X\geq -z)=1-P_X(X\leq -z)=1-F_X(-z)&\mbox{if} \ Y(z)<0 \\ F_X(z)+(1-F_X(-z))& \mbox{if}\ Y(z)\geq 0 \end{matrix}\right.$

But I dont know if it is right. Can someone help me? Thanks :)

Answer 1

$E(\exp(itZ))=E(\exp(itX)\mathbf{1}_{Y>0})+E(\exp(-itX)\mathbf{1}_{Y<0})$

Now $X$ and $-X$ are normally distributed(and hence have same characteristic functions) and independent of $Y$. and $E(\mathbf{1}_{Y>0})=P(Y>0)=\frac{1}{2}$

Hence the above is just $E(\exp(itZ))=\frac{1}{2}\phi(t)+\frac{1}{2}\phi(t)=\phi(t)$ where $\phi$ is the characteristic function of $X$.

Thus $Z$ and $X$ have same distribution.

In particular, in your method, you are not multiplying the probability($\frac{1}{2}$) of $Y$ being positive or negative. If you do, then you'll get the correct result.