Show that Z has a Standard Normal Distribution – Support Issue?

Question

I'm working through a practice in Casella & Berger:

Suppose $X$ and $Y$ are identically and independently distributed as $N(0,1)$. Define a new random variable $Z$ as follows:

$Z = X$ if $XY > 0$

$Z = -X$ if $XY < 0$

I am required to show that $Z$ also has a standard normal distribution.

I can show that $Z$ seems to have the same cumulative distribution function (CDF) as $N(0,1)$ by looking at the definition of the CDF. However, I'm confused about the support of $Z$. Since $Z$ is not defined for $XY = 0$, it seems that the support of $Z$ does not include $0$. This makes me doubt that $Z$ is actually a standard normal distribution, as its support appears different from $\mathbb{R}$.

Could someone please point out where I am going wrong in my reasoning?

Answer 1

$\DeclareMathOperator{\P}{\mathbb{P}}\DeclareMathOperator{\supp}{\mathrm{supp}}$ There is a bit of confusion about the support of random variables. The support of a random variable $Z$ is defined as the support of its distribution $\P(Z\in\cdot)$, which is the set of the points whose every open neighbohrood $N$ has the property that $\P(Z\in N)>0$, that is $$\supp(Z):=\{z:\mathbb P(Z\in N)>0,\forall N\text{ neighbohrood of $z$}\}.$$ Thus, acting as the function $g(x,y)$ from $\mathbb R^2$ to $\mathbb R$, $Z$ doesn't take the value $z=0$; but, acting as the random variable $Z=g(X,Y),$ $Z$ does take the value $z=0$, in the sense that $0\in\supp(Z)$. To see that, we'll compute the distribution of $Z$ and we'll find out that $$\mathbb P(Z\in(-\varepsilon,\varepsilon))>0, \: \forall\varepsilon>0.$$ Define $X':=-X$. Conditioning on the sign of $XY$, we get \begin{align} F_Z(z)&=\P(Z\le z)\\&=\P(XY>0)\P(Z\le z\mid XY>0) \\&+ \P(XY<0)\P(Z\le z\mid XY<0)\\&=\P(XY>0)\P(X\le z\mid XY>0) \\&+ \P(XY<0)\P(-X\le z\mid XY<0)\\&=\P(XY>0)\P(X\le z\mid XY>0) \\&+ \P(X'Y>0)\P(X'\le z\mid X'Y>0)\\&=2\P(XY>0)\P(X\le z\mid XY>0)\\&=2\P(X\le z, XY>0). \end{align} The penultimate equality holds because $(X,XY)$ and $(X',X'Y)$ are identically distributed. Conditioning on the sign of $Y$, we get \begin{align} F_Z(z)&=2\P(X\le z, XY>0)\\&=2\P(Y>0)\P(X\le z, XY>0\mid Y>0)\\&+2\P(Y<0)\P(X\le z, XY>0\mid Y<0)\\&=\P(X\le z, XY>0\mid Y>0)\\&+\P(X\le z, XY>0\mid Y<0)\\&=\P(X\le z, X>0\mid Y>0)\\&+\P(X\le z, X<0\mid Y<0)\\&=\P(X\le z, X>0)\\&+\P(X\le z, X<0). \end{align} The last equality holds because $X$ and $Y$ are independent. We have \begin{align} F_Z(z)&=\P(X\le z, X>0)+\P(X\le z, X<0)\\&= \begin{cases} 0+\P(X\le z),&\text{if $z\le0$,}\\ \P(0\le X\le z)+\P(X\le 0),&\text{if $z\ge0$,} \end{cases}\\&=F_X(z). \end{align} Thus, $Z$ is $\mathrm{N}(0,1)$-distributed too. Finally note that $0\in\supp(Z)$ because $$\mathbb P(Z\in(-\varepsilon,\varepsilon))>0, \: \forall\varepsilon>0.$$

Answer 2

Perhaps a simpler proof is like that: $Z=|Z|\cdot sign(Z)$, and $|Z|=|X|$, i.e. the absolute value of the $N(0,1)$ distribution. At the same time for any $a\ge 0$, since $X$ and $Y$ are independent, $$ \mathbb P(|Z|>a,sign(Z)=+1)=\mathbb P(|X|>a,sign(XY)=+1)= \mathbb P(|X|>a,sign(X)=1, sign(Y)=+1)+\mathbb P(|X|>a,sign(X)=-1, sign(Y)=-1)= \mathbb P(|X|>a,sign(X)=1)\mathbb P(sign(Y)=+1)+\mathbb P(|X|>a,sign(X)=-1)\mathbb P(sign(Y)=-1)= \frac12\left[\mathbb P(|X|>a,sign(X)=1)+\mathbb P(|X|>a,sign(X)=-1) \right]=\frac12 \mathbb P(|X|>a)=\mathbb P(|Z|>a)\cdot \mathbb P(sign(Z)=+1) $$ yielding that $|Z|$ and $sign(Z)$ are independent. So $Z=+|X|$ with probability $1/2$ and $Z=-|X|$ with probability $1/2$, which means that $Z$ is $N(0,1)$ from the symmetry of the normal distribution.