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

Question

Hey I want to check my solutions for this problem:

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

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

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

2. Calculate the correlation $Corr(X,Z)$ between $X$ and $Z$. Is $(X,Z)$ multivariate normally distributed? Explain.

3. Calculate the probabilities $P[X + Z] = 0$.

What I have:

1. 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:

$F_Z(z)=P(Z \leq z)=P(Z \leq z, Y \geq 0)+P(Z \leq z, Y > 0)=P(X \leq z)+P(-X \leq z)=P(X \leq z)+P(X \geq -z)=P(X \leq z)+(1-(P(X \leq -z))=\Phi(z)+(1-\Phi(-z))=\Phi(z)+(1-(1-\Phi(z)))=2\Phi(z)$

2. To calculate the $Corr(X,Z)$ I need to calculate $Cov(X,Z)=E[XZ]-E[X]E[Z]$ For $E[X]=0$ since $X$ is normal distributed So I need only to calculate $E[XZ]$

My idea was (but I don't know if it is correct) to calculate:

We know that $f_Z(z)=d/dz 2\Phi(z)=2\varphi(z)=$ Also $2 \times$ the distribution of $X$ so we have that $Z=2X$

$E[XZ]=E[2X^2]=2E[X^2]=\int_{-\infty}^{\infty}2z^2\frac {e^{-z^{2}/2}}{\sqrt {2\pi }}=2$

(I had to do this calculation with an online integral calculator. Is there a way to solve the integral faster?) We have $Var(Z)=Var(2X^2)=4Var(X^2)=$

So we have

$Corr(X,Z)=\frac{Cov(X,Z)}{\sigma(Z) \sigma(X)}=\frac{2}{\sqrt{4Var(X^2)Var(X)}}=\frac{1}{\sqrt{Var(X^2)Var(X)}}$

For 3. We can use the Convolution of probability distributions, right? $P^{X+Z}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}2\varphi(x+z)\varphi(z)dxdz$

Am I right? Where are my errors? Thanks for any help.

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.

$E(XZ)=E(X^{2}\mathbf{1}_{Y>0})-E(X^{2}\mathbf{1}_{Y<0})=\frac{1}{2}(E(X^{2})-E(X^{2}))=0$

If $(X,Z)$ was multivariate gaussian, then their $0$ covariance would have implied independence of $X$ and $Z$ which they are clearly not.

$P(X+Z=0)=P(2X=0,Y>0)+P(Y<0)=0+\frac{1}{2}=\frac{1}{2}$ . Note that this also proves that $(X,Z)$ cannot be multivariate Gaussian as otherwise, $aX+bZ$ would have Gaussian distribution. But here, $X+Z$ cannot even have continuous distribution as it gives half of it's mass to $0$.