Let X be a normal random variable with parameters μ = 0 and σ2 = 1, and let I, independent of X, be such that P{I = 1} = 1 2 = P{I = 0}. Now define Y by Y = X if I = 1 −X if I = 0 In words, Y is equally likely to equal either X or $-X.$

(a) Are X and Y independent?

(b) Are I and Y independent?

(c) Show that Y is normal with mean $0$and variance $1$.

(d) Show that $Cov(X,Y)=0$

Given in the question that $X$and $Y$are independent.

Suppose that $X=5.$Then, $Y$becomes a discrete random variable where it can assume values $5$and $-5$ with the same probabilities. Without any condition, $Y$is a continuous random variable.

No,$X$and$Y$ are not independent.

Given in the question that $I$and $Y$are independent.

If $I=1,$we have that $Y=X$and if $I=0$, we have $Y=-X$.

Since $X$is symmetric about zero, we have that $X$and $-X$are equally distributed.

So, in both cases, $Y$has posterior distribution equal to the prior distribution.

Yes,$I$and$Y$ are independent.

Given in the question that$Y$is normal with mean $0$and variance $1$.

We are going to prove that CDF of $Y$is$\mathrm{\Phi }$

Take any $y\in \mathrm{ℝ}.$Using the law of the total probability, we have that

$P(Y\le y)=P(Y\le y\mid I=1)P(Y=1)+P(Y\le y\mid I=0)P(I=0)$

$=\frac{1}{2}\left(P\right(X\le y)+P(-X\le y\left)\right)=\frac{1}{2}\left(P\right(X\le y)+P(X\ge -y\left)\right)$

$=\frac{1}{2}\left(\mathrm{\Phi }\right(y)+1-\mathrm{\Phi }(-y\left)\right)=\frac{1}{2}\cdot 2\mathrm{\Phi }\left(y\right)=\mathrm{\Phi }\left(y\right)$

We proved that CDF of $Y$is $\Phi$.

So we have that $Y~\mathcal{N}(0,1)$

Given in the question that$Cov(X,Y)=0$

Using the law of the total covariance, we have that

$\mathrm{Cov}(X,Y)=E\left(\mathrm{Cov}\right(X,Y\mid I\left)\right)+\mathrm{Cov}\left(E\right(X\mid I),E(Y\mid I\left)\right)$

Observe that,

$\mathrm{Cov}(X,Y\mid I)=E(XY\mid I)-E(X\mid I)E(Y\mid I)$

$=E\left(X^2\right)\times I-E\left(X^2\right)(1-I)-E\left(X\right)E\left(Y\right)$

$=X^2(2I-1)$

So applying the expectation, we have that
$E\left(Cov(X,Y\mid I)=E\left(X^2\right)E(2I-1)=0\right.$

On the other hand, we have that $E(X\mid I)=E\left(X\right)=0$and $E(Y\mid I)=E\left(X\right)\times I+E(-X)(1-I)=0$,

so we have that

We have that$\mathrm{Cov}\left(E\right(X\mid I),E(Y\mid I\left)\right)=0$which yields$\mathrm{Cov}(X,Y)=0$