Probability Theory – Problem 12: determine the distribution of ;

Let the random variable $X$ follow the standard normal distribution $N(0,\ 1)$, and let $I$ satisfy $P(I=1)=\frac{1}{2}=P(I=-1)$, with $X$ and $I$ independent. For the random variable $Y=IX$: (1) determine the distribution of $Y$; (2) discuss the independence of $I$ and $Y$; (3) discuss the independence of $X$ and $Y$.

Step 1. Express $Y$ via conditional distributions.

Given $X\sim \text{N}(0,1)$, $P(I=1)=\frac{1}{2}=P(I=-1)$, and $X,I$ are independent. Define $Y=IX$.

Conditioning on $I=1$, $$Y\mid I=1 = 1\times X = X,$$ so $$Y\mid I=1 \sim \text{N}(0,1).$$

Conditioning on $I=-1$, $$Y\mid I=-1 = -1\times X = -X.$$ Since the standard normal distribution is symmetric about $0$, when $X\sim \text{N}(0,1)$ we have $-X\sim \text{N}(0,1)$, and therefore $$Y\mid I=-1 \sim \text{N}(0,1).$$

Step 2. Compute the density of $Y$ by the law of total probability.

Denote the standard normal density by $$\varphi(y)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{y^{2}}{2}}.$$

Then $$f_{Y\mid I=1}(y)=\varphi(y),\quad f_{Y\mid I=-1}(y)=\varphi(y).$$

By the law of total probability, for every real number $y$, $$f_{Y}(y) =\sum_{i\in\{1,-1\}}P(I=i)f_{Y\mid I=i}(y) =\frac{1}{2}\varphi(y)+\frac{1}{2}\varphi(y) =\varphi(y).$$

Step 3. Conclude the distribution of $Y$.

Since $f_{Y}(y)=\varphi(y)$, which coincides with the standard normal density, we have $$Y\sim \text{N}(0,1).$$

Step 4. Verify independence of $I$ and $Y$ via the joint distribution.

To determine whether $I$ and $Y$ are independent, it suffices to check that for every Borel set $B\subset\mathbb{R}$, $$P(I=i,\ Y\in B)=P(I=i)P(Y\in B),\quad i=\pm 1.$$

First compute $P(I=1,\ Y\in B)$. When $I=1$, $Y=X$, so $$\{I=1,\ Y\in B\} =\{I=1,\ X\in B\}.$$

Because $X$ and $I$ are independent, $$P(I=1,\ Y\in B) =P(I=1,\ X\in B) =P(I=1)P(X\in B) =\frac{1}{2}P(X\in B).$$

By the conclusion of Step 3, $Y\sim \text{N}(0,1)$, which has the same distribution as $X$, so $$P(Y\in B)=P(X\in B).$$

Therefore $$P(I=1,\ Y\in B) =\frac{1}{2}P(X\in B) =\frac{1}{2}P(Y\in B) =P(I=1)P(Y\in B).$$

Step 5. Verify the same for $I=-1$.

When $I=-1$, $Y=-X$, so the event $$\{I=-1,\ Y\in B\} =\{I=-1,\ -X\in B\} =\{I=-1,\ X\in -B\},$$ where $-B=\{-x:x\in B\}$. Hence $$P(I=-1,\ Y\in B) =P(I=-1,\ X\in -B) =P(I=-1)P(X\in -B) =\frac{1}{2}P(X\in -B).$$

Since $X$ has the standard normal distribution, $P(X\in -B)=P(-X\in B)$. Moreover, $-X\sim \text{N}(0,1)$, which has the same distribution as $Y$, so $P(-X\in B)=P(Y\in B)$.

Therefore $$P(I=-1,\ Y\in B) =\frac{1}{2}P(Y\in B) =P(I=-1)P(Y\in B).$$

Step 6. Conclude the independence of $I$ and $Y$.

For both $i=1$ and $i=-1$ we have $$P(I=i,\ Y\in B)=P(I=i)P(Y\in B),$$ and therefore $I$ and $Y$ are independent.

Step 7. Construct an event revealing the dependence between $X$ and $Y$.

By definition $Y=IX$. Observe that if $X\neq 0$, then $$Y=X \Longleftrightarrow IX=X \Longleftrightarrow I=1.$$

Consider the event $\{X=Y\}$. Since $X$ has a continuous distribution, $P(X=0)=0$, so in the almost-sure sense the event $$\{X=Y\}$$ coincides with $$\{I=1\}$$ (they differ only on the null set $\{X=0\}$).

Therefore $$P(X=Y)=P(I=1)=\frac{1}{2}.$$

Step 8. Derive the consequence of independence for $P(X=Y)$.

If $X$ and $Y$ were independent, and since $Y$ also has a continuous distribution (as established in Step 3, $Y\sim \text{N}(0,1)$), then for every real number $x$, $$P(Y=x)=0.$$

Expanding via conditional probability, $$P(X=Y) =\int_{-\infty}^{+\infty}P(Y=x\mid X=x)f_{X}(x)\,dx.$$

If $X$ and $Y$ were independent, then $$P(Y=x\mid X=x)=P(Y=x)=0,$$ so $$P(X=Y)=\int_{-\infty}^{+\infty}0\times f_{X}(x)\,dx=0.$$

Step 9. Contradiction and conclusion.

On the one hand, from the explicit relation $Y=IX$ we obtained $$P(X=Y)=\frac{1}{2};$$ on the other hand, under the assumption that $X$ and $Y$ are independent we derived $$P(X=Y)=0.$$

These two results contradict each other, which proves that $X$ and $Y$ are not independent.

1. $Y\sim \text{N}(0,1)$, i.e., $Y$ has the same distribution as $X$. 2. $I$ and $Y$ are independent. 3. $X$ and $Y$ are not independent.