Probability Theory – Problem 59: Compute the distribution of ;

The random variable $X$ follows the standard normal distribution $N(0,\ 1)$, $I$ satisfies $P(I=1)=\frac{1}{2}=P(I=-1)$, and $X,I$ are independent. For the random variable $Y=I X$: (1) Compute the distribution of $Y$; (2) Discuss the independence of $I$ and $Y$; (3) Discuss the independence of $X$ and $Y$.

1. Representing $Y$ 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 $-X\sim \text{N}(0,1)$, so $$Y\mid I=-1 \sim \text{N}(0,1).$$

2. Computing the density of $Y$ via the law of total probability

Let $\varphi(y)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{y^{2}}{2}}$ denote the standard normal density. 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 each real number $y$: $$f_{Y}(y) =\sum_{i\in\{1,-1\}}P(I=i)f_{Y\mid I=i}(y) =\frac{1}{2}\varphi(y)+\frac{1}{2}\varphi(y) =\varphi(y).$$

3. Conclusion for the distribution of $Y$

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

1. Verifying independence via joint distributions

To determine whether $I$ and $Y$ are independent, we check whether for any 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 $$\{I=1,\ Y\in B\} =\{I=1,\ X\in B\}.$$ Since $X$ 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 result from Part 1, $Y\sim \text{N}(0,1)$, which has the same distribution as $X$, so $$P(Y\in B)=P(X\in B).$$ Therefore $$P(I=1,\ Y\in B) =\frac{1}{2}P(X\in B) =\frac{1}{2}P(Y\in B) =P(I=1)P(Y\in B).$$

2. Verification for $I=-1$

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

Since $X$ follows 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 $P(-X\in B)=P(Y\in B)$. Hence $$P(I=-1,\ Y\in B) =\frac{1}{2}P(Y\in B) =P(I=-1)P(Y\in B).$$

3. Conclusion on independence of $I$ and $Y$

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

1. Constructing an event that reveals dependence

By definition $Y=IX$. Note 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, $$\{X=Y\}$$ and $$\{I=1\}$$ are the same event (differing only on the zero-probability set $\{X=0\}$). Therefore $$P(X=Y)=P(I=1)=\frac{1}{2}.$$

2. What independence would imply for $P(X=Y)$

If $X$ and $Y$ were independent, and $Y$ also has a continuous distribution (as shown in Part 1, $Y\sim \text{N}(0,1)$), then for any 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 $Y$ are independent, then $$P(Y=x\mid X=x)=P(Y=x)=0,$$ so $$P(X=Y)=\int_{-\infty}^{+\infty}0\times f_{X}(x)\,dx=0.$$

3. Contradiction and conclusion

On one hand, from the specific relationship $Y=IX$, we obtained $$P(X=Y)=\frac{1}{2};$$ on the other hand, under the assumption that $X$ and $Y$ are independent, we derived $$P(X=Y)=0.$$ This is a contradiction, so $X$ and $Y$ are not independent.

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