Probability Theory – Problem 32: 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 $I$ mutually independent. For the random variable $Y=I X$: (1) Determine 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 that $X\sim \text{N}(0,1)$, $P(I=1)=\frac{1}{2}=P(I=-1)$, and $X,I$ are mutually 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 $$Y\mid I=-1 \sim \text{N}(0,1).$$

2. Computing the density of $Y$ via 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 $y$, $$f_{Y}(y) =\sum_{i\in\{1,-1\}}P(I=i)f_{Y\mid I=i}(y) =\frac{1}{2}\varphi(y)+\frac{1}{2}\varphi(y) =\varphi(y).$$

3. Conclusion on the distribution of $Y$

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

1. Verifying independence 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 $$\{I=1,\ Y\in B\} =\{I=1,\ X\in B\}.$$

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 of 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 the case $I=-1$

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

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

3. Conclusion on the 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 mutually independent.

1. Constructing an event that reveals dependence

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

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

2. Consequence of the independence assumption for $P(X=Y)$

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

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

Under the independence assumption, $$P(Y=x\mid X=x)=P(Y=x)=0,$$ so $$P(X=Y)=\int_{-\infty}^{+\infty}0\times f_{X}(x)\,dx=0.$$

3. Contradiction and conclusion

On the one hand, from the explicit relation $Y=IX$ we obtained $$P(X=Y)=\frac{1}{2};$$ on the other hand, the independence assumption yields $$P(X=Y)=0.$$

These two results contradict each other, proving 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 mutually independent. 3. $X$ and $Y$ are not independent.