Prove that $X$ and $Y$ are not independent.

Question

The Question is,

Let $X$ and $Z$ be independent, with $X \sim N(0,1)$, and with $P(Z=1)=P(Z=-1)=1/2$. Let $Y=XZ$.

I approached like below to show $P(X \cap Y) \neq P(X)P(Y)$.

$P(X \cap Y)=P(X\in B, Y \in B)$ $=P(X \in B, Y \in B |Z=1)P(Z=1)+P(X \in B, Y\in B|Z=-1)P(Z=-1)$ $=P(X \in B)P(Z=1) + P(X \in B, -X \in B)P(Z=-1)$ $={1 \over 2}P(X \in B)+{1\over2}P(X \in B, -X \in B)$

But I don't know how to go further... Thanks for any help.

What's $B$? Hint: Knowing that, say, $|X|<1$ tells you that $|Y|<1$. — lulu, Oct 19 '17 at 14:51

score 2 · Answer 1 · edited Oct 19 '17 at 16:32

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Let us calculate $$ \operatorname E[X^2Y^2]=\operatorname E[X^4Z^2]=\operatorname EX^4\operatorname EZ^2=3\sigma^2=3. $$ However, $$ \operatorname EX^2\operatorname EY^2=\left(\operatorname EX^2\right)^2\operatorname EZ^2=1. $$ Hence, we see that $$ \operatorname E[X^2Y^2]\ne\operatorname EX^2\operatorname EY^2. $$ If $X$ and $Y$ were independent, $X^2$ and $Y^2$ would be independent too, which would imply that they are uncorrelated. But we see that they are correlated. Hence, $X$ and $Y$ cannot be independent.

edited Oct 19 '17 at 16:32

Did

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answered Oct 19 '17 at 14:59

Cm7F7Bb

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Thank you for your help! Should $EX^4$ be calculated by mgf, or are there other ways to calculate $EX^4$? – Siri Oct 19 '17 at 17:24
@Y.Kim Another way is given here. – Cm7F7Bb Oct 19 '17 at 18:20

drhab · Answer 2 · 2017-10-19T15:10:31.560

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Hint: if $X$ and $Y$ are independent then also $X^2$ and $Y^2$ are independent.

Also observe that $Y^2=X^2$ a.s.

edited Oct 19 '17 at 15:10

answered Oct 19 '17 at 15:04

drhab

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Probably the simplest argument (as long as the OP is clear in the "then also" part). – Clement C. Oct 19 '17 at 16:56

score 1 · Answer 3 · answered Oct 19 '17 at 15:16

There's a basic point here which should be made. Events can be independent, as can random variables. You are right in saying that if $X$ and $Y$ are not independent then $P(X\cap Y)\neq P(X)P(Y)$, but this formula is for events $X$ and $Y$, not random variables. You need to look up the definition of independent random variables.

score 1 · Answer 4 · answered Oct 19 '17 at 15:43

Two random variables $X,Y$ are independent if for all intervals $\mathrm A,\mathrm B$ in their supports, we have $\mathsf P(X\in \mathrm A, Y\in \mathrm B) = \mathsf P(X\in\mathrm A)\mathsf P(Y\in\mathrm B)$. Conversely, they are not independent if there exist some intervals where this is not true.

Tweaking your formula you wish to find some $\mathrm A,\mathrm B$ where $$\mathsf P(X\in \mathrm A)\mathsf P(Y\in \mathrm B) \neq \tfrac 12\mathsf P(X\in \mathrm A\cap \mathrm B)+\tfrac 12\mathsf P(X\in\mathrm A\cap (-\mathrm B))$$

And well, when $\mathrm A=\mathrm B$, and $\mathrm A\subsetneq (0;\infty)$ then...

$~$

$(-\mathrm B)=\{-x: x\in\mathrm B\}$

Prove that $X$ and $Y$ are not independent.

4 Answers4