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Consider two random variables $X$ and $Y$. If X and Y are independent random variables, then it can be shown that: $$E(XY) = E(X)E(Y).$$

Let $X$ be the random variable that takes each of the values $-􀀀1\!\!\!$, $0$, and $1$ with probability $1/3$. Let $Y$ be the random variable with value $Y = X^2$.

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

Prove that $E(XY) = E(X)E(Y)$.


I understand that $E(XY) = E(X^3)$ since $Y = X^2$ so that makes each side of the equation equal to zero.

But I am not sure how to go about proving that $X$ and $Y$ are not independent.

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    $\begingroup$ Since Y is a function of X, they are not independent. i.e. if you know X, then you know Y as well. So, they are not independent. $\endgroup$ – user146290 Apr 5 '15 at 22:35
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    $\begingroup$ @user146290 it not always necessarily true if $Y$ is a function of X then $X$ and $Y$ are independent not independent. if the function is injective this will always be the case, but what if $X$ only took values 1, -1 with probability 1/2 and $Y=X^2$ here Y and X are independent. The intuition is that the value of X has to tell us information about what may be the value of Y to be not independent which is not necessarily true for all functions $\endgroup$ – Kamster Apr 6 '15 at 1:20
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$P(X=-1) = P(X=0) = P(X=1) =\frac{1}{3}$

$Y = X^2$ so $P(Y=1) = \frac{2}{3}$ and $P(Y=0) = \frac{1}{3}$ . $Y$ equals zero iff $X$ equals 0. But $Y$ equals 1 if $X$ is $1$ or $-1$.

$$E[X] = -1.\frac{1}{3} + 0.\frac{1}{3} + 1.\frac{1}{3} = 0$$ $$E[Y] = 1.\frac{2}{3} + 0.\frac{1}{3} = \frac{2}{3}$$

$$E[XY] = E[X^3] = E[X] = 0$$

The last equality holds because $X$ takes only values in $[-1.0.1]$ Thus, $$E[XY] =E[X]E[Y]=0$$

But, are $X,Y$ independent?

For $X,Y$ to be independent $P(X=x, Y=y) = P(X=x) P(Y=y)$ where $x \in [-1,0,1]$ and $y \in [0,1]$.

Let's consider $x=1 \implies y=1$ so $P(X=1, Y=1)=\frac{1}{3}$ while $P(X=1)P(Y=1) = \frac{1}{3}.\frac{2}{3} = \frac{2}{9} \neq P(X=1, Y=1)$

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Hint: By the definition of independence, two discrete random variables $X$ and $Y$ are independent if the joint probability mass function $P(X = x \text{ and } Y = y )$ satisfies $$ P(X = x \text{ and } Y = y ) = P(X = x) \cdot P(Y = y) $$ for all $x$ and $y$.

What are the possible values $x$ for the random variable $X$ in your example? What are the possible values $y$ for $Y$? If your $X$ and $Y$ are not independent, then you should be able to find an $x$ and $y$ pair that violates the above condition.

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Let $M_X(t)$ and $M_Y(t)$ be the moment generating functions of $X$ and $Y$, respectively. Then

$$ \begin{align*} M_X(t) &= \mathbb E[e^{tX}]\\ &= e^{-t}\mathbb P(X=-1) + \mathbb P(X=0) + e^t\mathbb P(X=1)\\ &= \frac13 e^{-t} + \frac13 + \frac 13 e^t\\ &= \frac13(e^{-t} + 1 + e^t) \end{align*} $$ and $$ \begin{align*} M_Y(t) &= \mathbb E[e^{tY}]\\ &= \mathbb P(Y=0) + e^t\mathbb P(Y=1)\\ &= \frac13 + \frac 23 e^t\\ &= \frac13(1 + 2e^t). \end{align*} $$ Let $Z=X+Y$. Then $$ \begin{align*} \mathbb P(Z=n) &= \mathbb P(X+Y= n)\\ &=\mathbb P(X+X^2 = n)\\ &=\mathbb P(X(X+1)=n)\\ &=\begin{cases} \frac23,& n=0\\ \frac13,& n=2. \end{cases} \end{align*} $$ The moment generating function of $Z$ is $$ \begin{align*} M_Z(t) &= \mathbb E[e^{tZ}]\\ &= \frac23 + \frac13 e^{2t}\\ &= \frac13(2 + e^{2t}) \end{align*} $$ while $$ \begin{align*} M_X(t)M_Y(t) &= \frac13(e^{-t}+1+e^t)\frac13(1+2e^t)\\ &= \frac19(e^{-t}+3+3e^t+2e^{2t}). \end{align*} $$ Since $M_Z(t)\ne M_X(t)M_Y(t)$, we conclude that $X$ and $Y$ are not independent.

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