I am having trouble coming up with a solution to this problem. This is a recommended exercise for an upcoming midterm (Not for marks, just for practice).

$\textbf{The question is:}$

Suppose $X_{1}$ and $X_{2}$ are independent random variables with $P(X_{1}=1)=P(X_{1}=-1)=\frac{1}{2}$ and $P(X_{2}=1)=1-P(X_{2}=-1)=p$ with $0<p<1$. Let $Y=X_{1}X_{2}$. Show that $X_{2}$ and $Y$ are independent.

$\textbf{My attempt at a solution}$

Since $X_{1},X_{2}$ are independent, we have $P(x_{1},x_{2})=P(x_{1})P(X_{2})$. Therefore we have the following:





Which is in fact a pdf. Now, since $Y=X_{1}X_{2}$, we have $(X_{1},X_{2}) \longrightarrow Y$

$(-1,-1) \longrightarrow Y=1$

$(-1,1) \longrightarrow Y=-1$

$(1,-1) \longrightarrow Y=-1$

$(1,1) \longrightarrow Y=1$

Therefore, $P_Y(1)=\frac{1}{2}$ and $P_Y(-1)=\frac{1}{2}$.

$\textbf{I am not entirely sure where to go from here (If any of this is even right at all)}$


1 Answer 1


After finding pmf of $Y$, let $x,y\in \{1,-1\}$,

\begin{align} P(X_2 = x, Y=y) &= P(X_2=x, X_1=x_2y) \\ &= P(X_2=x)P(X_1=x_2y)\\ &= P(X_2=x) \cdot \frac12\\ &= P(X_2=x)P(Y=y) \end{align}

Hence, they are independent.


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