Showing Independence of Two Gaussian Random Variables

Question

Let $U$ be a uniform r.v. on $\{1,2,3,4\}$ and let $X, Y$ be independent standard Gaussian r.v.s. Define $A_1 = 1$ when $U \in \{1,2\}$ and $A_1 = -1$ when $U \in \{3,4\}$. Define $A_2 = 1$ when $U \in \{1,3\}$ and $A_1 = -1$ when $U \in \{2,4\}$.

I'm trying to show that $A_1 |X|$ and $A_2 |Y|$ are independent Gaussian variables. My reasoning is that $A_1 |X|$ is a left half-normal distribution half of the time, and a right half-normal distribution half of the time, and the same is true for $A_2 |Y|$. Since $X, Y$ are independent, then $A_1 |X|$ and $A_2 |Y|$ must be independent.

I'm pretty sure I'm missing some things, however. Any help would be appreciated.

Answer 1

Note that $A_1$, $A_2$ are themselves random variables, so if $X$ and $Y$ are independent, it does not necessarily follow in general that $A_1|X|$ and $A_2|Y|$ must also be independent. For example, suppose $B_1 = A_1$, but $B_2 = -A_1$. Then $B_1|X|$ and $B_2|Y|$ are no longer independent: knowledge about the sign of $B_1$ from knowing $B_1|X|$ immediately restricts the values that $B_2|Y|$ can take on.

Answer 2

Presumably you mean $X$, $Y$ and $U$ are independent, not just $X$ and $Y$ are independent.

It suffices to show that for intervals $I$ and $J$, $P(A_1 |X| \in I, A_2 |X| \in J) = P(X \in I) P(Y \in J) = P(A_1 |X| \in I) P(A_2 |Y| \in J)$. WLOG you can assume $I$ and $J$ are each on one side of $0$. Then e.g. if $I \subseteq [0,\infty)$ and $J \subseteq [0,\infty)$, $$P(A_1 |X| \in I, A_2 |Y| \in J) = P(U = 1, |X| \in I, |Y| \in J) = \dfrac{1}{4} P(|X| \in I) P(|Y| \in J) = P(X \in I) P(Y \in J)$$ and similarly in the other cases.

Answer 3

We should also start with the assumption that $X$ and $Y$ are independent of $U$ as well, but I think this is implicit. I'm going to write $A_1(U)$ and $A_2(U)$ so that we realize that $A_1$ and $A_2$ depend on $U$; in reality, you have $A_1 = 1_{\{1,2\}}(U) - 1_{\{3,4\}}(U)$ and $A_2 = 1_{\{1,3\}}(U) - 1_{\{2,4\}}(U)$; if you aren't comfortable with this "indicator" notation, no worries. Let $i,j \in \{-1,1\}$ and consider $P(A_1(U) = i, A_2(U) =j)$. You can check that whatever we choose for $i,j$, we get $P(A_1(U) = i, A_2(U) =j) = 1/4$. Also, $P(A_1(U) = i) = 1/2$ and $P(A_2(U)=j) = 1/2$ regardless of $i,j \in \{-1,1\}$. Therefore, $A_1(U)$ and $A_2(U)$ are independent. Now, you have $A_1(U)|X|$ is independent of $A_2(U)|Y|$ since they are products of independent variables.