# Determining if these random variables are independent

Let $X$ and $Y$ be iid $N(0, \sigma^2)$ variables.

1. Show that $X^2+Y^2$ and $X/ \sqrt{X^2 + Y^2}$ are independent. Hint: Compute the joint distribution of the first and the square of the second expression.

Since $X \sim N(0, \sigma^2)$, we know that $X/\sigma \sim N(0,1)$. So $X^2/\sigma^2 \sim \chi^2(1)$. But this means that $X^2 = \sigma^2(X^2/\sigma^2) \sim \chi^2(\sigma^2)$. Of course, we have the same result for $Y$.

Let $W_1 = X^2/(X^2 + Y^2)$ and $W_2 = X^2 + Y^2$.

We get the joint distribution,

$$f_{W_1, W_2}(w_1, w_2) = \left[ \left( \frac{1}{\Gamma(\sigma^2/2)2^{\sigma^2/2}} \right)^2 (w_1(1-w_1))^{\sigma^2/2 - 1} \right] \left[ (w_2)^{\sigma^2 - 2} e^{-w_2} \right]$$

There is a theorem in my textbook (Introduction to Mathematical Statistics By Hogg) which says

Let the random variables $X_1$ and $X_2$ have supports $S_1$ and $S_2$ respectively, and have the joint pdf $f(x_1, x_2)$. Then $X_1$ and $X_2$ are independent if and only if $f(x_1, x_2)$ cna be written as a product of a nonnegative function of $x_1$ and a nonnegative function of $x_2$. That is $$f(x_1, x_2) \equiv g(x_1)h(x_2),$$ where $g(x_1) > 0$, $x_1 \in S_1$, zero elsewhere, and $h(x_2) > 0$, $x_2 \in S_2$, zero elsewhere.

So, by this theorem, $W_1$ and $W_2$ are independent. (We know that $\left( \frac{1}{\Gamma(\sigma^2/2)2^{\sigma^2/2}} \right)^2 (w_1(1-w_1))^{\sigma^2/2 - 1}$ is positive because $w_1 = \frac{x^2}{x^2 + y^2} \leq 1$.)

1. Show that $X+Y$ and $X-Y$ are independent. Hint: Use a standard property of the normal distribution.

$X+Y \sim N(0,2)$ and $X-Y \sim N(0,2)$. Let $U_1 = X+Y$ and $U_2 = X-Y$.

We have,

$$f_{U_1, U_2}(u_1, ,u_2) = \left[ \left( \frac{1}{4 \pi} \right) e^{-u_1^2}{4} \right] \left[ e^{ (-u_2^2)/4} \right]$$

By the above theorem, $X+Y$ and $X-Y$ are independent.

Do you think my answers are correct?

• @whuber Well, in this case, I want to know 1) if "Since $X \sim N(0, \sigma^2)$, we know that $X/\sigma \sim N(0,1)$. So $X^2/\sigma^2 \sim \chi^2(1)$. But this means that $X^2 = \sigma^2(X^2/\sigma^2) \sim \chi^2(\sigma^2)$." is correct. and 2) if the fact that I used the transformation $W_1 = X^2/(X^2 + Y^2)$ and $W_2 = X^2 + Y^2$ (and $U_1 = X + Y$ & $U_2 = X - Y$ for the second one) and used that to apply the theorem I mentioned, is correct. You don't need to check the details, just the approach that I used. I guess I will try to write my answers clearer/simpler from now on... – Artus Jan 12 '14 at 19:01