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?

Thanks you in advance

  • 1
    $\begingroup$ In this question and previous ones, you are asking your readers for the mathematical equivalent of a complete code review. Although you have done a great job presenting your work, and that is much appreciated, ultimately such detailed reviews are not what this site is about. If you have concerns about particular steps, please identify and isolate them and present them as directly and simply as possible for evaluation. (This will help you find your mistakes yourself.) For more guidance concerning how to ask answerable questions, please consult our help center. $\endgroup$ – whuber Jan 12 '14 at 17:29
  • $\begingroup$ @whuber Thanks a lot. But my main concern here is not for people to check every detail. I just want to know if my understanding of the question/answer is correct. In other words, I just want to know if there are serious problems with my answer or not (regardless of the computational details). The reason why I'm posting them here is because there is no way I can check my answers. I'm studying for an exam...so I just want to know if my overall answer is correct. $\endgroup$ – Artus Jan 12 '14 at 18:32
  • $\begingroup$ Your answer in each case is a sequence of a dozen or so assertions or manipulations. I'm sure you are not in doubt about most of them. Since you are doubtful about the answer overall, that would seem to suggest there are a small number of steps that give you pause. Which ones? The more you can do to help your readers understand the issues of most concern to you, the better your chances will be of getting good answers in a short time. Focusing your question is one way to make it more accessible and more easily answered. $\endgroup$ – whuber Jan 12 '14 at 18:41
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    $\begingroup$ @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... $\endgroup$ – Artus Jan 12 '14 at 19:01
  • $\begingroup$ Those changes really focus your question (and I'm pretty sure they address important issues in your solutions). Why don't you edit your question to emphasize those steps? $\endgroup$ – whuber Jan 12 '14 at 21:15

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