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$(X,Y)$ is distributed uniformly on the unit disk.

The transformations are:

$$ Z = {X + Y \over \sqrt{2}}\,,\qquad W = {X - Y \over \sqrt{2}} $$

I solved these equations in terms of $X$ and $Y$ and got:

$X$ = $(Z+W)/\sqrt(2)$

$Y$ = $(Z-W)/\sqrt(2)$

which has Jacobian equal to 1. Joint distribution of $Z$ and $W$ should be

$f(z, w)$ = $1/2\pi$ but with range -$\sqrt(2)$ $<$ $z$ $<$ $\sqrt(2)$. I'm not sure how to restrict the range of W.

The questions of interest are:

(a) What is the distribution of Z? Find $E[Z]$ and $Var[Z]%|$.

It seems fairly easy to see that $E[Z] = 0$ since the expectation of $X$ and $Y$ are both zero (both random variables are centered around zero). However, how would we compute $E[X^2]$ in order to compute the variance?

(b) Are Z and W uncorrelated? And are they independent?

I think that they are uncorrelated, but that they are not independent. I think this part might get easier after figuring out the exact range of the joint distribution.

(c) What is the distribution of $X/Y$ (the ratio of X and Y)?

Making the transformation $U$ = $X/Y$ and $V$ = $Y$ we have Jacobian 1. I'm confused as to what the range of joint distribution should be.

(d) What is the distribution of $Z/W$?

I'd imagine this part would get easier after figuring out part (c)

Felix Marin
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John Smith
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  • First things first: why do you think the range of $(Z,W)$ is the square ${(z,w)\mid-\sqrt2\lt z,w\lt\sqrt2}$? Even before that, I do not see the density of $(X,Y)$ in your question, did you write it down? – Did Nov 10 '13 at 09:10
  • Sorry! I was really sleepy when I wrote the question. I didn't think range of (Z,W) was a square. I meant that one of the variables range from $-\sqrt(2)$ to $\sqrt(2)$ and I was having trouble getting the range of the other variable. Also the density of (X,Y) is $1/\pi$ and their marginals should be Y|X ~ Uniform from $(-\sqrt(1-x^2), \sqrt(1-x^2))$ and X|Y also uniform based on symmetricity. – John Smith Nov 10 '13 at 18:21
  • The density of $(X,Y)$ is not $1/\pi$ since the function $1/\pi$ is not integrable on $\mathbb R^2$. I believe the most useful thing you can do here is to write down correctly the density of $(X,Y)$, so... what is this density? – Did Nov 10 '13 at 21:29
  • Thank you for your reply. Why wouldn't the joint density be $1/\pi$? The area of region should be just the area of circle which is $\pi$, is that not right? – John Smith Nov 10 '13 at 22:39
  • No, if you say the density is $1/\pi$ then the density is $1/\pi$ everywhere, and that is impossible. Or you mean that the density is $1/\pi$ somewhere and $0$ elsewhere--and then the density is not $1/\pi$ but... – Did Nov 11 '13 at 00:21
  • Well the density of (X,Y) would be $1/\pi$ inside the unit disk and 0 outside it. I thought that was obvious from saying that (X,Y) is uniform on the unit disk, but probably still better to be clearer. – John Smith Nov 11 '13 at 00:47
  • Right, $f_{X,Y}(x,y)=(1/\pi)\mathbf 1_{x^2+y^2\leqslant1}$ for every $(x,y)$. Now, $(Z,W)$ is what you know. Does this yield the density $f_{Z,W}$? – Did Nov 11 '13 at 09:02

1 Answers1

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The transformation you have chosen is a rotation of coordinates, and so the joint density of $W$ and $Z$ is the same as the joint density of $X$ and $Y$, that is, having value $\pi^{-1}$ inside the unit disc and value $0$ outside. While it is true that the maximum values of $X$ and $Y$ are $1$, they cannot take on value $1$ simultaneously and so your range $(-\sqrt{2}, \sqrt{2})$ is incorrect.

(a) To find $\displaystyle E[X^2] = \int_{-\infty}^\infty \int_{-\infty}^\infty x^2 f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$, change to polar coordinates. You will also need $\operatorname{cov}(X,Y)$ which can also be found by computing $E[XY]$ via the change to polar coordinates.

(b) Uncorrelated is easy to do since $E[ZW] = \frac{1}{2}E[X^2-Y^2] = 0$. Why? They are not independent for the same reason that $X$ and $Y$ are not independent: the support of the joint density is not a product set and so the joint density does not factor into the product of the marginal densities over the entire plane.

(c) Don't use Jacobians here, just ask yourself how you might compute $P\{Y/X \leq \alpha\}$ directly. Drawing a picture will help tremendously here. But if not, look at this answer.

(d) Since the joint density of $Z$ and $W$ is the same as the joint density of $X$ and $Y$, ....

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Dilip Sarwate
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