# Sums and products of random variables

• May 16th 2011, 04:35 PM
dizzle1518
Sums and products of random variables
Hi all,

Can anyone help me with the below question?

for each of the following pairs of random variables X,Y, indicate
a. whether X and Y are dependent or independent
b. whether X and Y are positively correlated, negatively correlate or uncorrelated

i. X and Y are uniformly distributed on the disk {(x,y) in R^2: 0<=x^2+y^2<=1}
since this is a circle with R=1 the area is pi. Since x and y are uniform then fxy(xy) is 1/area = 1/pi. In order to see if x and y are independent i need to computer marginal densities and multiply to see if i get fxy(xy). This is where I am stuck. I know that to get fx i need to integrate fxy(xy) with respect to dy and for fy integrate with respect to dx. But what am i integrating? if fx = integral from -1 to 1 of 1/pi dy? and fy = to integral from -1 to 1 1/pi dx? For correlation i first need COV(XY) which equals = E(XY)-E(X)E(Y). To get E(XY) i integrate xy*fxy(xy) so the double integral from -1 to 1 of xy/pi? Is this correct? Now, since x and y are uniform then E(X) and E(Y) are just the interval over two so since each is between -1 and 1 we get 2/2=1=E(X)=E(Y). The VAR(X) and VAR(Y) is 2^/12, the radical of which gives standard deviation and from these value we can compute rho.

ii. X and Y are uniformly distributed on the parallelogram {(x,y) in R^2: x-1<=y<=x+1, -1<=x<=1}. the area of the parallelogram is 4, so fxy(xy) is 1/4. to get fx i integrate 1/4 from x-1 to x+1 with respect to dy which yields 1/2. To get fy i integrate 1/4 from -1 to 1 with respect to dx which also yields 1/2. since fx*fy=1/4=fxy(xy) then x and y are independent and thus uncorrelated. correct?

X and Y are uniformly distributed on the diamond {(x,y) in R^2: |x|+|y|<=1, |x|<=1}. i am not sure about this one. what will the graph of this one look like? is it a diamond on -1<=x<=1 and -1<=y<=1?
• May 24th 2011, 10:32 PM
matheagle
uniformly distributed on a circle means that they are dependent.
It's not a rectangle.

Since $\displaystyle P(.3<X<.5|Y=.5)\ne P(.3<X<.5|Y=.99)$
which both would have to be equal to $\displaystyle P(.3<X<.5)$

The marginals are easy to compute

$\displaystyle f_X(x)=\int^{\sqrt{1-x^2}}_{-\sqrt{1-x^2}}{dy\over \pi}$

$\displaystyle ={2\sqrt{1-x^2}\over \pi}$

on -1<x<1 of course
• May 24th 2011, 10:43 PM
bryangoodrich
Quote:

Originally Posted by matheagle
uniformly distributed on a circle means that they are dependent. It's not a rectangle.

Could you elaborate on this? I recall my instructor rambling something about geometry to the chalkboard, but she never actually explained it to the class. I sat in on a couple of classes with a professor that did mention it, but I didn't stay around long enough to gain an intuition with it. I suspect that if it is rectangular, then you can make a movement in one dimension (y-ward) without a change in the other dimension (x-ward). However, on a circle you cannot do that. For every y-ward movement you're also making an x-ward movement. Is this what you are getting at?
• May 24th 2011, 10:49 PM
matheagle
For two rvs to indep it is necessary and suff that their joint density factors and they are defined on some rectangle.

Look at the example of the unit circle.
Draw a line where Y=0 (x-axis) and Y=.95.
The probablity of X in the same interval,given these Y's, say .95<X<1 isn't the same.
Just like sets, for X and Y to be independent
Since $\displaystyle P(.95<X<1|Y=0)= P(.95<X<1)$....