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Topic review (newest first)

2013-04-24 06:00:15


I agree the answer is 2.

I am not aware of any definition of the median of a multivariable PDF.

2013-04-24 05:21:06

It has to do with the e^(-x-y) function. Consider the integral from 0 to infinity of e^(-x) which equals 1.

2013-04-23 18:37:25


For a) doesn't there need to be an upper bound?
You have X>=0 and Y>=0, for lower bounds.

If none is provided then why isn't the answer 2?

2013-04-23 16:25:18

This is not a statistics/probability course its multivariable calculus but the section in my book is called probability theory. My teacher writes his own homework problems, and he didnt really go over how to calculate most of these stats.

The point (X, Y ) is randomly distributed in the region x ≥ 0, y ≥ 0 according to the following probability density:
f(x,y) = e^(−x−y)

(a) Compute the average (= expected value) of X + Y .
(b) Compute the median of X/Y .
(c) Compute the probability that |X − Y | ≤ 1.
(d) Compute the average value of 1/(X + Y ), for example, by changing variables according to x = u and y = uv.

I computed a) to be 2 using the double integral from 0-->inf (x+y)(f(x,y))dxdy.

I don't know how to set up to calculate the median in part b) and am not exactly sure how to do d) we are still kinda learning changing variables techniques.

Thanks for any help!

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