Mark M.

# Need help determining PDF, please.

Given are two random variables X and Y for which the joint probability density function fX,Y (x, y) is given by:

fX,Y (x, y) = 0.5xy for 0 ≤ x ≤ y ≤ 2

fX,Y (x, y) = 0 otherwise.

Let Z = Y − X. Determine the probability density function fZ(z) for 0 ≤ z ≤ 2 and the expectation E(Z) of the random variable Z.

I was trying to solve this by using a double integral, and then taking the derivative to obtaint the PDF. But this did not seem to work. Is there a way to solve this using a double integral? If so what would the limits have to be? Could anyone please show me how to solve it I have been stuck for hours on this question.

Karen A.

tutor
I get x goes from 0 to 2-z and y = x+z so I Have a single integral where x goes from 0 to 2-z and y = x+z so if the constant of .5 is correct ( which I’m not sure) you get .5 * the integral from 0 to 2-z of x(x+z) dx I’ll try to write it up in Latex\
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05/23/20

Karen A.

tutor
But thats just if z is positive
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05/23/20

Karen A.

tutor
But thats just if z is positive
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05/23/20

Karen A.

tutor
if z is negative then x goes from -z to 2 and again y = x+z
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05/23/20

Karen A.

tutor
I can't figure out how to put a nice Latex version here. Let me know if you know how to do it or if you want more explanation above.
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05/23/20

Mark M.

I got it, thanks!
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05/24/20

## 2 Answers By Expert Tutors

By:

Tom K.

tutor
The domain is [0,2] Note how f(0) = 4/3; we get 0 when x=y, and when we integrate 1/2y^2 from 0 to 2, we get 4/3. f(2) = 0 and f is monotone decreasing on the interval, which we should expect.
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05/24/20

Karen A.

tutor
I am getting a volume of 2 which would mean the probability distribution should really f(x,y) = .25xy on the interval 0<=x,y<= 2
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05/23/20

Mark M.

Thanks for the answer, I am sure the joint pdf is 0.5xy for 0≤x≤y≤2.
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05/23/20

Mark M.

When I calculate the double integral of y from 0 to 2 and x from 0 to y I get a value of 1. So I think it is valid
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05/23/20

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