Density function

barabbas

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Feb 4, 2020
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Hi, would be thankful to get some help here.

The joint density function for X and Y is given by

f(x,y) = cxe^2, 0 ≤ x ≤ 1, y ≥ 0,

c = constant.

a ) Find constant c that makes f(x,y) a density function.

Here I did a regular double integral where y goes from 0(a) to ∞(b) (is that the correct way of doing it?) and X from 0(a) to 1(b). Put the function f(x,y) = 1. Get that c = 2.

b) Find P(Y < X).
Here I'm not sure what to do.
P(Y < X) = 1 - P(X ≤ Y). So I take 1 - double integral of f(x,y). To what should I set the boundaries? Y from 0(a) to ∞(b). And X from 0(a) to Y(b)?
 
Are you sure you have f(x,y) properly defined? Maybe a "y" missing from somewhere?

Contemplate the line y = x to determine your boundaries.
 
Yes I messed up. It’s. f(x,y) = cxe^(-y)
You want to solve for \(c\) in
\(\displaystyle c\int_0^1 {\int_0^\infty {{e^{ - y}}\ x \ dy } dx}=1\)..........................................edited
 
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Are you sure you have f(x,y) properly defined? Maybe a "y" missing from somewhere?

Contemplate the line y = x to determine your boundaries.

Okay so when calculating P(Y<X) I get,

I take double integral of f(x,y) = 2xe^(-y) dydx. Y from 0(a) to x(b) and X from 0(a) to 1(b). P(Y<X) = (4/e) - 1

Any thoughts?
 
0 < 4/e - 1 = 0.4715 < 1

Do you believe it?

I mean I think so? Or maybe not since Y goes to infinity. The probability should be a lot lesser than 0.4715? that X is bigger than Y since X only goes to 1.
 
I mean I think so? Or maybe not since Y goes to infinity. The probability should be a lot lesser than 0.4715? that X is bigger than Y since X only goes to 1.
Don't let me talk you out of it. Part of what you should be learning is confidence. Check your work. Can you reproduce it? Did you make any errors on purpose? Just do a little reality check before you think you are done.
 
All my errors are done on purpose. OK, just some. Well maybe none.
 
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