Probability random vector, transformation

[Laura1321412]

Homework Statement

Let f(x,y) = e^(-x-y), 0<x< infinity, 0<y<infinity, zero elsewhere, be the pdf of X and Y. Then if Z = X + Y, compute P(Z<=0), P(Z,<=6), and, more generally P(Z<=z), for 0<z<infinity. What is the pdf of Z?

Homework Equations

P(x,y) = ∫∫(f(x,y) dxdy

The Attempt at a Solution

so, P(Z<= 0 ) is pretty obviously 0

P(Z<=6) = P(X+Y <= 6)
=P(X<=6-Y)

into equation

∫from (0 to infinty) ∫ from (0 to 6-Y ) e^(-x-y) dx dy

-e^(-x-y) eval from 0 to 6-Y
= (-e^-6 ) + e^-y

∫from (0 to infinity) (-e^-6 ) + e^-y dy
= -ye^(-6) -e^-y eval from (0 to infinity)
= -infinity -0 - ( 0 - 1)
= - infinity

so, I am definately going wrong somewhere because a probability of negative infinity makes no sense... I know this could be done using a different method of tranformations, but i think I am supposed to do something along these lines because that's what is taught in the preceeding chapter.

 

[mighty2000]

Hello, thank you for your post. Your attempt at solving this problem is on the right track, but there are a few errors that I would like to point out.

First, P(Z<=0) is not necessarily zero. Remember that Z = X + Y, so in order for Z to be less than or equal to 0, both X and Y would have to be 0. This means that P(Z<=0) would be equal to f(0,0), which is not necessarily zero.

Second, when calculating P(Z<=6), you are correct in setting up the integral as P(X<=6-Y). However, when evaluating the integral, you need to take into account the range of values for Y. Since Y can range from 0 to infinity, you need to split the integral into two parts: one from 0 to 6, and one from 6 to infinity. This is because when Y is greater than 6, the integral will evaluate to 0.

Lastly, when calculating the integral for P(Z<=z), you need to take into account the range of values for both X and Y. This means that the integral should be set up as ∫∫(f(x,y) dxdy, with the limits of integration being 0 to z for both x and y.

In general, the pdf of Z can be found by convolving the pdfs of X and Y. This means that the pdf of Z would be f(z) = ∫f(x,z-x)dx, with the limits of integration being 0 to z. This can also be written as f(z) = ∫f(z-y,y)dy, with the limits of integration being 0 to z.

I hope this helps clarify any confusion you may have had. Keep up the good work!