probability question

setrajonas

Tomorrow's my Probability/Statistics final, and I was just going over some review questions, wherein I got stonewalled by this one:

Let X1 and X2 be independently and identically distributed with a common probability density function f(x)=(lamba)[e^(-lamba * x)] for x>0 and lamba>0. Let U = X1 + 2 * X2 and V = 2 * X1 + X2. What is the joint probability density function g(u,v) for u>0 and v>0?

I have no clue how to proceed from here, seeing as U and V look exactly identical.

Daric

Ok, I am very bad at math but doesn't the order of operations work on this one?

So you would do 2 * X2 and then add X1 for U.

And you would do 2 * X1 and then add X2 for V.

snowkissed

setrajonas wrote: »

I have no clue how to proceed from here, seeing as U and V look exactly identical.

But they're not exactly identical simply because X1 and X2 are different, independent variables. It may turn out sometimes that X1 and X2 have the same values and thus U and V are identical, but this won't always happen.

I have to think about this a little more as it's been a while since I did stats, but I thought I'd mention the above.

Shazkar Shadowstorm

I should be able to help you because I also have my prob/stats final coming up.. something like this was on my midterm. Allow me to look it up and I shall return.

I should know this. Ruh oh.

Folken Fanel

Are you familiar with moment generating functions?

Moment generating functions (mgf) are derived from the functional
m(theta) = E[e^(theta*x)] = Expectation of (e^(theta*x))

In this case, the mgf is

m(theta) = E[e^(theta*(x1 + 2x2))]
= E[e^(theta*x1) * e^(theta*x2) * e^(theta*x2)]
= E[e^(theta*x1)]*E[e^(theta*x2)]*E[e^(theta*x2)] (since x1 and x2 are iid)
= (lambda/(lambda-theta))^3
= mgf of erlang pdf where n=3

Use a similar process for part 2 of your question.

setrajonas

Folken Fanel wrote: »

Are you familiar with moment generating functions?

Moment generating functions (mgf) are derived from the functional
m(theta) = E[e^(theta*x)] = Expectation of (e^(theta*x))

In this case, the mgf is

m(theta) = E[e^(theta*(x1 + 2x2))]
= E[e^(theta*x1) * e^(theta*x2) * e^(theta*x2)]
= E[e^(theta*x1)]*E[e^(theta*x2)]*E[e^(theta*x2)] (since x1 and x2 are iid)
= (lambda/(lambda-theta))^3
= mgf of erlang pdf where n=3

Use a similar process for part 2 of your question.

I could follow that all the way up to "= E[e^(theta*x1)]*E[e^(theta*x2)]*E[e^(theta*x2)] (since x1 and x2 are iid)", then I'm not sure how you got to the next step. Also, we didn't learn the Erlang pdf in this course, so I somehow doubt that it's part of the answer, heh.

Daric

Apparently my skill at algebra 2 is not enough.