Our new Indie Games subforum is now open for business in G&T. Go and check it out, you might land a code for a free game. If you're developing an indie game and want to post about it, follow these directions. If you don't, he'll break your legs! Hahaha! Seriously though.

Our rules have been updated and given their own forum. Go and look at them! They are nice, and there may be new ones that you didn't know about! Hooray for rules! Hooray for The System! Hooray for Conforming!

setrajonas
Registered User regular

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.

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.

0

## Posts

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

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

DariconI 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.

snowkissedonI should know this. Ruh oh.

Shazkar Shadowstormonm(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.

Folken FanelonTwitter:Folken_fgcSteam:folken_XBL:flashg03PSN:folken_PASFV: folken_Avatar shamelessly stolen from rieytails deviant art

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.

setrajonasonDaricon