Expectation with 3 variables?

• Dec 15th 2011, 02:09 PM
Wevans2303
Expectation with 3 variables?
I have this question:

http://i44.tinypic.com/r0bclk.jpg

I know I need to find E[X1], E[X2] and E[X3] and then fiddle a bit to get the answers, but when I integrate do I just treat the two variables I am not interested in as constants?
• Dec 16th 2011, 01:45 AM
chisigma
Re: Expectation with 3 variables?
By definition the expected value of an $\displaystyle g(x_{1},x_{2},x_{3})$ is...

$\displaystyle E(G)= \int_{1}^{3} \int_1}^{2} \int_{0}^{1} g(x_{1},x_{2},x_{3})\ \frac{x_{1}\ x_{2}\ x_{3}}{3}\ d x_{1}\ d x_{2}\ d x_{3}$ (1)

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 16th 2011, 01:06 PM
Moo
Re: Expectation with 3 variables?
Hello,

To be more precise on what chisigma wrote above, you'd be looking for the expected value of $\displaystyle g(X_1,X_2,X_3)$ (in general, capital letters point to random variables, whereas small letters point to variables).

But here, it's much better to calculate the first 2 moments of $\displaystyle X_1,X_2,X_3$, then say that since the joint pdf can be written as functions of only x1, x2 or x3, the three variables are independent. And hence this simplifies quite a lot the calculations, especially for the last 2 expectations.

You'd be able to treat the other rv's as constants if you're considering a conditional expectation with respect to some given rv or sigma-algebra. But given the kind of problem you're working on, it's not something you've studied yet, maybe soon :)
• Dec 18th 2011, 12:39 AM
matheagle
Re: Expectation with 3 variables?
the marginal densities are obvious by inspection.
You need not integrate out the other variables.

Clearly the density of X1 is cX1 where $\displaystyle c\int_1^3 xdx=1$
• Dec 19th 2011, 09:09 AM
Wevans2303
Re: Expectation with 3 variables?
Okay understood, thanks guys. :D
• Dec 26th 2011, 08:29 PM
longchampshop
Re: Expectation with 3 variables?
IT's useful,thanks
Longchamp Classic Large
• Jan 5th 2012, 11:41 AM
Wevans2303
Re: Expectation with 3 variables?
Just an update.

I am using chisigmas method for this question as a means of practice..

for the first one E[3X1+5X2+4X3] is there a shorthand alternative other than calculating the first 2 moments? As opposed to integrating:

E[3X1+5X2+4X3] * (X1X2X3)/3

?
• Jan 5th 2012, 11:46 AM
matheagle
Re: Expectation with 3 variables?
the density of X3 by inspection, no integration, is $\displaystyle 2X_3$ on (0,1)
that's a beta distribution, the mean is alpha/(apha+beta)=2/3
you can do the same with the other two random variables, even though they aren't beta's

Then $\displaystyle E(4X_3)=4(2/3)$
• Jan 5th 2012, 12:00 PM
Wevans2303
Re: Expectation with 3 variables?
Quote:

Originally Posted by matheagle
the density of X3 by inspection, no integration, is $\displaystyle 2X_3$ on (0,1)
that's a beta distribution, the mean is alpha/(apha+beta)=2/3
you can do the same with the other two random variables, even thougt they aren't beta's

Then $\displaystyle E(4X_3)=4(2/3)$

Ah it makes perfect sense now, thanks for that.