Thread: Expectation with 3 variables?

I have this question:



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?

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)



Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$

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

the marginal densities are obvious by inspection.
You need not integrate out the other variables.

Clearly the density of X1 is cX1 where $\displaystyle c\int_1^3 xdx=1$

Okay understood, thanks guys.

IT's useful,thanks
Longchamp Classic Large

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

?

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

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.