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