Questions about Linear Combinations of Random Variables

[ken15ken15]

Homework Statement

Homework Equations

Y=1/2*(X1-X3)^2+1/14*(X2+2X4-3X5)^2

The Attempt at a Solution

For (a) part, I have only learned to find the moment-generating function of Y, but not finding the p.d.f.
Moreover, the examples I have seen only involves random variables Xi to the power 1, but not to higher power.

For (b) part, the difficulty for me is just similar to part (a).

 

[dirk_mec1]

Hint: $$(Z_1^2 + ... +Z_n^2) +(Z_1^2 + ... +Z_m^2) = (Z_1^2 + ... +Z_n^2 +...+ Z_m^2)$$

 

[ken15ken15]

Hint: $$(Z_1^2 + ... +Z_n^2) +(Z_1^2 + ... +Z_m^2) = (Z_1^2 + ... +Z_n^2 +...+ Z_m^2)$$

sorry but...what property is it? I haven't learned this before...

 

[Ray Vickson]

Homework Statement

Homework Equations

Y=1/2*(X1-X3)^2+1/14*(X2+2X4-3X5)^2

The Attempt at a Solution

For (a) part, I have only learned to find the moment-generating function of Y, but not finding the p.d.f.
Moreover, the examples I have seen only involves random variables Xi to the power 1, but not to higher power.

For (b) part, the difficulty for me is just similar to part (a).

$Z_1 = X_1 - X_3$ has an $N(0,a)$ distribution, $Z_2 = X_2+2X_4-3X_5$ is $N(0,b)$, and $Z_1, Z_2$ are independent. So $Y$ involves a weighted sum of squares of independent, mean-0 random variables.

 

[ken15ken15]

$Z_1 = X_1 - X_3$ has an $N(0,a)$ distribution, $Z_2 = X_2+2X_4-3X_5$ is $N(0,b)$, and $Z_1, Z_2$ are independent. So $Y$ involves a weighted sum of squares of independent, mean-0 random variables.

So do you mean given the random variables are independent, the sum or difference of them will also have the same distribution of them? Also, how can a duel with the square of the random variables? Do I need to consider it with transformation of variables?