Question

Hard Stats Question Help!

Answer 1

read attachment. :)

Answer 2

Okay, since they are giving you the variance, can you tell me some equations they gave you regarding variance and expected value?

Answer 3

as well as covariance?

Answer 4

Okay so for (a) we use the fact that: \[
E[cX] = cE[X],\quad E[X+Y] = E[X]+E[Y]
\]Thus \[
E[Z] = E\left[\frac{X+Y}{2}\right] = \frac{1}{2}E[X+Y] = \frac{1}{2}(E[X]+E[Y]) = \frac{\mu_X+\mu_Y}{2}
\]

Answer 5

u still there

Answer 6

For (b), by saying they are independent, they're essentially telling us that \(Cov(X,Y)=0\).

Answer 7

for part a is E[z] just μ

Answer 8

We want to use \[
Var(aX) = a^2 Var(X),\quad Var(X+Y) = Var(X)+Var(Y)+2Cov(X,Y)
\]

Answer 9

@sall1234 Yeah, I'm not saying it isn't.... remember that \(\mu_X = \mu_Y = \mu\)? Just use that fact with the formula I gave.

Answer 10

is part b σ²/2 and is part c (3/4)σ²

Answer 11

Anyway for (b) we get \[
Var(Z) = Var\left(\frac{X+Y}{2}\right) = \frac{1}{4}Var(X+Y) = \frac{Var(X)+Var(Y)+0}{4} = \frac{\sigma_X^2+\sigma_Y^2}{4}
\]

Answer 12

For (c) we get \[
\frac{\sigma_X^2+\sigma_Y^2+2\cdot 0.5\sigma^2}{4} = \frac{\sigma_X^2+\sigma_Y^2+\sigma^2}{4}
\]

Answer 13

@sall1234 That looks right... why you ask these questions?

Answer 14

I wasnt sure I was right? Now i know for sure! :)