# Moments of linear combinations of normally distributed variables

So I have this problem I have to solve for my homework for my Linear Regression Class that is due tomorrow and I honestly dont know where to start with this problem.

"Suppose to have $X\sim N(0,1)$ $Y=3-2X$ and $Z=2-X$

Find $E(Y)$ $E(Z)$ $Var(Y)$ $Var(Z)$"

I think the fact that it is a normal distribution might have something to do with solving this but other than that I am confused...

• Everybody who posts a question here needs help with a statistics question. Please choose a title that relates to the specific statistical problem you have. The normality of $X$ is irrelevant; the problem deals with basic properties of expectation and variance only. Sep 3 '15 at 23:00

Since this is homework, I'll just give hints. Remember that E means "expected value". Now, think about what the expected value for 3 - 2X and 2 - X would be.

Further hint: The normal distribution is not a key point here.

• so would E(Y) = 3-2(0)=2 and E(Z)= 2-0=2 since E(X) = 0 ? and based off that the var(y) = 4 and var(z)=1 Sep 3 '15 at 22:07
• Slight mistake 3-2(0)= 3 ;)
– user30490
Sep 3 '15 at 22:34

"Suppose to have X~N(0,1) Y=3-2X and Z=2-X

Find E(Y) E(Z) Va(Y) Var(Z)"

I'll show you how to solve one of each and then you should be able to do the rest.

\begin{align} E(Y)&=E(3-2X)\\ &=E(3)-E(2X)\\ &=3-2\times E(X)\\ &=3-2\times0\\ &=3 \end{align}

Likewise \begin{align} Var(Y)&=Var(3-2X)\\ &=Var(3)+Var(-2X)\\ &=0+(-2)^2\times Var(X)\\ &=0+(-2)^2\times1\\ &=0+4\times1\\ &=4 \end{align}

The calculations for E(Z) and Var(Z) follow similarly.