If X1, X2, and X3 are independent and have the means 4, 9, and 3 and the vaiance 3,7, and 5, find the mean and the variance of:

a. Y = 2X1 -3X2 +4X3

b. Z = X1 -2X2 -X3

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- May 17th 2008, 09:45 AMhaleyGeorgemean and variance
If X1, X2, and X3 are independent and have the means 4, 9, and 3 and the vaiance 3,7, and 5, find the mean and the variance of:

a. Y = 2X1 -3X2 +4X3

b. Z = X1 -2X2 -X3 - May 17th 2008, 09:50 AMMoo
Hello,

The mean is a linear application.

Therefore : $\displaystyle E(aX)=aE(X)$ and $\displaystyle E(X+Y)=E(X)+E(Y)$, E designing the mean.

From the definition, $\displaystyle var(X+Y)=var(X)+var(Y)+2cov(X,Y)$

But if X and Y are independent, then $\displaystyle cov(X,Y)=0 \implies var(X+Y)=var(X)+var(Y)$

Plus, $\displaystyle var(X)=E(X^2)-(E(X))^2 \implies var(aX)=E((aX)^2)-(E(aX))^2=a^2 var(X)$

From there, for the first one for example :

$\displaystyle E(2X_1-3X_2+4X_3)=2E(X_1)-3E(X_2)+4E(X_3)$

And $\displaystyle var(2X_1-3X_2+4X_3)=4var(X_1)+9var(X_2)+16var(X_3)$ - May 21st 2008, 12:50 PMhaleyGeorge
I know this is probably a really stupid question, but do I just plug in the numbers given for mean and variance into the equation? Or what is my next step?

- May 21st 2008, 12:51 PMMoo
- May 21st 2008, 01:17 PMhaleyGeorge
ok, then for the first one I got E(Y) = -7 and var(Y) = 155. Is that correct? Thanks for all your help!

- May 21st 2008, 01:23 PMMoo
- May 21st 2008, 01:31 PMhaleyGeorge
E(Z) = -17 and var(Z) = 36