# mean and variance

• May 17th 2008, 09:45 AM
haleyGeorge
mean 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 AM
Moo
Hello,

Quote:

Originally Posted by haleyGeorge
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

The mean is a linear application.

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

From the definition, $var(X+Y)=var(X)+var(Y)+2cov(X,Y)$
But if X and Y are independent, then $cov(X,Y)=0 \implies var(X+Y)=var(X)+var(Y)$

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

From there, for the first one for example :

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

And $var(2X_1-3X_2+4X_3)=4var(X_1)+9var(X_2)+16var(X_3)$
• May 21st 2008, 12:50 PM
haleyGeorge
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 PM
Moo
Quote:

Originally Posted by haleyGeorge
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?

Yes, you just plug them in it :)
• May 21st 2008, 01:17 PM
haleyGeorge
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 PM
Moo
Quote:

Originally Posted by haleyGeorge
ok, then for the first one I got E(Y) = -7 and var(Y) = 155. Is that correct? Thanks for all your help!

Yes !!! (Clapping)

How much do you find for Z ? :)
• May 21st 2008, 01:31 PM
haleyGeorge
E(Z) = -17 and var(Z) = 36