Independent Random Variables

• Nov 29th 2010, 06:15 PM
demo12
Independent Random Variables
Hi,

I can't seem to get the same answer as the back for this question. Can someone show me how to properly do it?

Let X and Y be two independent random variables where E(X) = 4, E(X^2)= 20 and V(Y)= 21. Find:

(a) V (X + 3Y)
(b) the standard deviation of (X+ + 3Y)

The answers at the back are 193 and 13.9.

Thanks
• Nov 29th 2010, 06:58 PM
harish21
for independent variables, the covariance will be 0. thus:

\$\displaystyle Var(X+3Y)= Var(X)+Var(3Y) = Var(X)+3^2\;Var(Y)=Var(X)+9Var(Y)=....\$

you have E[X] and E[X^2], from which you can find Var(X).

Finding Var(X+3Y) means that you are a step away from finding the s.d. Do that!
• Nov 29th 2010, 07:31 PM
demo12
Quote:

Originally Posted by harish21
for independent variables, the covariance will be 0. thus:

\$\displaystyle Var(X+3Y)= Var(X)+Var(3Y) = Var(X)+3^2\;Var(Y)=Var(X)+9Var(Y)=....\$

you have E[X] and E[X^2], from which you can find Var(X).

Finding Var(X+3Y) means that you are a step away from finding the s.d. Do that!

Thanks for the prompt response. Good news is I got the right answer, however, I don't understand why the 3 turns into 3^2. Can you please explain to me why this is? The meaning of covariance isn't very clear to me either.

Thanks!
• Nov 29th 2010, 07:44 PM
harish21
Well, you should have a textbook and/or classnotes with you. And you should have been given these concepts before you are made to work on the problems.

Anyways, the three turns to 9 because it is a well known property.

For a random variable X with finite variance:

\$\displaystyle Var(aX) = E[(aX)^2] - E[aX]^2\$ [Note: a is a constant]

The mean of \$\displaystyle aX\$ is a times the mean of the random variable X.

\$\displaystyle E[aX] = aE[X] \; and \; E[(aX)^2] = E[a^2 (X^2)] = a^2 E[X^2]\$

\$\displaystyle \therefore Var(aX) = a^2 E[X^2] - [aE(X)]^2\$

\$\displaystyle = a^2 (E[X^2] -(E[X])^2)\$

\$\displaystyle = a^2 (Var(X))\$

More properties???

Look HERE and

go here for Covariance
• Nov 29th 2010, 07:56 PM
demo12
That cleared things up a bit. Thanks!