Statistics question (variance)

  • Thread starter hanelliot
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Studying for an intro course test and I have no one to compare it to right now.. any help would be appreciated.

Here is the question.

Q. Suppose X and Y are random variables such that p(X,Y)=1/3, Var(X) = 9 and Var(Y) = 1. Compute Var(X-2Y).

Since X and Y are not independent, we are using this formula: Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y), correct?
So, Var(X-2Y) = Var(X) - 4Var(Y) - 2Cov(X,Y)? How do I proceed from here?
 

mathman

Science Advisor
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Studying for an intro course test and I have no one to compare it to right now.. any help would be appreciated.

Here is the question.

Q. Suppose X and Y are random variables such that p(X,Y)=1/3, Var(X) = 9 and Var(Y) = 1. Compute Var(X-2Y).

Since X and Y are not independent, we are using this formula: Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y), correct?
So, Var(X-2Y) = Var(X) - 4Var(Y) - 2Cov(X,Y)? How do I proceed from here?
First you have the get the correct equation:

Var(X-2Y) = Var(X) + 4Var(Y) - 4Cov(X,Y)
You have Var(X) and Var(Y) already.
I'll assume that p(X,Y) is the correlation function. If so Cov(X,Y)=p(X,Y)(Var(X)Var(Y))1/2.
 

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