Statistics - variance question

[hanelliot]

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?

 

[Defennder]

Hi, you may want to clarify what exactly p(X,Y) means here.

 

[Mene]

I understand that studying for an intro course test can be challenging, especially when you don't have anyone to compare your work to. However, I am happy to assist you with this question on variance.

First, you are correct in using the formula Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y) since X and Y are not independent. To proceed, we need to calculate the covariance (Cov) of X and Y.

Cov(X,Y) = E[(X-E[X])(Y-E[Y])] where E[X] and E[Y] are the expected values of X and Y respectively. Since p(X,Y) = 1/3, we can calculate the expected values as follows:

E[X] = 1*(1/3) + 2*(1/3) + 3*(1/3) = 2
E[Y] = 1*(1/3) + 2*(1/3) + 3*(1/3) = 2

Now, we can calculate the covariance using the formula:

Cov(X,Y) = E[(X-2)(Y-2)] = E[XY - 2X - 2Y + 4] = E[XY] - 4E[X] - 4E[Y] + 16

To calculate E[XY], we can use the fact that X and Y are independent, so E[XY] = E[X]E[Y] = 4. Therefore, Cov(X,Y) = 4 - 4(2) - 4(2) + 16 = 4.

Now, we can plug this value into the formula for Var(X-2Y):

Var(X-2Y) = Var(X) - 4Var(Y) - 2Cov(X,Y) = 9 - 4(1) - 2(4) = 1

Therefore, the variance of X-2Y is 1. I hope this helps and best of luck on your test! Remember to always check your calculations and seek help if needed.