Covariance of Binomial Random Variables

[Obraz35]

Homework Statement

Let X be the number of 1's and Y be the number of 2's that occur in n rolls of a fair die. Find Cov(X, Y)

Homework Equations

Cov(X,Y) = E(XY) - E(X)E(Y)

The Attempt at a Solution

Both X and Y are binomial with parameters n and 1/6. Thus it is easy to find E(X) and E(Y), but for some reason finding E(XY) is really tripping me up. How does one go about finding expected values of products of binomial variables?

 

[mighty2000]

Hi there! It's great to see that you are tackling this problem and trying to find the covariance between X and Y. Let me help you out with finding E(XY).

First, let's define the random variables X and Y more clearly. X is the number of 1's that occur in n rolls of a fair die, and Y is the number of 2's that occur in n rolls of a fair die.

Now, let's break down the expected value of XY into smaller parts. We know that XY = 1*2 = 2 when X=1 and Y=2, and that XY = 0 for all other combinations of X and Y.

So, we can write E(XY) as:

E(XY) = 2*P(X=1 and Y=2) + 0*P(all other combinations)

Since we are dealing with a fair die, we know that the probability of getting a 1 is 1/6 and the probability of getting a 2 is also 1/6. Therefore, the probability of getting a 1 and a 2 in a single roll is (1/6)*(1/6) = 1/36.

Substituting this into our equation for E(XY), we get:

E(XY) = 2*1/36 + 0*P(all other combinations)

= 1/18

Therefore, we can conclude that E(XY) = 1/18.

Now, using the formula for covariance, we can find the covariance between X and Y:

Cov(X,Y) = E(XY) - E(X)E(Y)

= 1/18 - (n/6)*(n/6)

= 1/18 - n^2/36

= (1-2n^2)/36

I hope this helps in your understanding of finding expected values of products of binomial variables. Let me know if you have any further questions. Good luck with your problem!