$$E[X_1] = 1 ,Var[X_1]= 2 \ \text{ and }\ E[X_2] = 2 , Var[X_2] = 3,$$ where $X_1,X_2$ are independent random variables, and

$$Y_1 = X_1+X_2 ,\ \ \ Y_2 = X_1-X_2.$$

I have found $E[Y_1] = 3 ,Var[Y_1] = 5\ \text { and }\ \ E[Y_2] = -1 , Var[Y_2] = 5.$

But am not sure how to find covariance of $(Y_1,Y_2)$ to use in the formula. Any help is appreciated.


1 Answer 1


Covariance has a property known as bilinearity. In this example, this means that $$\text{cov}(X_1+X_2,X_1-X_2) = \text{cov}(X_1,X_1)-\text{cov}(X_1,X_2)+\text{cov}(X_2,x_1)-\text{cov}(X_2,X_2).$$

From here, you should be able to fill in the information you have to compute your final value. In particular, keep in mind that $X_1, X_2$ are independent (this tells us that their covariance is 0), and note that in general, $\text{cov}(Z,Z) = \text{var}(Z)$.


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