Question

Let X and Y be any random variables and let a,b,c, and d be real numbers.

a. Show that Cov(aX+b,cY+d) = acCov(X,Y).

Answer 1

To show that Cov(aX+b,cY+d) = acCov(X,Y), substitute the given variables into the equation and use the properties of covariance.

Step-by-step explanation:

To show that Cov(aX+b,cY+d) = acCov(X,Y), we need to use the properties of covariance. First, let's recall the definition of covariance:

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

Now, let's substitute the given variables:

Cov(aX+b,cY+d) = E((aX+b - E(aX+b))(cY+d - E(cY+d)))

Expanding the equation:

Cov(aX+b,cY+d) = E(a(X - E(X)))E(c(Y - E(Y)))

Since a and c are constants, we can pull them out of the expectation:

Cov(aX+b,cY+d) = acE((X - E(X))(Y - E(Y)))

Which is equivalently:

Cov(aX+b,cY+d) = acCov(X,Y)