Is the Calculation of Cov(X1 + X2, X2 + X3) Correct?

[BookMark440]

Homework Statement

Suppose X1 , X2 , X3 , and X4 are independent with a common mean 1 and common variance 2. Compute Cov( X1 + X2 , X2 + X3).

Homework Equations

Cov (X,Y) = E[(X-u)(Y-v)] = E[XY] - uv, where u and v are the means of X and Y
E[XY] = E[X]E[Y]
E[X+Y] = E[X] + E[Y]

The Attempt at a Solution

Since there is a common mean=1, Cov(X,Y) = E[XY] -1 = E[X]E[Y] - 1
Substituting for E[XY]= E[X1+X2]E[X2+X3]
Substituting for E[X+Y]= (E[X1]+E[X2])(E[X2]+E[X3])
Expanding (E[X1]+E[X2])(E[X2]+E[X3]) = E[X1]E[X2] + E[X1]E[X3] + E[X2]^2+ E[X2]E[X3]
Substituting for mean=1: (1)(1) + (1)(1) + 1^2 +(1)(1) = 4

Cov (X,Y) = E[XY] - 1 = 4 -1 = 3

Can someone check this for me? I think it is wrong. If the variables are independent with identical means and variances, shouldn't the covariance equal the variance?

 

[mighty2000]

Your solution is correct. The covariance between two variables is equal to the product of their variances when they are independent. In this case, the covariance between X1+X2 and X2+X3 is equal to the product of their variances, which is 2*2 = 4. Therefore, the covariance is equal to 4-1 = 3.