# Thread: variance and covariance equivalence

1. ## variance and covariance equivalence

hello. I'm working on this problem. I hope someone can help me. I have worked a solution but I'm not confident about it.

Problem:
Express var(X+Y), var(X-Y) and cov(X+Y,X-Y) in terms of the variances and covariance of X and Y

My attempt:

Looking through references I found this identity:

var(X+Y)=var(X) + var(Y) + 2cov(X,Y)

This seems clear enough, although I don't know how to arrive at that solution, so it really doesn't solve my problem. I think that var(X-Y) and cov(X+Y,X-Y) can be derived from the above formula using substitutions. Can anyone point me in the right direction? I'm not sure I am understanding the question.

2. Hello,

Originally Posted by gablar
hello. I'm working on this problem. I hope someone can help me. I have worked a solution but I'm not confident about it.

Problem:
Express var(X+Y), var(X-Y) and cov(X+Y,X-Y) in terms of the variances and covariance of X and Y

My attempt:

Looking through references I found this identity:

var(X+Y)=var(X) + var(Y) + 2cov(X,Y)

This seems clear enough, although I don't know how to arrive at that solution, so it really doesn't solve my problem. I think that var(X-Y) and cov(X+Y,X-Y) can be derived from the above formula using substitutions. Can anyone point me in the right direction? I'm not sure I am understanding the question.
Try to work it out by using the definitions of the covariance and the variance with respect to the mean.

$var(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2$

$cov(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)$

Remember that the mean $\mathbb{E}$ is a linear application, so that $\mathbb{E}(X+Y)=\mathbb{E}(X)+\mathbb{E}(Y)$.
But $\mathbb{E}(XY)\neq \mathbb{E}(X)\mathbb{E}(Y)$ unless X and Y are independent..

3. Thank you very much, I got now.