Results 1 to 3 of 3

Math Help - variance and covariance equivalence

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    7

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by gablar View Post
    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..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    7
    Thank you very much, I got now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. expectation, variance, covariance
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: July 30th 2011, 01:04 AM
  2. variance & covariance of orthogonal contrasts
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: June 13th 2011, 08:13 AM
  3. Covariance equal to variance
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 6th 2011, 08:23 PM
  4. Covariance/variance with two random variables
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 2nd 2009, 01:12 PM
  5. variance and covariance need to clarify!
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: January 12th 2009, 01:06 AM

Search Tags


/mathhelpforum @mathhelpforum