CT3 April 2017 Q 3

Discussion in 'CS1' started by Molly, Mar 18, 2023.

  1. Molly

    Molly Ton up Member

    Hi all,

    I guess a little confusion with covariances here.
    the solution suggests that E(Z+Z_)=E(Z+*Z-), even though the two are not independent. I didnt know this was allowed. Can we always compute the joint expectation this way?

    Thanks
    Molly
     
  2. Andrea Goude

    Andrea Goude ActEd Tutor Staff Member

    Hi Molly

    In this question the random variables are not independent but are uncorrelated, cov(Z-Z+)=0.

    E(Z-Z+)=E[(X-Y)(X+Y)]=E(X^2-XY+XY-Y^2)=E(X^2-Y^2)=E(X^2)-E(Y^2)=0 for these random variables of Z+ and Z-.

    This is part of the simplified covariance formula cov(Z+Z-)=E(Z-Z+)-E(Z-)*E(Z+).

    Thanks Andrea
     
    Molly likes this.
  3. Molly

    Molly Ton up Member

    ah thank you :)
     

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