summary with active recall: - the reason that if you add + b to the r.v. the reason covariance is not affected (which is supposed to measure variation wrt other variables), then it doesn't really change how it varies, since an offset does nothing - Conv(X,Y+Z) = Cov(X,Y) + Cov(X,Z), so covariance behaves linearly with of the arguments - cov(X,X) = Var(X)\ - cov(aX+b, Y) = a*cov(X, Y) - since an offset b doesn't change the variation (and covariance is a measure of that sort) but a does, since some values become larger, then the jumps on average for X will be larger. But only 1 increase since only one term changed - cov(X,Y+Z) = cov(X,Y) + cov(Y,Z) - hints at the fact cov measure linear behaviours since adding to one r.v. made the cov increase by 1 cov. - cov(X,Y) = E[XY] - E[X]E[Y]
@4:24 I think there's an error if I'm not mistaken. If we assume 0 means for both X and Y then both expectations will be 0 and the cov(X, Y) will also be 0. As a matter in fact that's true also even if only one of the r.v.s has 0 mean and the other not.
hmmm cov is zero if the variables are indepedent. If centering made covariance zero then covariance would be a really bad measure of "co variation". Offsets, should make a difference since it affects everything the same. Covariance should measure something like how X varies as Y varies (and in reversem hence why cov has a joint distribution)
summary with active recall:
- the reason that if you add + b to the r.v. the reason covariance is not affected (which is supposed to measure variation wrt other variables), then it doesn't really change how it varies, since an offset does nothing
- Conv(X,Y+Z) = Cov(X,Y) + Cov(X,Z), so covariance behaves linearly with of the arguments
- cov(X,X) = Var(X)\
- cov(aX+b, Y) = a*cov(X, Y)
- since an offset b doesn't change the variation (and covariance is a measure of that sort) but a does, since some values become larger, then the jumps on average for X will be larger. But only 1 increase since only one term changed
- cov(X,Y+Z) = cov(X,Y) + cov(Y,Z)
- hints at the fact cov measure linear behaviours since adding to one r.v. made the cov increase by 1 cov.
- cov(X,Y) = E[XY] - E[X]E[Y]
Thanks a lot for clarifying the issue for us
At 4:29 why did professor write that E[xy] is a covariance of x and y in the conclusion.
cov( X,Y ) = E [ X,Y ] - E [ X ] E [ Y ], since E [ X ] and E [ Y ] both equal zero (assume zero means), cov( X,Y ) = E [ X,Y ] .
@4:24 I think there's an error if I'm not mistaken. If we assume 0 means for both X and Y then both expectations will be 0 and the cov(X, Y) will also be 0. As a matter in fact that's true also even if only one of the r.v.s has 0 mean and the other not.
hmmm cov is zero if the variables are indepedent. If centering made covariance zero then covariance would be a really bad measure of "co variation". Offsets, should make a difference since it affects everything the same. Covariance should measure something like how X varies as Y varies (and in reversem hence why cov has a joint distribution)
this dude doesn't do mistakes.
Nice!! Thanks man, that was really helpful!!
hocam çok güzel anlatmışsınız emeğinize sağlık :)
life saver! thank you!
3:25
3:22