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Covariance

Anonymous Thu 10 Jun 15:45:19 2021 No.13261654 View Reply Original Report

Quoted By: >>13261669 >>13261681 >>13261869 >>13261885 >>13261897

Suppose I have Cov(x,z) and Cov(y,z). Is then Cov(x+y,z)=Cov(x,z)+Cov(y,z)?

Anonymous

Anonymous Thu 10 Jun 2021 15:48:52 No.13261669 Report

Quoted By:

>>13261654
Yes.

Anonymous

Anonymous Thu 10 Jun 2021 15:51:51 No.13261681 Report

Quoted By:

>>13261654
you can find all the reasoning you need on wikipedia

Anonymous

Anonymous Thu 10 Jun 2021 16:00:26 No.13261710 Report

Quoted By: >>13261732

I dont understand why, but for now its enough that it works kek. Thanks then

Anonymous

Anonymous Thu 10 Jun 2021 16:06:46 No.13261732 Report

Quoted By:

>>13261710
You can convince yourself this is true by writing down the covariance in its expectation form and rearranging terms.

Anonymous

Anonymous Thu 10 Jun 2021 16:51:19 No.13261869 Report

Quoted By:

>>13261654
Cov(x,z) = E[xz] - E[x]E[z]

Cov(y,z) = E[yz] -E[y]E[z]

Cov(x+y, z) = E[xz + yz] -E[x+y]E[z] = E[xz] + E[yz] - E[x]E[z] - E[y]E[z] = E[xz] - E[x]E[z] + E[yz] - E[y]E[z] = Cov(x,z) + Cov(y,z)

Anonymous

Anonymous Thu 10 Jun 2021 16:54:18 No.13261885 Report

Quoted By: >>13261897

>>13261654
$ Cov(x,z) = E[xz] - E[x]E[z] Cov(y,z) = E[yz] -E[y]E[z] Cov(x+y, z) = E[xz + yz] -E[x+y]E[z] = E[xz] + E[yz] - E[x]E[z] - E[y]E[z] = E[xz] - E[x]E[z] + E[yz] - E[y]E[z] = Cov(x,z) + Cov(y,z) $

Anonymous

Anonymous Thu 10 Jun 2021 16:56:14 No.13261897 Report

Quoted By:

>>13261885
>>13261654
$ Cov(x,z) = E[xz] - E[x]E[z] \\ Cov(y,z) = E[yz] -E[y]E[z] \\ Cov(x+y, z) = E[xz + yz] -E[x+y]E[z] = E[xz] + E[yz] - E[x]E[z] - E[y]E[z] = E[xz] - E[x]E[z] + E[yz] - E[y]E[z] = Cov(x,z) + Cov(y,z) $

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