Want to see if this covariance formula I'm thinking of is correct?

Question

In my book, in the section about multiple random variables

I am told that the Covariance of random variables $X_1$ and $X_2$ is Cov($X_1,X_2$) = E($X_1X_2)-\mu_1\mu_2$

My question is, is an equivalent form of the above:

Cov($X_1,X_2$) = E($X_1$)E($X_2$) - $\mu_1\mu_2$?

No. The expected value function is only linear with respect to addition and multiplication by a constant factor. See here: https://math.stackexchange.com/questions/253546/expected-value-linearity — K.defaoite, Oct 21 '20 at 10:21

Graham Kemp · Accepted Answer · 2020-10-21T04:01:44.577

0

Not always. That will only occur when $\mathsf E(X_1X_2)=\mathsf E(X_1)\mathsf E(X_2)$, which is the case when $X_1$ and $X_2$ are linearly uncorrelated.

Since, $\mu_1=\mathsf E(X_1)$ and $\mu_2=\mathsf E(X_2)$, then $\mathsf E(X_1)\mathsf E(X_2)-\mu_1\mu_2=0$.

edited Oct 21 '20 at 04:01

answered Oct 21 '20 at 04:00

Graham Kemp

129,094

I see so only when independent, got it, thank you! – Oct 21 '20 at 04:01
No. They do not need to be independent to be uncorrelated. – Graham Kemp Oct 21 '20 at 04:02
Oh, what are the other conditions then? – Oct 21 '20 at 04:02
$X_1,X_2$ are uncorrelated exactly when $\mathsf Cov(X_1,X_2)=0$. Therefore $\mathsf E(X_1X_2)=\mathsf E(X_1)\mathsf E(X_2)$ exactly when $X_1,X_2$ are uncorrelated. – Graham Kemp Oct 21 '20 at 04:03
I see, my book seems to just focus on the independence aspect of it though. Thanks a lot for your help. – Oct 21 '20 at 04:05
Independent random variables are always uncorrelated. But uncorrelated random variables do not need to be independent. – Graham Kemp Oct 21 '20 at 04:06

Want to see if this covariance formula I'm thinking of is correct?

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