Mixture of normals

Asked Feb 13 '22 at 12:19

Active Feb 13 '22 at 12:19

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Suppose that n pairs $(x_i, y_i), i = 1,,,, n,$ are independently drawn from the following mixture of normals distribution:

where $0<p<1$

I have a solution given as: $$E[y]=pE[y|A]+(1-p)E[y|\bar{A}]=4p$$ $$E[x]=pE[x|A]+(1-p)E[x|\bar{A}]=4-4p$$ $$E[xy]=pE[x|A]E[y|A]+(1-p)(E[x|\bar{A}]E[y|\bar{A}]+C[x,y|\bar{A}])=0$$ and lastly $$C[x,y]=E[xy]-E[x]E[y]=-16(1-p)$$

My question is why $c(x,y)$ can't be calculated by $pC(x,y|A) +(1-p)C(x,y|\bar{A})=0$ in this case because their covarinces in distribution are both 0?

asked Feb 13 '22 at 12:19

Maybeline Lee

1

Why do you think you should be able to compute it that way? What rule are you referring to? – frank Feb 13 '22 at 12:40
Based on the given solution that says E() can be computed in that way, why can't c() be computed like this as well? – Maybeline Lee Feb 13 '22 at 13:36
1

The proposed formula neglects the effects of the means on the overall covariance. – whuber Feb 13 '22 at 16:08

Mixture of normals

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