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So, the problem gives me this facts (for a Normal bivariate distribution X,Y)

$$Var(Y|X=x) = 5$$ $$E(Y|X=x) = 2 + x$$

It asks me to find $$E[Y^2|X=7]$$

I tried this: using the conditional variance $$Var[Y|X=x] = E[Y^2|X=x] + E[Y|X=x]^2 $$ So, $$ E[Y^2|X=x] = Var[Y|X=x] - E[Y|X=x]^2$$ $$ E[Y^2|X=x] = 5 - (2 + x)^2$$ Evaluating in X = 7, $$ E[Y^2|X=7] = 5 - (9)^2$$ $$ E[Y^2|X=7] = -76$$

Am I wrong? where?

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  • $\begingroup$ but $\mathsf{var}(Y) = E(Y^2) - E^2(Y)$ $\endgroup$ – Ahmad Bazzi May 21 at 9:56
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Simple mistake you made: $Var(Z)=E[Z^2]-(E[Z])^2$ (and not with a $+$).

So, $E[Z^2]=Var(Z)+(E[Z])^2$.

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The second term in the second step has a plus sign.

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