# Can Conditional Expected Value be negative in normal distribution?

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?

• but $\mathsf{var}(Y) = E(Y^2) - E^2(Y)$ – Ahmad Bazzi May 21 at 9:56

Simple mistake you made: $$Var(Z)=E[Z^2]-(E[Z])^2$$ (and not with a $$+$$).
So, $$E[Z^2]=Var(Z)+(E[Z])^2$$.