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x1,x2,..,xn are i.i.d Normal(θ,1), I would like to find the var( x̄ ^ 2). By CLT, sqrt(n)(x̄ - θ) ~ Normal(0,1), so n((x̄-θ)^2) would be Gamma(1/2,2) (chi square with d.f = 1).

so, I have Var(n*((x̄-θ)^2)) = 2.

=> Var((x̄-θ)^2) = 2/(n^2).

=> Var(x̄^2 - 2x̄θ + θ^2) = 2/(n^2)

=> Var(x̄^2) + 4θ^2*Var(x̄) = 2/(n^2)

=> Var(x̄^2) + 4θ^2/n = 2/(n^2)

=> Var(x̄^2) = 2/(n^2) - 4θ^2/n, which apparently to wrong since it may less than 0.

Could anyone let me know which step I am wrong?

Thank for your help.

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    The mistake is in the step between 2nd and 3rd arrows. $Var(X+Y) = Var(X) + Var(Y)$ only if $X$ and $Y$ are independent. Clearly, $\overline{x}^2$ and $\overline{x}$ are not independent! :) I guess you should use the definition of $Var$ in the very first line and use expectations instead, since expectations are linear always. Good luck! – 2017-02-05

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Unlike means, variances are not in general additive, so you'll need a different strategy. You also need your final answer to be $\theta$-independent, since translating $X$ translates $\bar{X}$ without changing its variance. We may assume without loss of generality that $\theta=0$. Since $\bar{X}^2$ is Gamma-distributed, your challenge is finding the variance of a Gamma distribution.