For $X = (X_1,X_2,...,X_n)$ i.i.d. $N(0,\theta),$ show $ T = \bar X $ is not complete for $\theta$.

Question

For $X = (X_1,X_2,...,X_n)$ i.i.d. $N(0,\theta),$ show $ T = \bar X $ is not complete for $\theta$.

Definition: Let f (t, θ), θ ∈ Θ be a family of distributions for a statistic T (X). The family is called complete if $E_θg(T) = 0$ for all θ ∈ Θ implies $P_θ(g(T) = 0) = 1$ for all θ ∈ Θ. Equivalently, T(X) is called a complete statistic for θ.

I let $g(T)=t=\bar X$.

$E(g(T))=E(\bar X)=0 \quad $ (fairly obvious)

But how do I show g(T) is not identically zero everywhere.

Answer 1

$\bar X$ is a (scaled) sum of i.i.d. normally distribtued variables, so is also normally distributed. In particular, the probability that it takes the value $0$ is $0$, not $1$ as required.