# How to show that this statistic is complete

Suppose that $$S, {f_θ : θ ∈ Θ})$$ is a statistical model, corresponding to an observed random vector $$\mathbf X = (X_1, . . . , X_n).$$

Let $$\theta_1(\mathbf X)$$ and $$\theta_2(\mathbf X)$$ be unbiased estimators for $$\theta$$ . Define $$\theta_3(\mathbf X) = a \theta_1(\mathbf X) +(1-a)\theta_2(\mathbf X)$$.

I have deduced that $$\theta_3(\mathbf X)$$ is an unbiased estimator of $$\theta$$:

$$a\Bbb E[ \theta_1(\mathbf X)] + \Bbb E[ \theta_2(\mathbf X)] - a \Bbb E[ \theta_2(\mathbf X)] =$$

$$= a \theta + \theta - a \theta = \theta$$.

Hence $$\Bbb E[ \theta_3(\mathbf X)] = \theta$$ and it is an unbiased estimator for $$\theta$$

Now I have the statistic $$T(\mathbf X)= (\theta_1(\mathbf X), \theta_2(\mathbf X))$$. I'm trying to show whether it's a complete statistic or not, I tried to show that $$T(\mathbf X)$$ is a complete statistic using the following definition:

Suppose that $$T=T(\mathbf X)$$ is a statistic taking values in a set $$\Theta$$. Then $$T$$ is a complete statistic for $$\theta$$ if for any function $$g: S \rightarrow \Bbb R$$ :

$$\Bbb E_{\theta}[g(T)] = 0$$ for all $$\theta \in T \Rightarrow \Bbb >P_{\theta}[g(T)=0]=1$$ for all $$\theta \in \Theta$$

So from this point i'm kind of stuck. Also, should I maybe use the fact that $$\theta_3(\mathbf X)$$ is unbiased? I have a feeling that constructing $$g$$ for $$T(\mathbf X)$$ based on $$\theta_3(\mathbf X)$$ being unbiased would help?

Any help is appreciated

• You can't choose a function $g$. It says "for any function $g$". Commented Feb 3, 2020 at 13:48
• well then how do I prove it "for any function $g$"?
– user634512
Commented Feb 3, 2020 at 13:49
• A more approriate response than to ask the question again in a comment would be to correct the question. Commented Feb 3, 2020 at 13:50
• @joriki ok, i will correct it
– user634512
Commented Feb 3, 2020 at 13:51
• @NCh oh... yes, you're right :D. It's not. Thanks. Я был настолько зацыклен показать что оно равно одному, что забыл что если оно не равно, то тоже сойдёт за ответ :D
– user634512
Commented Feb 3, 2020 at 14:00

There exists $$g(x,y)=x-y$$ such that $$\mathbb E[g(\theta_1(\mathbf X),\theta_2(\mathbf X))]=\mathbb E[\theta_1(\mathbf X)]-\mathbb E[ \theta_2(\mathbf X)]=0$$ for any $$\theta$$, and $$\mathbb P(g(\theta_1(\mathbf X),\theta_2(\mathbf X))=0)\neq 1$$ since the estimators do not coincide a.s., then $$T(\mathbf X)=(\theta_1(\mathbf X),\theta_2(\mathbf X))$$ is not complete by definition.
• almost sure, or with probability $1$.