# Doubt in proof for Complete Statistic for Uniform Distribution

A statistic $$T(X)$$ is called complete statistic for a parameter $$\theta$$, if $$E_{\theta}g(T) = 0$$ for all $$\theta$$ implies $$P_{\theta}(g(T) = 0) = 1$$ for all $$\theta$$.

I interpret $$P_{\theta}(g(T) = 0) = 1 \> \forall \theta \quad$$ as

$$\> g(t) = 0 \> for \> almost \> every \> t \in T \enspace \forall \theta\quad$$ (for continuous distribution)

$$\> g(t) = 0$$ $$\forall t \in T \enspace \forall \theta\quad\quad\quad\quad\quad$$ (for discrete distribution)

Note: for the following continuous case(unifrom distribution), I've abused this notation a bit and have written $$'\forall t'$$ instead of $$\>'for \> almost\> every\> t'$$

In the book Statistical inference(2nd ed.) by Berger and Casella in Example 6.2.23, to prove that $$T(X) = \max_i X_i$$ is a complete statistic for random sample $$X_1 , X_2 , ... X_n$$ following Uniform distribution $$f(x;\theta) = 1/\theta,\; 0\leq x \leq \theta$$
We assume a function $$g(t)$$ satisfying $$E_{\theta}g(T) = 0 \> \forall \theta$$ and finally arrive at the condition that $$g(\theta) = 0 \> \forall \theta$$. I've understood the proof till here but couldn't understand that how from this condition, can we conclude that $$T$$ is a complete statistic.

Shouldn't we need to show that $$g(t) = 0 \> \forall t \> \forall \theta$$ to conclude that $$T$$ is a complete statistic. For example if we consider a function $$g(t) = t-\theta$$ ,then $$g(\theta) = \theta-\theta = 0 \>\forall \theta$$ but that need not necessarily mean that $$g(t) = 0 \> \forall t \> \forall \theta$$.

I saw a similar proof here but couldn't get how $$g = 0$$. I think here also it meant that $$g(\theta) = 0\> \forall \theta$$ and not $$g(t) = 0 \> \forall t \> \forall \theta$$.

I couldn't understand where I am going wrong or is my interpretation above is wrong. Please help.

• Function $g(t)$ does not depend on $\theta$. When we substitute $T$ instead of $t$, we get a r.v. with some distribution which depends on $\theta$. So, $\mathbb P(g(T)=0)$ depends on $\theta$. We need that for every $\theta$, $g(T)=0$ a.s. If you get somewhere $g(\theta)=0$ for all $\theta$, you can denote variable as you wish: $g(x)=0 \forall x$, $g(t)=0 \forall t$ and so on. – NCh Feb 6 at 14:15