Verifying whether $X$ is a complete statistic

The pmf of $$X$$ is as follows:

$$X = -1 \rightarrow p(x)= \theta$$

$$X = 0 \rightarrow p(x)= \theta^2$$

$$X = 1 \rightarrow p(x)= 1-\theta-\theta^2$$

I know that to show whether $$X$$ is complete it is only necessary to prove that if $$E[g(X)] = 0 \rightarrow P(X=0) = 1$$

I have done the following:

$$E[g(X)] = \sum_{t=-1,0,1} g(x)p(x)$$, but $$p(t)>0$$ then $$g(x) = 0\, \forall x \in \text{Support}$$.

However I don't know if I can assure such statement since I do not know the value of $$\theta$$.

• What does g refer to? Also, what do you mean by complete? – roundsquare Sep 4 at 16:26
• Your characterization of a complete statistic is erroneous: somehow, "$g(T)$" disappeared and was replaced by "$X.$" See en.wikipedia.org/wiki/Completeness_(statistics)#Definition. Using the correct definition will help you complete this exercise. – whuber Sep 4 at 16:51

Assuming the parameter space is $$\Omega=(0,1)$$.
\begin{align} \operatorname E_{\theta}[g(X)]&=\theta g(-1)+\theta^2 g(0)+(1-\theta-\theta^2)g(1) \\&=\theta^2(g(0)-g(1))+\theta(g(-1)-g(1))+g(1)\quad,\,\forall\,\theta\in\Omega \end{align}
Observe that $$\operatorname E_{\theta}[g(X)]=0$$ implies that the coefficients of $$\theta^2$$ and $$\theta$$ are zero alongwith $$g(1)=0$$.