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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 $\theta$.

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  • $\begingroup$ What does g refer to? Also, what do you mean by complete? $\endgroup$ – roundsquare Sep 4 at 16:26
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    $\begingroup$ 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. $\endgroup$ – whuber Sep 4 at 16:51
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Assuming the parameter space is $\Omega=(0,1)$.

You have

\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 $g(1)=0$.

Hence conclude using the correct definition of completeness.

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