I am trying to determine if the Hypergeometric Distribution is a complete family of distributions.
The problem states that we have an urn with $m$(a known value) white balls and $\theta$(unknown value) black balls. Sample $m$ from the urn without replacement. Let $X$ be the number of white balls sampled from the $m$ pulled from the urn.
Given the information it sounds to me like this is a hypergeometric distribution with the pdf:
$$f(X;\theta,m,m)=\frac{{m \choose x}{\theta \choose m-x}}{{m+\theta \choose m}}$$
To show if complete will show equality to zero of the equation below implies $T=0$:
$$\sum_{t=0}^{m}T(t)\frac{{m \choose t}{\theta \choose m-t}}{{m+\theta \choose m}}=0$$
equivalently:
$$\sum_{t=0}^{m}T(t){m \choose t}{\theta \choose m-t}=0$$
Note: $${m \choose t}{\theta \choose m-t}>0; \forall t\in\{0,1,2...,m\}$$
Cannot figure out a $T$ that does not equal to zero that will make this true. Am I correct in concluding that $T=0$ and hence the hypergeometric is complete?
The fact that the coefficients are positive doesn’t tell us anything, since $T(t)$ could be negative.
The binomial coefficients $\binom\theta{m-t}$ are polynomials of degree $m-t$ in $\theta$. The polynomial resulting from the sum must be identically $0$, so the coefficient of every power of $\theta$ must be zero. The only term of degree $m$ is the one with $T(0)$, so $T(0)=0$. Then the only remaining term of degree $m-1$ is the one with $T(1)$, so $T(1)=0$. And so on.