0
$\begingroup$

I am trying to prove that a statistic $T$ is complete, where $T \sim \text{Gamma}(n, e^\theta)$. I am stuck at deducing that $g(T)$ must be zero almost surely, just by looking at the expectation: $$ E[g(T)] = \int_0^\infty g(t) t^{n-1} e^{-te^\theta} \frac{e^{-n \theta}}{\Gamma(n)} dt= 0 $$ The above shoulld imply that $g(t)=0$ almost surely.

Why is $g(t)=0$ if $$\int_0^\infty g(t) t^{n-1} e^{-te^\theta} dt = 0$$ for all $\theta$? I know that $t^{n-1} e^{-te^\theta}$ is a positive function, but $g(t)$ does not have to be? Is it because the integral must be zero for the whole gamma family?

$\endgroup$
2
  • 2
    $\begingroup$ It might help to recognize your last integral as the Laplace transform of the function $t\to g(t)t^{n-1}$ evaluated at $e^\theta\gt 0.$ $\endgroup$ – whuber Nov 28 '18 at 18:53
  • 1
    $\begingroup$ Thank you for the hint, I was not familiar with Laplace transformations, and I managed to find an answer here, but, I think I'm out of my league.. math.stackexchange.com/questions/47507/… $\endgroup$ – alekdimi Nov 28 '18 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.