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Let $X\sim N(0,\theta)$ for $\theta>0$. Show that $X^2$ is complete and sufficient for $\theta$. I assume this is referring to $\theta$ as the variance of $X$.

I'm unsure of how to show sufficiency in this context. I assumed that I would take the likelihood function of $X$ and divide it by the PDF of $X^2$ (where $X^2\sim GAM(\frac{1}{2},2\theta)$), but when I take that ratio $\theta$ is still leftover in the expression, suggesting that $X^2$ is NOT sufficient; but since that's what I'm supposed to show, clearly I'm doing something wrong. So I'm not sure how I'm supposed to show sufficiency. I'm sure it's simple, but I'm just not seeing it.

jippyjoe4
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The pdf of $X$ is $\sqrt{2\pi\sigma}\exp(x^2/\theta),$ which only depends on the data $X$ through $x^2,$ implying that $X^2$ is sufficient. This follows from the Fisher-Neyman factorization theorem, which it almost sounds as though you are appealing to (but you would not be dividing by the pdf of the candidate sufficient statistic). You can also see sufficiency directly from the definition. $X$ is symmetric about $0$ and so conditional on $X^2=y$, $X$ is equidistributed on $\sqrt{y}$ and $-\sqrt{y}$, regardless of the value $\theta$.

However, there is a more general result applying to exponential families, giving easy conditions for complete and sufficient statistics. It sounds like your exercise wants you to do that, since you still have to show completeness.

Hasse1987
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  • I know that a gamma random variable is in the linear exponential family, but I don't understand exactly what other conditions I have to satisfy in order to conclude that $X^2$ is both sufficient and complete for $\theta$ in particular. – jippyjoe4 Mar 03 '18 at 07:02
  • The result runs something like: if the density of an exponential family density can be parametrized as $exp(\theta^Tx)h(\theta)g(x)$, then $x$ is complete and sufficient for $\theta.$ – Hasse1987 Mar 03 '18 at 15:40