Suppose we are estimating $\tau(\theta)=\theta e^{-\theta}$ from $X_1,...,X_n \sim \mathrm{G}(\theta,r)$ (G is the gamma distribution) then it is easily shown that $T=\sum_{i=1}^n \ln(X_i)$ is sufficient or $\exp(T)$ also. I'm wondering, is a continuous function of a sufficient statistic also sufficient and if $T$ is sufficient for $\theta$ is it also sufficient for $\tau(\theta)$?


Let's back up a minute and remind ourselves what the definitions of these things are. First, we have a class of models $$ \{f_\theta(x): \theta \in \Theta\}, $$ Let $\mathcal L_\theta(X \mid T(X))$ denote the conditional distribution of $X$ given $T(X)$ and a fixed $\theta$. Then $T(X)$ is sufficient if $\mathcal L_\theta(X \mid T(X)) = \mathcal L_{\theta'}(X \mid T(X))$ for all $\theta, \theta' \in \Theta$.

Any one-to-one measurable function $g(\cdot)$ of $T$ will also be sufficient, because then conditioning on $g(T)$ is the same as conditioning on $T$ (intuitively, if you know one then you know the other). Just continuity, howevever, isn't enough. If $X \sim \mathcal N(\mu, 1)$ then $X$ is sufficient for $\mu$ but $X^2$ is not because one cannot distinguish $\mu$ from $-\mu$. In this case, $\mathcal L_\mu(X \mid X^2)$ would have most of its mass concentrated at $\sqrt {X^2}$ if $\mu > 0$ and $-\sqrt{X^2}$ if $\mu < 0$.

What does it mean for $T(X)$ to be "sufficient for $\tau(\theta)$"? Before addressing this, I don't think being "sufficient for $\tau(\theta)$" is a useful concept in your particular example; you would just say $T$ is sufficient for $\theta$ and that would give you all the information you would need. If $T$ is sufficient then $T$ in some sense tells you everything you need to know about $\theta$, and the value of $\tau(\theta)$ is something about $\theta$.

Being sufficient for a sub-parameter isn't covered by the original definition, but you can try to squeeze it in. We could maybe decompose $\theta = h(\tau, \eta)$ where $h$ is one-to-one and rewrite the model $$ \{f_{\tau, \eta}(x): \eta \in \mathbf H, \tau \in \mathbf T \} $$ and say $T(X)$ is "sufficient for $\tau$" if $\mathcal L_{\eta, \tau}(X \mid T(X)) = \mathcal L_{\eta, \tau'}(X \mid T(X))$ for all $\eta, \tau, \tau'$. In which case, if $T(X)$ is sufficient then it is trivially sufficient for $\tau$ - just take $\theta = h(\tau, \eta)$ and $\theta' = h(\tau', \eta)$ and observe $$ \mathcal L_{\eta, \tau}(X \mid T(X)) = \mathcal L_\theta(X \mid T(X)) = \mathcal L_{\theta'}(X \mid T(X)) = \mathcal L_{\eta, \tau'}(X \mid T(X)). $$

  • $\begingroup$ Ok thanks, I found it kind of weird to find sufficient statistic for a function of a parameter. I think it should be just about finding the sufficient statistic for $\theta$ and then evaluate different things about the function $\tau(\theta)$. $\endgroup$ – Raxel Nov 17 '13 at 0:12

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