Is $T(X) = X$ a trivial sufficient statistic

probabilityrandom variablessolution-verificationstatistical-inferencestatistics

I know that the First Theorem for Sufficient Statistics states the following:

  • For a given statistical model for the random vector X $=(X_1, . . ., X_n)$ with pdf/pmf $f_{\theta}$, then $T(X)$ (with pdf/pmf $\tilde{f}_{\theta}$) is a sufficient statistic if and only if for every $x \in S$, the ratio $\frac{f_{\theta}}{\tilde{f}_{\theta}}$ does not depend on $\theta$.

Therefore, is it reasonable for me to conclude that if $T(X)=X$, then $f_{\theta}=\tilde{f}_{\theta}$, and so the ratio $\frac{f_{\theta}}{\tilde{f}_{\theta}}=1$?

And since this is always independent of $\theta$, we can conclude that $T(X)=X$ will always be a sufficient statistic irrespective/independently of the statistical model / distribution of the random vector X.

In practice I don’t believe that $T(X)=X$ is of any use to consider, however, I was wondering if this is, nevertheless, a sufficient statistic, by the argument I have presented.


Edit: I was able to find one comment on the MSE that referenced this being a degenerate case, but haven’t found anything definitive as of yet.

Best Answer

The original sample is always a sufficient statistic. No data reduction occurs, but it is sufficient because, tautologically, the original sample contains all the information about any unknown parameters that was present in the original sample.

See also:

Maximum sufficient statistics?

Ways of characterizing sufficient statistics

Is every estimator a sufficient statistic?

Sufficient statistic for $\theta$ in $N(\theta,\theta)$ model