# Mathematical Statistics – How to Determine if a Statistic is Sufficient for Variance?

mathematical-statisticssufficient-statisticsvariance

I have $$X_1,\dots,X_n,X_{n+1}\overset{iid}{\sim}F_X(x)$$, where $$F_X$$ has a finite mean $$\mu$$ and variance $$\sigma^2$$.

If I calculate $$\bar X_n = \dfrac{1}{n}\sum_{i=1}^n$$ and $$S^2_n = \dfrac{1}{n-1}\sum_{i=1}^n\left(x_i – \bar x_n\right)^2$$ based on the first $$n$$ observations, I am able to use those, along with $$n$$ and $$X_{n+1}$$, to calculate $$S^2_{n+1} = \dfrac{1}{(n+1)-1}\sum_{i=1}^{n+1}\left(x_i – \bar x_n\right)^2$$ based on all $$n+1$$ observations.

Does this make $$(\bar X_n, S^2_n, n, X_{n+1})$$ a sufficient statistic for $$\sigma^2?$$ If not, is my function of those four values a sufficient statistic for $$\sigma^2?$$

Intuitively, I say this should be the case, since I have as much information to estimate $$\sigma^2$$ by having $$(\bar X_n, S^2_n, n, X_{n+1})$$ as I do from having all of the $$X_i$$ values, but I struggle to formally prove this or even begin to prove it.

With i.i.d. samples from the non-parametric family you specify, the order statistic $$(X_{(1)}, \ldots, X_{(n)})$$ is minimal sufficientâ€”only the order of the observations lacks information about the distribution from which they arise. It's also complete: consequently, the sample mean and variance, while not sufficient themselves, as functions of the order statistic are not only unbiased estimators of their population analogues, but the unique uniformly minimum-variance unbiased estimators.
If you know $$(\bar X_n, S^2_n)$$ is sufficient for a sample of size $$n$$, then $$(\bar X_{n+1}, S^2_{n+1})$$ is sufficient for a sample of size $$n+1$$. If you can show the latter statistic is a function of $$(\bar X_n, S^2_n, X_{n+1})$$, which is trivial, it follows that $$(\bar X_n, S^2_n, X_{n+1})$$ is also sufficient, as @whuber points out.