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Given sample data $x_1, \ldots, x_n$ generated from a probability distribution $f(x|\theta)$ ($\theta$ being an unknown parameter), a statistic $T(x_1, \ldots, x_n)$ of the sample data is called sufficient if $f(x|\theta, t) = f(x|t)$.

However, I'm always kinda confused by this definition, since I think of a sufficient statistic as a function that gives just as much information about $\theta$ as the original data itself (which seems a little different from the definition above).

The definition of Bayesian sufficiency, on the other hand, does mesh with my intuition: $T$ is a Bayesian sufficient statistic if $f(\theta|t) = f(\theta|x)$.

So why is the first definition of sufficiency important? What does it capture that Bayesian sufficiency doesn't, and how should I think about it?

[Note: I believe that every sufficient statistic is also Bayesian sufficient, but not conversely (the reverse implication doesn't hold in the infinite-dimensional case, according to Wikipedia).]

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interesting question...btw, the paper cited in wikipedia with example of Bayes but not classically sufficient is online --… – Yaroslav Bulatov Sep 6 '10 at 0:35
Cool, thanks for the link. Nice to learn that Bayes sufficiency comes from Kolmogorov. – grautur Sep 6 '10 at 19:40

this response addresses part of grautur's question: how to think about the property of being "sufficient".

ra fisher suggested the terminology "sufficient" for a statistic $t$ that satisfies the [heuristic] requirement $$\kern{-2.5in}(1)\kern{2.5in} f(x|\theta, t) = f(x|t).$$

[there is nothing heuristic about this way of writing the requirement if $x$ is discrete. but if $x$ is continuous, the pdf notation $f(x|t)$ is usually heuristic. an example of the latter is when $x = (x_1,\dots x_n)$, where the $\{x_i\}$ are $iid$ N($\theta$,1). here the sample mean $\bar x$ is a sufficient statistic and $(1)$ requires considering the conditional pdf $$\kern{-.5in} (2)\kern{.5in} f(x_1,\dots, x_n|\bar x) \equiv f(x_1 - \bar x,\dots, x_n - \bar x | \bar x) = f(x_1 - \bar x,\dots, x_n - \bar x).$$ the last equality in $(2)$ holds since the sample deviations are independent of the sample mean in the normal case. (this fact gives an instant proof that the sample mean and variance are independent in the normal case.) here the "pdf" on the RHS of (2) does not exist as an $n-$dimensional pdf since the joint distribution of the sample deviations is singular - they sum to zero. replacing "$f$" in (1) by "dist", removes its heuristic nature.]

at any rate, the idea of (1) is that, as the conditional distribution of $x|t$ does not depend on $\theta$, one can [in principle] generate a new $x$ - call it $x^*$, say, from the known conditional distribution of $x|t$, so that $x^* \sim x$ for all $\theta$.

as an illustration, consider the normal example above. here, obtaining an $x^*$ is particularly easy since the deviations are independent of $\bar x$. thus one can generate $n$ iid N(0,1) variables $z_1,\dots, z_n$ and let $x^* = (x_1^*,\dots. x_n^*)$, where $x_i^* = z_i - \bar z + \bar x : 1\le i\le n$. clearly $x \sim x^*$ for all values of $\theta$, so that $x^*$ is just as "good" a sample from the population as the original $x$ for learning about $\theta$. clearly no one would actually want to use the sample deviations for $x^*$ in addition to its sufficient statistic $\bar x^* = \bar x$, as those deviations, the $\{z_i - \bar z\}$, were obtained by a completely extraneous random experiment having nothing to do with the actual data process. one then should agree that the original sample deviations $\{x_i - \bar x\}$ should also not be used, since they can be considered as being generated in the same way, by an extraneous random experiment, where now nature, rather than the statistician, did the generating.

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Suppose, we can factorize the pdf as follows:

$f(x|\theta) = h(x) g(t(x),\theta)$

If we can achieve the above factorization then it is obvious that the estimate of $\theta$ will depend only on the value of $t(x)$. In other words, our estimate of $\theta$ will be the same for two sets of data ($x_1$, $x_2$) if $t(x_1) = t(x_2)$.

The above feature of $t(x)$ is conceptualized by calling $t(x)$ as a sufficient statistic. It is sufficient in the sense that once we know $t(x)$ we implicitly have an estimate of $\theta$ and is termed as a statistic as it is a function of the data $x$.

I am a bit unsure about the relationship with bayesian sufficiency but will update my answer if I have some relevant thoughts on that issue.

By the way your definition of sufficient statistic does not seem to make sense to me as the right handside should depend on $\theta$.

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I'm not sure what you mean by your last sentence -- the definition of sufficient statistic I listed comes from Wikipedia (and the point is exactly that $f(x|\theta, t)$ should not depend on $\theta$). – grautur Sep 6 '10 at 19:39
I am not sure if the wiki's notation is correct. If you see some examples on the wiki you would note that every example of sufficient statistic is such that the pdf factors into two functions. One which depends only on the data and one which depends only on the data through t(x) and the parameter theta. For example, see the discussion on sufficient statistic for the possion distribution. – user116 Sep 6 '10 at 19:57
Your factorization is right. However, to comment on why the RHS does not depend on $\theta$: the formula is saying that $\theta$ provides no additional information beyond $t$. – Neil G Sep 14 '10 at 19:15

I think I see where your confusion is rooted. The notation f(x|θ) is chosen in the first place because there's a implication of a causal relationship between theta (the inferred cause) and x (the resulting observations.) And also because we usually have simple pdf representing the density on x given θ. This function is necessarily invertible, so nothing forces us to write things this way.

However, in the problem you presented, we are given x and inferring θ. So, f(x|θ) is read backwards. It could have been written L(θ|x) (the likelihood of θ given x). And then your intuition is almost prefect, except for one flaw discussed below.

We can write that a statistic t is sufficient iff L(θ|x) = L(θ|t,x), or equivalently, L(θ|t,x) = L(θ|t).

The flaw with your equation is that you need to condition on both t and x. Because even if L(θ|x) = L(θ|t), t could be based on information other than (just) x. For example, if x comprises n examples, θ could be the result of a different n examples. In that case, the combination of t and x would be like seeing 2n examples, and would almost always yield a different result.

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