By the Fisher's factorization theorem, a statistics is a sufficient statistic if (and only if) the joint density,

$$ f(x_1, x_2, x_3, \dots x_n; \theta) $$

can be factorized into two functions, $ g(s; \theta) $ and $ h(x_1, x_2, \dots x_n) $.

What is the intuition behind this theorem?

I know that a proof exists for this theorem, and I also understand the original definition for a sufficient statistic (that the conditional distribution of the sample, given the statistic should not depend on the parameter).

  • $\begingroup$ But then, to find the actual probability $ f(x_1, x_2, x_3, ... x_n; \theta) $, we'd still need the proportionality constant right? $\endgroup$
    – WorldGov
    Commented Nov 15, 2019 at 8:09

1 Answer 1


Fisher's factorisation theorem is one of several ways to establish or prove that a statistic $S_n(X_1,\ldots,X_n)$ is sufficient. The meaning of sufficiency remains identical through all these manners of characterising it though, namely that the conditional distribution of the sample $X_1,\ldots,X_n$ conditional on $S_n(X_1,\ldots,X_n)$ is constant in $\theta$, i.e., is the same distribution for all values of $\theta$.

For instance, given an iid sample $X_1,\ldots,X_n$, the order statistic $S_n(X_1,\ldots,X_n)=(X_{(1)},\ldots,X_{(n)})$ is sufficient because the distribution of $X_1,\ldots,X_n$ given $S_n(X_1,\ldots,X_n)$ is the uniform distribution on the permutations of $\{1,\ldots,n\}$: \begin{align*}\mathbb P_\theta((X_1,\ldots,X_n)&=(y_1,\ldots,y_n)|S_n(X_1,\ldots,X_n)=(x_{(1)},\ldots,x_{(n)})\\&=\frac{\mathbb I_{(x_{(1)},\ldots,x_{(n)})}(y_{(1)},\ldots,y_{(n)})}{n!}\end{align*}

enter image description hereThe factorisation theorem thus does not modify the original distribution of the sample and does not particularly help in finding it. It operates the other way, as a mean of figuring out sufficient statistics by looking at the joint pmf or pdf. Once a sufficient statistic has been found, its own distribution is sufficient (in the sense of "enough") for creating a likelihood function in the parameter $\theta$. It will then differ from the "full" likelihood by a constant (in the parameter) equal to the function (of the sample) $h(x_1,\ldots,x_n)$ found in the factorisation theorem. But this constant has no relevance for statistical inference.

  • 1
    $\begingroup$ I agree with @Xi'an, here is a link talking about the factorization theorem. 'it is not always all that easy to find the conditional distribution of $X_1, X_2, ..., X_n$ given $Y$. Not to mention that we'd have to find the conditional distribution of $X_1, X_2, ..., X_n$ given $Y$ for every $Y$ that we'd want to consider a possible sufficient statistic!' newonlinecourses.science.psu.edu/stat414/node/283 $\endgroup$
    – Bill Chen
    Commented Nov 14, 2019 at 17:15

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