# Does Fisher's factorization theorem provide the pdf of the sufficient statistic?

From Wikipedia

Fisher's factorization theorem or factorization criterion provides a convenient characterization of a sufficient statistic. If the probability density function is $ƒ_θ(x)$, then $T$ is sufficient for $θ$ if and only if nonnegative functions $g$ and $h$ can be found such that $$f_\theta(x)=h(x) \, g_\theta(T(x)), \,\!$$ i.e. the density $ƒ$ can be factored into a product such that one factor, $h$, does not depend on $θ$ and the other factor, which does depend on $θ$, depends on $x$ only through $T(x)$.

I was wondering if $\frac{g_\theta(t)}{c}$, where $c := \int g_\theta(t) dt$, is the pdf of $T(X)$ when the pdf of $X$ is $ƒ_θ(x)$?

Thanks and regards!

• Don't you really mean $\int_{\{ x\colon T(x)=t \} } g_\theta(x)\; dx$ ? Oct 21, 2019 at 11:50

No. Just consider the case when $h(x)=1$, so that $c=\int\cdots\int f(x_1,\ldots, x_n)\,\mathrm{d} x_1 \ldots\,\mathrm{d} x_n=1$, & $g(t)$ itself would be your putative probability density function: it clearly won't (in general) be normalized.

Consider say two observations from an exponential distribution: $$f_{X_1,X_2}(x_1,x_2)=\lambda^2\mathrm{e}^{-\lambda(x_1+x_2)}=\lambda^2\mathrm{e}^{-\lambda t}$$

You would have it that $f_T(t)=\lambda^2\mathrm{e}^{-\lambda t}$, but that's not even a probability distribution - it doesn't integrate to one.

With $c$ now defined as an integral with respect to $t$, the pdf for $t$ would be $\lambda\mathrm{e}^{-\lambda t}$. That's a pdf, but for an exponential distribution!

The correct solution in this case is got by finding the joint distribution of $T$ & say $X_2$, then by integrating out $x_2$.

$$f_{T}(t)=\int_0^t f_{X_1,X_2}(t-x_2,x_2)\,\mathrm{d}x_2=\lambda^2t\mathrm{e}^{-\lambda t}$$

Which is a gamma distribution with shape parameter 2.

• Thanks, but in your case, why isn't $g_{\theta}(t)$ the density function of $T(X)$?
– Tim
May 6, 2013 at 1:13
• I saw your edit. but $c := \int g_\theta(t) dt$ not $c := \int g_\theta(T(x)) dx$. (sorry for mistake in my original post)
– Tim
May 6, 2013 at 1:48

This is not a complete answer whatsoever, but I think this question deserves some attention.

In the discrete case we can easily get that if $$T$$ is sufficient, then $$f_{\theta}(\mathbf{x})=g_{\theta}(T(\mathbf{x}))h(\mathbf{x})$$ where $$g_{\theta}$$ is the p.m.f of $$T$$. Indeed, \begin{equation} \begin{aligned} f_{\theta}(\mathbf{x})&:=P_{\theta}(\{\mathbf{x}\})=P_{\theta}(\{\mathbf{x}\}\cap(T=T(\mathbf{x})))\\&=P_{\theta}(\{\mathbf{x}\}|T=T(\mathbf{x}))P_{\theta}(T=T(\mathbf{x}))\text{, } \end{aligned} \end{equation} so, taking $$h(\mathbf{x}):=P_{\theta}(\{\mathbf{x}\}|T=T(\mathbf{x}))$$ and $$g_\theta(y):=P_{\theta}(T=y)$$ we have that $$g_{\theta}$$ is the p.m.f of $$T$$.

However, I found no evidence of an annalogous result being true for the absolutely continuous case (check this). The book "Statistical Inference" by Casella-Berger seems to assume that this is true at least in:

1. Theorem 8.2.4
2. Corollary 8.3.13

But also I found no complains about of this (at least) dubious assumptions in the internet. Also there is no mention about this in the several errata list I have checked to this day.

• Even if the factorization equation holds for the discrete case, it does not mean that $g$ is the pmf of the sufficient statistic. If we assume that $T$ is a sufficient statistic then one can always get a factorization where $g$ is the PMF of $T$ (discrete case). Jan 13 at 16:53