I am not very experienced with statistics, so I apologize if this is an incredibly basic question. A book I am reading (Examples and Problems in Mathematical Statistics - Zacks) makes the following claim that I cannot understand

  1. $X = (X_{1},\ldots, X_{n})$ has joint CDF belonging to $\mathcal{F}$ which is parametrized by $\theta\in \Theta$. $\mathcal{F}$ has some dominating measure through which all densities will be defined.
  2. A likelihood function defined over $\Theta$ is any function $L(\theta | X)$ that is equal to the density $f(x | \theta)$ up to multiplication by a function depending only on $x$. That is, $L(\theta | x) = \alpha(x)f(x | \theta)$. Let $L^{T}$ be a likelihood of a statistic $T(X)$.
  3. The Neyman Fischer factorization theorem says that $T(X)$ is sufficient iff there exists some nonnegative functions $a,b$ so that $f(x | \mu) = a(x)b(T(x),\mu)$
  4. Claim: If $T(X)$ is a sufficient statistic, then the likelihood ratio $$L(\theta | X)/L^{T}(\theta | T(X))$$ is constant with respect to $\theta$.

Unfortunately, I don't see why this is true. Let $h$ be the density of $T(X)$ (with respect to any appropriate dominating measure)

$$\frac{L(\theta | X)}{L^{T}(\theta | T(X))} = \alpha(x)\frac{f(x | \theta)}{h(t(x) | \theta)} = \alpha(x)\frac{a(x)b(T(x),\theta)}{h(T(x)|\theta)}$$

Why is this constant with respect to $\theta$ if I don't know the dependence of $b$ or $h$ on $\theta$?


1 Answer 1


If $T(\cdot)$ is sufficient, then $L(\theta)$ depends on $y$ only through $T(y)$. (This follows from the definition of sufficiency, as another $y'$ such that $T(y)=T(y')$ would lead to an equivalent likelihood $L(\theta;y')$.)

That is, there is a $g$ such that $$ L(\theta)\propto g(T(y);\theta) $$ From the definition of the likelihood, we also know that $L(\theta)\propto f(y;\theta)$.

Hence, $f(y;\theta)/g(T(y);\theta)=:h(y)$ does not depend on $\theta$. This is so because $$L(\theta)\propto f(y;\theta)\Rightarrow L(\theta)=c\cdot f(y;\theta)$$ and $$L(\theta)\propto g(T(y);\theta)\Rightarrow L(\theta)=c'\cdot g(T(y);\theta),$$ such that $c'\cdot g(T(y);\theta)=c\cdot f(y;\theta)$ or $$f(y;\theta)/g(T(y);\theta)=c'/c$$

  • $\begingroup$ I accept everything that you're writing, but I'm not sure how it gets the claim. My understanding of $L^{T}(\theta | t)$ is that it is the density (with respect to an arbitrary dominating measure) of $T(X)$ at $t$ when the underlying probability distribution is given by $\mu$. Could you clarify the relationship of $L^{T}(\theta)$ with the functions $f$ and $g$ appearing in your answer? $\endgroup$ Commented Jun 26, 2020 at 12:26
  • $\begingroup$ What I was trying to express is that your $L^T(\theta|t)$ is (proportional to) my $g$, in the sense that if consider the likelihood function for a given value $t$ of the sufficient statistic, any other sample leading to the same value of the sufficient statistic leads to the same likelihood function. Also, the likelihood function is proportional to the density function $f$. So the ratio of these two things amounts to the ratio of two constants (w.r.t. $\theta$), so does not depend on $\theta$. $\endgroup$ Commented Jun 26, 2020 at 13:04

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