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Let X $=X_1,...X_n$ be a random sample iid from the probability density function:

$$ f(x;\theta)=\frac{\Gamma(\theta)\sin(\pi\theta)\theta^{1-\theta}}{\pi}e^{-\theta x}x^{-\theta}$$ $x>0, 0<\theta < 1$

Is the distribution a member of the exponential family of probability distributions?

we know that the exponential family member distribution have the following form:

$$ f_x(x;\theta)=c(\theta)g(x)\exp \Big\{\sum_{j=1}^{l}Q_j(\theta)T_j(x) \Big\}$$

SOLUTION:

$c(\theta)=\frac{\Gamma(\theta)\sin(\pi\theta)\theta^{1-\theta}}{\pi}$

$h(x)=1$

$\nu(\theta)=-\theta$

$T(x)=x + \log(x)$

However i am not sure how these were obtained ? I am aware that $e^{-\theta x}x^{-\theta} = e^{-\theta(x+\log(x))}$. Then $T(x)$ contains all expression containing x and $c(\theta)$ everything containing $\theta$. What is the purpose of $h(x)$ and $\nu(\theta)?$ I am also interested in finding the sufficient statistic for parameter $\theta$.

Now the joint density of $X_1,...,X_n$ is

$$ f(x;\theta)= c(\theta)^n e^{-\theta \sum (x_1 + \log x_i)}$$

which comes from the fisher factorization.

Therefore $\sum (x_1 + \log x_i)$ is sufficient statistic for $\theta$.

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  • $\begingroup$ All the functions of $\theta$ and $x$ you defined are part of the general one parameter exponential family setup. We are trying to find whether our family is in the exponential family by matching the density structure. Naturally, we have to take all those functions into account. $\endgroup$ – StubbornAtom Aug 12 '18 at 17:36
  • $\begingroup$ Take a look at my edited answer to see if your queries are answered. $\endgroup$ – StubbornAtom Aug 13 '18 at 18:01
  • $\begingroup$ Where you have $f_x(x;\theta)$ you should have $f_X(x;\theta).$ The two symbols $X$ and $x$ refer to two different things; otherwise one would not be able to understand $\Pr(X\le x). \qquad$ $\endgroup$ – Michael Hardy Aug 13 '18 at 18:18
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Using the reflection formula $$\Gamma(\theta)\Gamma(1-\theta)=\frac{\pi}{\sin \theta\pi}\quad,0<\theta<1$$ your population density is simply

$$f_{\theta}(x)=\frac{e^{-\theta x}x^{-\theta}\theta^{1-\theta}}{\Gamma(1-\theta)}\mathbf 1_{x>0}\quad,0<\theta<1$$

(This 'simplification' is obviously not needed for the given problem)


Let $f_{\theta}(x)$, with $\theta\in\Theta$ and $x\in\mathfrak X$, be the pdf of a random variable $X$. Assume that

  • The parameter space $\Theta$ is an open interval.
  • The support $\mathfrak X=\{x:f_{\theta}(x)>0\}$ is independent of $\theta$.

We say that the joint density $f_{\theta}(\mathbf x)$ of $(X_1,\cdots,X_n)$ is a member of the one-parameter exponential family if it can be written as

\begin{align} f_{\theta}(\mathbf x)=\exp\left[A(\theta)T(\mathbf x)+B(\theta)+C(\mathbf x)\right] \end{align}

, where $A(\theta)$ and $B(\theta)$ are real valued functions of $\theta$ only, and $C(\mathbf x)$ and $T(\mathbf x)$ are real valued functions of $\mathbf x$ only.

All the functions mentioned above come from a simple analysis of the case of equality in the Cramer-Rao lower bound, from which the one-parameter exponential family can be derived.


Yes, the family of distributions $\{f_{\theta}:0<\theta<1\}$ is a member of the one-parameter exponential family (the first two assumptions are seen to be valid) since the joint density of the sample $\mathbf X=(X_1,X_2,\cdots,X_n)$ is of the form

\begin{align} f_{\theta}(\mathbf x)&=\prod_{i=1}^nf_{\theta}(x_i) \\&=\left(\frac{\theta^{1-\theta}}{\Gamma(1-\theta)}\right)^n\exp\left(-\theta\sum_{i=1}^nx_i\right)\left(\prod_{i=1}^nx_i\right)^{-\theta}\mathbf1_{x_1,\cdots,x_n>0} \\&=\exp\left[-\theta\sum_{i=1}^n(\ln x_i+x_i)+n\ln \left(\frac{\theta^{1-\theta}}{\Gamma(1-\theta)}\right)+\ln (\mathbf 1_{\min x_i>0})\right] \end{align}

Thus we have expressed the joint density in the general structure. That is, for some functions $A,B,C$ and $T$, we have expressed the joint density as

\begin{align} f_{\theta}(\mathbf x)&=\exp\left[A(\theta)T(\mathbf x)+B(\theta)+C(\mathbf x)\right] \\&=g(\theta, T(\mathbf x))h(\mathbf x) \end{align}

, where $g(\theta,T(\mathbf x))=\exp\left[A(\theta)T(\mathbf x)+B(\theta)\right]$ depends on $\theta$ and on $x_1,\cdots,x_n$ through $T$, and $h(\mathbf x)=\exp(C(\mathbf x))$ is independent of $\theta$.

So it is justified that a sufficient statistic for $\theta$ by Factorization theorem is

$$T(\mathbf X)=\sum_{i=1}^n(\ln X_i+X_i)$$

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    $\begingroup$ Actually, $$h(\mathbf x)=-\infty\cdot\mathbf 1_{\min x_i\leqslant0}$$ $\endgroup$ – Did Aug 12 '18 at 13:10
  • $\begingroup$ @Did Let me rectify then. $\endgroup$ – StubbornAtom Aug 12 '18 at 13:14

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