Showing that $f_\varphi(x)$ is a member of the one-parameter exponential family and $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$

Question

Let $X_1, \dots, X_n$ denote a random sample from the PDF

$$f_{\varphi}(x)= \begin{cases} \varphi x^{\varphi - 1} &\text{if}\, 0 < x < 1, \varphi > 0\\ 0 &\text{otherwise} \end{cases}$$

I want to show that $f_\varphi(x)$ is a member of the one-parameter exponential family. Furthermore, I want to show that $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$.

Chapter 9.13.3 Exponential Families of the textbook All of Statistics: A Concise Course in Statistical Inference by Larry Wasserman says the following:

Most of the parametric models we have studied so far are special cases of a general class of models called exponential families. We say that $\{ f(x; \theta) : \theta \in \Theta \}$ is a one-parameter exponential family if there are functions $\eta(\theta)$, $B(\theta)$, $T(x)$ and $h(x)$ such that $$f(x; \theta) = h(x) e^{\eta(\theta) T(x) - B(\theta)}.$$ It is easy to see that $T(x)$ is sufficient. We call $T$ the natural sufficient statistic.

I calculate the likelihood to be

$$\begin{align} L(\varphi; \mathbf{x}) &= \prod_{i = 1}^n \varphi x_i^{\varphi - 1} \mathbb{1}_{0 < x < 1, \varphi > 0} \\ &= \varphi^n x^{\sum_{i = 1}^n (\varphi - 1)} \prod_{i = 1}^n \mathbb{1}_{0 < x_i < 1, \varphi > 0} \\ &= \varphi^n x^{\sum_{i = 1}^n (\varphi - 1)} \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \end{align}$$

To get this in the appropriate form, I tried

$$\begin{align} \log\left[\varphi^n x^{\sum_{i = 1}^n (\varphi - 1)} \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] &= n\log(\varphi) + \sum_{i = 1}^n (\varphi - 1) \log(x_i) + \log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] \\&= \exp{ \left\{ n\log(\varphi) + (\varphi - 1) \sum_{i = 1}^n \log(x_i) + \log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] \right\} } \\ &= \exp{ \left\{ n\log(\varphi) + (\varphi - 1) \sum_{i = 1}^n \log(x_i) \right\} } \exp{\left\{ \log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] \right\}} \\ &= \exp{ \left\{ n\log(\varphi) + (\varphi - 1) \sum_{i = 1}^n \log(x_i) \right\} } \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \end{align}$$

And we then select

$$\begin{align} &\eta(\varphi) = ? \\ &T(\mathbf{x}) = ? \\ &B(\varphi) = ? \\ &h(\mathbf{x}) = ? \end{align}$$

I'm unsure how much of this is correct. The first thing that might be problematic is $\log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right]$. Specifically, I'm not sure that it's valid to log this identity matrix, but I'm not sure what else to do. As you can see, I later use an exponential to cancel out the log, but, as I said, I'm not sure that it was mathematically valid to apply log to it in the first place. The second is that I needed to show that $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$, but I'm not sure how to further factorize the final expression in order to get this.

Another issue is that I cannot find any definitions or theorems for this in this chapter. There's usually a theorem that formally and explicitly states what you need to do/show in order show something (in this case, showing that $f_\varphi(x)$ is a member of the one-parameter exponential family and that $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$). The excerpt that I posted above was simply part of the normal discussion of the chapter at the very beginning. Based on that, I presumed that, for showing that $f_\varphi(x)$ is a member of the one-parameter family, I just have to show that the likelihood can take the form $f(x; \theta) = h(x) e^{\eta(\theta) T(x) - B(\theta)}$. However, for the sufficient statistic part, it just says that "it is easy to see that $T(x)$ is sufficient", so I'm not exactly sure what I'm supposed to do with this (do I need to use the Fisher-Neyman factorization theorem, or does the exponential family form imply that $T(\mathbf{x})$ is sufficient, and so we don't need to do anything else?).

So, overall, what is the correct way to do this to show that $f_\varphi(x)$ is a member of the one-parameter exponential family and $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$?

Answer 1

Here is my attempt.

1.Show that $f(x; \varphi)$ is in the one-parameter exponential family.

Assume that $\varphi > 0$.

Rewriting $f(x; \varphi)$, we have that

\begin{align*} f(x; \varphi) &= \exp \left\{\log \left[ \varphi x^{\varphi - 1} \cdot \mathbb{I}(0 < x < 1) \right] \right\} \\ &= \exp \left\{ \log \varphi + (\varphi - 1) \log x + \log \mathbb{I}(0< x < 1)\right\} \\ &= \exp \left \{ \log \varphi - (1 - \varphi) \log x \right\} \mathbb{I}(0 < x < 1) \\ \end{align*}

Defining the quantities

\begin{align*} h(x) &= \mathbb{I}(0 < x < 1) \\ T(x) &= -\log x \\ \eta(\varphi) &= (1 - \varphi) \\ B(\varphi) &= - \log \varphi \\ \end{align*}

shows that $f(x; \varphi)$ is of the form $h(x) \exp(\eta(\varphi) T(x) - B(\varphi))$ and hence belongs to the one-parameter exponential family.

2. Show that $T = - \sum^n_{i=1} \log X_i$ is sufficient for $\varphi$.

Assuming the $X_1, \dots X_n$ are i.i.d., and using the above, the joint density $f(x_1, \dots, x_n; \varphi)$ is

\begin{align*} f(x_1, \dots, x_n; \varphi) &= \prod^n_{i=1} \exp \left \{ \log \varphi - (1 - \varphi) \log x_i \right\} \mathbb{I}(0 < x_i < 1) \\ &= \exp \left \{n \log \varphi + (1 - \varphi) \sum^n_{i=1} - \log x_i \right\} \prod^n_{i=1} \mathbb{I}(0 < x_i < 1) \\ \end{align*}

Define

\begin{align*} h(x_1, \dots, x_n) &= \prod^n_{i=1} \mathbb{I}(0 < x_i < 1) \\ g(T(x_1, \dots, x_n), \varphi) &= \exp \left \{n \log \varphi + (1 - \varphi) \sum^n_{i=1} - \log x_i \right\} \\ \end{align*}

Noting that $h$ does not depend on the parameter $\varphi$ and that $g$ depends on the data $X_1, \dots, X_n$ only through the statistic $T(X_1, \cdots, X_n) = \sum^n_{i=1} - \log X_i$, the factorisation theorem implies that $T$ is sufficient for $\varphi$.

---

Further comment.

In response to

It is easy to see that $T(X)$ is sufficient.

in the extract of Chapter 9.13.3 Exponential Families in All of Statistics: A Concise Course in Statistical Inference you have quoted, use the factorisation theorem to see this.

Answer 2

The trick here is that you can combine the basic properties of the logarith and exponentiation

$$a =\exp\left[\log(a)\right]$$

and

$$\log(a^b) = b \log(a)$$

into

$$a^b =\exp\left[\log(a^b)\right]= \exp\left[b \log(a)\right] $$

and that rule you can use to rewrite your distribution into the typical form of an exponential family.

$$\varphi x^{\varphi -1} = \varphi \exp\left[(\varphi-1) \cdot \log(x)\right] = \exp [\overbrace{(\varphi-1)}^{\eta(\varphi)} \cdot \overbrace{\log(x)}^{T(x)} + \overbrace{ \log( \varphi)}^{B(\varphi)}]$$