one dimensional sufficient statistic

Question

$X_1,X_2,...,X_n$ are iid random variables having pdf $$f_X(x|\theta) = (\theta + 1)\theta x(1-x)^{\theta-1}I_{(0,1)}(x)$$

Give a one-dimensional sufficient statistic.

I started to solve this by factoring $\theta$ away from x as much as possible and ended up with

$$f(x|\theta) = x/(1-x) I_{(0,1)}(x) \theta(\theta + 1)(1-x)^{\theta}$$

Is this the right way to go about this?

Answer 1

By the factorization criterion $$ \prod_{i=1}^n f_{X_i}(x; \theta) = (\theta + 1)^n\theta^n \prod_{i=1}^n ( 1 - x_i )^{\theta} \times \prod_{i=1}^n \frac{x_i}{(1-x_i)} $$ hence $h(X) = \prod_{i=1}^n \frac{x_i}{(1-x_i)}$ and $g(\theta; X) = \prod_{i=1}^n ( 1 - x_i )^{\theta}$, so the MSS is $f(X) = \prod_{i=1}^n ( 1 - x_i )$.

Answer 2

Your first issue is that you need to perform the factorization on the joint density of the sample. What you did instead was look at the density of a single observation.

So, your joint density of the sample $\boldsymbol x = (x_1, \ldots, x_n)$ comprising independent and identically distributed observations is simply the product of the densities of each observation: $$f(\boldsymbol x \mid \theta) = \prod_{i=1}^n f(x_i \mid \theta) = \prod_{i=1}^n (\theta+1)\theta x_i (1 - x_i)^{\theta - 1} \mathbb 1 (0 < x_i < 1).$$ Here I have used a slightly different notation for the indicator function, but its meaning should be clear. Now, our goal is to write this in the form $$f(\boldsymbol x \mid \theta) = h(\boldsymbol x) g(\boldsymbol T(\boldsymbol x) \mid \theta),$$ where $\boldsymbol T : \mathbb R^n \to \mathbb R^m$ is some function of the sample that satisfies the following:

1. $m$ is a positive integer satisfying $m \le n$ and represents the reduction in dimension of the sufficient statistic. In this case we are told to find $\boldsymbol T$ such that $m = 1$; i.e., $\boldsymbol T = T$, a function that maps a vector of dimension $n$ to a scalar.
2. $T$ does not depend on the parameter $\theta$, only on $\boldsymbol x$ and $n$.

Further, we require that $g$ depend on $\boldsymbol x$ only through $T$; that is to say, if we were given only $T(\boldsymbol x)$ instead of $\boldsymbol x$ itself, we could still compute $g$.

So, with this in mind, we write $$\begin{align*} f(\boldsymbol x \mid \theta) &= (\theta+1)^n \theta^n \prod_{i=1}^n x_i \biggl(\prod_{i=1}^n (1 - x_i)\biggr)^{\theta-1} \prod_{i=1}^n \mathbb 1 (0 < x_i < 1) \\ &= \underbrace{\biggl( \prod_{i=1}^n x_i \biggr) \mathbb 1 (0 < x_{(1)} \le x_{(n)} < 1)}_{h(\boldsymbol x)} \underbrace{(\theta+1)^n \theta^n \biggl(\underbrace{\prod_{i=1}^n (1 - x_i)}_{T(\boldsymbol x)}\biggr)^{\theta-1}}_{g(T(\boldsymbol x) \mid \theta)}. \end{align*}$$ A few important observations: first, $h$ contains the indicator function as well as the product $\prod_{i=1}^n x_i$. You cannot ignore the indicator, because the joint density has finite support on $(0,1)^n$.

Second, the choice of $T$ is not unique: you could also have chosen, for example, $$T(\boldsymbol x) = \frac{1}{n} \log \prod_{i=1}^n (1 - x_i) = \frac{1}{n} \sum_{i=1}^n \log (1 - x_i),$$ which is the mean of a suitably log-transformed sample. This is because the logarithm is one-to-one on the support of $X$.

Third, $h$ and $g$ could also have been chosen differently--these are not necessarily unique. What matters is that we must make the choice such that $g$ can be computed if all we know is $\theta$ and $T(\boldsymbol x)$.