# Complete sufficient statistic of non-identical distribution: $X_i \sim EXP(i\theta)$

## Problem

Suppose that $$X_1, \dots, X_n$$ are independent $$\mathrm{EXP}(i\theta)$$ random variables. Find a complete sufficient statistic for $$\theta$$.

## My Attempt

Since pdf of $$x_i$$ is $$$$f(x_i | \theta) = \frac{1}{i\theta}\exp[{-x_i/i\theta}],$$$$

the joint density is

\begin{align} f(\mathbf{x} | \theta) &= \prod_{i=1}^{n} \frac{1}{i\theta}\exp\left[{-x_i/i\theta}\right] \\ &= \frac{1}{n!\theta^n} \exp\left[\frac{1}{\theta}{\sum_{i=1}^{n}\frac{x_i}{i}}\right] \\ &= g(t | \theta)h(\mathbf{x}), \end{align}

where $$g(t|\theta) = \frac{1}{n!\theta^n} \exp[t/\theta]$$, $$h(\mathbf{x}) = 1$$, and $$t=\sum_{i=1}^{n} x_i / i$$. By factorization theorem, $$T=\sum_{i=1}^{n} X_i / i$$ is a minimum sufficient statistic.

## Question

How can I show that $$T$$ is a complete statistic? I think I'll have to use the definition of complete statistic because random variables are not identically distributed.

• Random variables are not identically distributed, but I think 'exponential family factorization' is still applicable which guarantees that $T$ is minimal complete sufficient. To use the definition of completeness, you can derive the distribution of $T$ (which is straightforward since $X_i/i$ are i.i.d Exponential). Apr 9 '21 at 15:09