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 \begin{equation} f(x_i | \theta) = \frac{1}{i\theta}\exp[{-x_i/i\theta}], \end{equation}

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.


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.

  • $\begingroup$ 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). $\endgroup$ Apr 9 '21 at 15:09

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