# How to find an unbiased estimator for reciprocal of scale parameter given an iid exponential sample?

For a random sample $$X_1, ..., X_n$$ from an exponential distribution with scale parameter $$\lambda$$, the density is given by $$f(x) = \frac{1}{\lambda}e^{-\frac{1}{\lambda}x}; \,x \geq 0,\, \lambda > 0$$.

How might you find an unbiased estimator $$W(\mathbf{x})$$ of $$\frac{1}{\lambda}$$?

Stop me if I'm wrong: The MLE for $$\lambda$$ turns out to be $$\bar{X}$$, so the MLE of $$\frac{1}{\lambda}$$ is $$\frac{1}{\bar{X}}$$. My approach is to calculate the expectation $$\text{E}\frac{1}{\bar{X}}$$, hope it differs by a constant $$C$$, and take the solution as $$\frac{1}{C\bar{X}}$$.

But taking the expectation of $$\frac{1}{\bar{X}}$$ requires knowing the distribution of $$\bar{X}$$, which has led me to the Erlang distribution, and after some work end up with E$$\frac{1}{\bar{X}} = \frac{n}{(n-1)\lambda}$$. So it would seem that $$C = \frac{n}{n-1}$$, and $$W(\mathbf{x}) = \frac{n-1}{n\bar{X}}$$.

I don't know whether this is correct, but in any case, is there a better way? This was rather laborious and required knowledge or derivation of the Erlang.

• There is no generic rule for finding an unbiased estimator. Your steps are working for a scale family such as the exponential distribution, in the sense that the expectation of $\bar X^\alpha$ is proportional to $\lambda^\alpha$ but it would not work for a location family for instance. Commented Dec 21, 2021 at 9:09
• en.wikipedia.org/wiki/…, along with a little algebra, gives us an unbiased estimator $(n-1)/[(n-2)\bar{x}]$. Commented Dec 30, 2023 at 20:53

So far this is more of a long comment. You are right that $$\bar{x}$$ have an Erlang distribution, that is, a Gamma distribution. (Refer to Wikipedia). I find the following density for $$\bar{x}$$ $$f(\bar{x}) = \frac{(n/\lambda)^n }{\Gamma(n)} \bar{x}^ {n-1} e^{-\frac{n}{\lambda} \bar{x}}$$ for $$\bar{x} >0$$. Then I find by integration the expectation $$\DeclareMathOperator{\E}{\mathbb{E}} \E\frac{1}{\bar{x}} = \frac{(\lambda/n)^{2n-1}}{n-1}$$