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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Commented Dec 21, 2021 at 9:09
  • $\begingroup$ en.wikipedia.org/wiki/…, along with a little algebra, gives us an unbiased estimator $(n-1)/[(n-2)\bar{x}]$. $\endgroup$
    – jbowman
    Commented Dec 30, 2023 at 20:53

1 Answer 1


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 $$ \DeclareMathOperator{\E}{\mathbb{E}} \E\frac{1}{\bar{x}} = \frac{(\lambda/n)^{2n-1}}{n-1} $$
but there is no obvious way that helps to get an unbiased estimator!


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