The times between arrivals at a fish shop queue can be described by an exponential distribution $X$ with parameter $\theta$ .
If $Y$ is the random variable given by the sample mean on $n$ inter-arrival times
Write down an expression for an unbiased estimator $T$ of $\theta$.
The answer is:
$T=\frac{1}{Y}(1-\frac{1}{n})$
However when I tested if it is unbiased I found $E(T)$ is $(n-1)θ$ instead of $θ$
What I have done is: Finding $E(\frac{1}{Y}) = nθ$, to make it $= θ$, I divide it by n, So I can set $T = \frac{1}{nY}$
In my view, from the given answer, $E(\frac{1}{Y})$ is supposed to be $[\dfrac{n}{n-1}]θ$, in order to make E(T) = θ, but I can't get it...
I don't know where the $(n-1)$ comes from.
Here is my working:
Since Y is the sample mean of n arrivals
$$E(Y) = n\sum_{i=1}^{n} Y_i = \frac{1}{n\theta}$$ therefore $$E(\frac{1}{Y})= n\theta$$
Naively, you would think that the rate parameter would be estimated by the reciprocal of the average time to wait. So you start with the estimator $\frac{1}{Y}$. Then you calculate its expected value. You will find that it is not $\theta$, but is proportional to $\theta$. Then divide the proportionality constant out to make an unbiased estimator.
If you need more help, you might look into the derivation of the usual unbiased estimator for the variance, namely $S^2 = \frac{1}{n-1} \sum_{i=1}^n \left ( x_i - \frac{1}{n} \sum_{j=1}^n x_j \right )^2$. Basically this goes by considering $S_1^2$ which has an $n$ instead of $n-1$, calculating $E[S_1^2]=\frac{n-1}{n} \sigma^2$, and then correcting it.