I've been given the following as an estimate for the parameter of a Poisson distribution, and must explain why it is a sensible estimate.
$$\log\left(\frac1n \sum_{i=1}^n 2^{x_i} \right) $$
I'm pretty confused because I've covered method of moments estimators and the MLE, but this looks like neither of those. Any advice as to what I'm missing would be much appreciated.
Note: The log is the natural logarithm - they like to use that instead of ln for clarity and say that ln can easily confuse some expressions.
First off, I don't believe this estimator is "good". The sample mean is the best unbiased estimator for $\lambda$. But I can offer you some intuition behind why this estimator is "reasonable".
Let $Y_i = 2^{X_i}$. Then the estimator you propose is basically: $ln(\bar{Y})$.
Now we consider $\mathbb{E}[\bar{Y}]$.
$\mathbb{E}[\bar{Y}] = \mathbb{E}[Y_i] = \mathbb{E}[2^X] = \sum_{x=0}^\infty \frac{2^xe^{-\lambda} \lambda^x}{x!} = \frac{e^{-\lambda}}{e^{-2\lambda}}\sum_{x=0}^\infty \frac{(2\lambda)^xe^{-2\lambda}}{x!} = e^{-\lambda + 2\lambda} = e^\lambda$
If the expected value of $\bar{Y} = \frac{1}{n}\sum_{i=1}^n2^{x_i}$ is $e^\lambda$, then it makes sense to take the natural log of $\bar{Y}$.
Here is an example setting $\lambda = 10$. Sampling distributions are plotted for the sample mean (black), the estimator described above (blue) and the estimator if you use $log_2$ instead of $ln$ (green).