# Bias correction for MLE of mean of geometric random variable

Parameter estimation [ edit] For both variants of the geometric distribution, the parameter $$p$$ can be estimated by equating the expected value with the sample mean. This is the method of moments, which in this case happens to yield maximum likelihood estimates of $$p .{ }^{[7][8]}$$ Specifically, for the first variant let $$k=k_{1}, \ldots, k_{n}$$ be a sample where $$k_{i} \geq 1$$ for $$i=1, \ldots, n$$. Then $$p$$ can be estimated as $$\hat{p}=\left(\frac{1}{n} \sum_{i=1}^{n} k_{i}\right)^{-1}=\frac{n}{\sum_{i=1}^{n} k_{i}} .$$

For either estimate of $$\hat{p}$$ using Maximum Likelihood, the bias is equal to $$b \equiv \mathrm{E}\left[\left(\hat{p}_{\mathrm{mle}}-p\right)\right]=\frac{p(1-p)}{n}$$ which yields the bias-corrected maximum likelihood estimator $$\hat{p}_{\mathrm{mle}}^{*}=\hat{p}_{\mathrm{mle}}-\hat{b}$$

I think their bias correction is wrong.

• Can you elaborate? Why do you think the bias correction is wrong?
– Sycorax
Dec 10, 2021 at 20:48

$$n/\sum k_i=g(\sum k_i)\approx g(n/p) + \Big(\sum k_i-n/p\Big)\frac{dg(\sum k_i)}{d\sum k_i}\Bigg|_{\sum k_i=n/p}$$ $$\hspace{50mm} + \frac{1}{2!}\Big(\sum k_i-n/p\Big)^2\frac{d^2g(\sum k_i)}{d(\sum k_i)^2}\Bigg|_{\sum k_i=n/p}$$
$$E\Big[n/\sum k_i\Big]\approx p + \text{Var}\Big[\sum k_i\Big]\frac{n}{(n/p)^3}$$
$$\hspace{25mm}=p + \frac{n(1-p)}{p^2}\frac{n}{(n/p)^3}$$
$$\hspace{25mm}=p + \frac{p(1-p)}{n}$$
$$E\Big[n/\sum k_i-p\Big]\approx\frac{p(1-p)}{n}$$