# MLE of the Geometric Distribution

Suppose that $$X_{1},X_{2},...,X_{n}$$ are independently and identically distributed as $$Ge(\theta)$$.

(i) Find the maximum likelihood estimator of $$\theta$$

My solution:

$$\theta = \frac{n}{\sum_{i=1}^{n}x_{i}}$$

Therefore, $$E(\hat\theta) = \frac{1}{\theta}$$

(ii) Hence show that the maximum likelihood estimator of $$\psi = \frac{(1-\theta)}{\theta}$$ is the sample mean $$(\bar X)$$.

Try as I might, I can't re-arrange the answer to question 1 into the form shown in question 2. Please may someone help me?

Regrettably, there are two distributions that are called geometric [1], the classical one, taking values in $$1,2,\ldots$$ and the shifted variant that takes values in $$0,1,2,\ldots$$. The problem you posed was probably referring to the latter, where the likelihood of the sample $$X_1,\ldots,X_n$$ is $$L(\theta)=\prod_{j=1}^n \theta (1-\theta)^{X_j} \,.$$ By differentiating $$\log L$$, we see that $$L(\theta)$$ is maximized at a parameter $$\hat{\theta}$$ that satisfies $$\sum_{j=1}^n \Bigl(\frac{1}{\hat{\theta}}-\frac{X_j}{1-\hat{\theta}}\Bigr)=0 \,.$$ Dividing by $$n$$, we infer that the MLE $$\; \hat{\theta}$$ satisfies $$\frac{1}{\hat{\theta}}=\frac{\bar{X}}{1-\hat{\theta}} \,,$$ so $$\hat{\theta} =\frac{1}{1+\bar{X}} \quad \text{and} \quad \frac{1-\hat{\theta}}{\hat{\theta}}=\bar{X} \,.$$