# Geometric Waiting Time MLE

If the time is measured in discrete periods, a model that is often used for the time $$X$$ failure of an item is:$$P_{\theta}[X=k]=\theta^{k-1}(1-\theta), k=1,2,...$$ where $$0<\theta<1$$. Suppose that we only record the time of failure, if failure occurs on or before time $$r$$ and otherwise just note that the item has lived at least $$r+1$$ periods. Hence, we observe $$Y_1,...,Y_n$$, which are iid and have common frequency function: $$f(k,\theta)=\theta^{k-1}(1-\theta), k=1,...,r$$ $$f(r+1,\theta)=1-P_\theta[X\leq r]=1-\Sigma_{k=1}^r(1-\theta)=\theta^r$$ The $$r+1$$ notation, means survival for at least $$r+1$$ periods. Let $$M=$$number of indices i such that $$Y_i=r+1$$. I want to derive that the MLE of $$\theta$$ based on $$Y_1,...,Y_n$$ is: $$\hat{\theta}(\mathbf{Y})=\frac{\Sigma_{i=1}^nY_i-n}{\Sigma_{i=1}^nY_i-M}$$ but I'm pretty confused about it.

Note: This process is known as Censored Geometric Waiting Time.

Any help would be appreciated!

The likelihood is defined to be the probability to observed your data under a given parameters : $$P(Y_1,...,Y_n|\theta)$$

Let's consider that all event are iid:

$$P(Y_1,...,Y_n|\theta) = \prod_1^n{P(Y_i|\theta})$$

Now let's split the case where $$Y_i = r+1$$ from the rest:

$$P(Y_1,...,Y_n|\theta) = \prod_{Y_i

Now we know the expression of the probability for both of these product:

$$P(Y_1,...,Y_n|\theta) = \prod_{Y_i=Y_i|\theta)}$$

$$P(Y_1,...,Y_n|\theta) = \prod_{Y_i

$$P(Y_1,...,Y_n|\theta) = \prod_{Y_i

$$P(Y_1,...,Y_n|\theta) = \prod_{Y_i

From Now on it is easier to take the log : $$L(\theta) = log(P(Y_1,...,Y_n|\theta))$$

$$L(\theta) = \sum_{Y_i

$$L(\theta) = log(\theta) * [\sum_{Y_i

$$L(\theta) = log(\theta) * [\sum_{Y_i

Now we take the derivative with respect to $$\theta$$:

$$L'(\theta) = \frac{\sum_{Y_i

We then look for $$\theta$$ such that $$L'(\theta)=0$$

$$\frac{\sum_{Y_i

$$\frac{\sum_{Y_i

$$\frac{1-\theta}{\theta} = \frac{n-M}{M*r+\sum_{Y_i

$$\frac{1-\theta}{\theta} = \frac{n-M}{M*(r+1)+\sum_{Y_i

$$\frac{1-\theta}{\theta} = \frac{n-M}{\sum_{Y_i=r+1}Y_i+\sum_{Y_i

$$\frac{1-\theta}{\theta} = \frac{n-M}{\sum_{1}^nY_i - n}$$

$$\frac{1-\theta}{\theta} = \frac{n-M}{\sum_{1}^nY_i - n}$$

$$\frac{1}{\theta} = \frac{n-M}{\sum_{1}^nY_i - n} + 1$$

$$\frac{1}{\theta} = \frac{n-M+\sum_{1}^nY_i - n}{\sum_{1}^nY_i - n}$$

$$\frac{1}{\theta} = \frac{\sum_{1}^nY_i - M}{\sum_{1}^nY_i - n}$$

$$\theta = \frac{\sum_{1}^nY_i - n}{\sum_{1}^nY_i - M}$$

Done !