I am trying to solve the following exercise:
A state has several districts. Suppose that each district has the same proportion of persons in favor of a legislative project about gun control. En each of eight districts chosen at random, we count the number of persons that have to be asked until finding the first one in favor of the project (we define $X$ as this number). The results are : $3,8,9,6,4,5,3,2$ (e.g. in the first district the two first persons were against the project and the third one was in favor). Calculate the MLE of $P(X \geq 5)$ using the given data.
My attempt at a solution:
First, notice that $X$ is a geometric distribution with unknown parameter $\theta$. I've previously calculated the maximum likelihood estimator for a random sample $X_1,\ldots,X_n$ (in this case we have $n=8$), and I've arrived to $\theta_\text{MLE}=\dfrac{1}{\overline{X_n}}$.
Using the given data, we have $$\theta_\text{MLE}=\dfrac{8}{x_1+\cdots+x_8}$$$$=0.2$$
So, I think that the idea is to calculate $P(X \geq 5)$ using the estimated value $0.2$ as an approximation of $0.2$. If this is the case, we have $$P(X \geq 5)= 1-(P(X=1)+P(X=2)+P(X=3)+P(X=4))$$$$=1-(0.2+0.8\times0.2+(0.8)^2\times0.2+(0.8)^3\times0.2)$$
I would appreciate if someone could take a look at my solution and make any necessary corrections or post the correct answer in case mine is wrong. Thanks in advance.
Best Answer
What people call the "invariance" (which is really equivariance) of MLEs gives you the solution.
If $\widehat\theta$ is the MLE of $\theta$, the $g(\widehat\theta)$ is the MLE of $g(\theta)$. That's equivariance.
The random variable is the number of trials needed to get one success with probability $\theta$ of success on each trial. The probability $\Pr(X\ge5)$ is the probability of failure on all of the first four trials. That probability is $(1-\theta)^4$. If the MLE of $\theta$ is $0.2$ then the MLE of $(1-\theta)^4$ is $(1-0.2)^4.$
Your solution would be correct if you put parentheses in the right places, i.e. it's not $1-P(X=1)+P(X=2)+P(X=3)+P(X=4),$ but rather $1-\Big(P(X=1) + P(X=2) + P(X=3) + P(X=4)\Big).$
However, there is a simpler way to express it, namely (as shown above) as $(1-0.2)^4.$