Let $X_1,..,X_n$ be an i.i.d. sample of geometric(p) random variables with unknown parameter $0<p<1$. I woluld like to find the Maximum-Likelihood estimate of p.

With the pmf $P(X=k)=p(1−p)^k $ for $k∈{1,2,3,…}$ and $0<p<1$ my result is the following: $\hat{p}=\left( \frac{1}{1+\frac{1}{n}\sum_{j=1}^n x_j} \right)$ if I assume that $\sum_{j=1}^n x_j\neq0$. Now, my question is: what happens if $\sum_{j=1}^n x_j=0$ and what is $\hat{p}$ in this case?

  • $\begingroup$ You will have $\hat{p}=1$ or as close to $p$ as your precision will allow $\endgroup$ – Henry May 1 at 13:03
  • $\begingroup$ Why is it a problem if the sum is zero? It seems to me that your calculation is problematic if the sum is $n$. $\endgroup$ – A. Pongrácz May 1 at 13:03
  • $\begingroup$ @Henry is it possible that $\hat{p}=1$ while $p\in (0,1)$? MLE is very new for me, sorry if my question is stupid. $\endgroup$ – tommy_m May 1 at 13:08
  • $\begingroup$ @A.Pongrácz Oh yes, thanks for your hint! $\endgroup$ – tommy_m May 1 at 13:10
  • $\begingroup$ If the sum is n, is $\hat{p}=0$? $\endgroup$ – tommy_m May 1 at 13:15

You have to choose between the two formulations of the geometric distribution

  • $0$ is a possible outcome with probability $p$. If so, then $k \in \{0,1,2,\ldots\}$ with $\mathbb P(X=k) = p(1-p)^k$. The sum of $n$ iid geometric random variables can be $0$ and the maximum likelihood estimator is $\hat{p}= \dfrac{1}{1+\frac{1}{n}\sum_i x_i}$

  • $0$ is not a possible outcome. If so, then $k \in \{1,2,3,\ldots\}$ with $\mathbb P(X=k) = p(1-p)^{k-1}$. The sum of $n$ iid geometric random variables must be at least $n$ and the maximum likelihood estimator is $\hat{p}= \dfrac{1}{\frac{1}{n}\sum_i x_i}$

In either case, if the obervation is the lowest possible ($0$ or $n$ respectively) then the maximum likelihood estimate is $\hat{p}=1$. I would say this even if you start with an open interval $(0,1)$; the alternative would be to say that $\hat{p}$ is arbitrarily close to $1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.