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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?

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  • $\begingroup$ 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
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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 $1$

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