I have a sample of geometric random variables $X_1,… ,X_n$ and I am interested in finding the maximum likelihood estimator for such a sample.
Hence we have the corresponding PMF $$f_X(x|\theta)=\frac{1}{\theta}(1- \frac{1}{\theta})^{x-1}$$ whereby $x\geq1$ and $\theta \geq 1$.
Im looking to find the maximum likelihood estimator so first I calculate the likelihood function and the log likelihood function.
$$\begin{aligned}\mathcal{L}_X(\theta) &= \prod^n_{i=1}\left[\frac{1}{\theta}\left(1- \frac{1}{\theta} \right)^{x_i-1}\right] \\ &= \theta^{-n} \left(1- \frac{1}{\theta} \right)^{n\bar{x}-n} \end{aligned}$$
And the log likelihood as
$$\begin{aligned}\ell_X(\theta) &= -n\log(\theta)+(n\bar{x}-n) \log\left(\frac{\theta-1}{\theta} \right) \\ &= -n\log(\theta)+(n\bar{x}-n)\log(\theta-1)-(n\bar{x}-n)\log(\theta) \end{aligned}$$
Then I differentiate the log likelihood function and equate it to $0$ to get critical points of the funcion
$$\ell'_X(\theta)=\frac{n\bar{x}-n}{\theta-1}-\frac{n\bar{x}}{\theta}=0$$ from here I arrive at $$\begin{aligned} \frac{n\bar{x}-n}{\theta-1} &= \frac{n\bar{x}}{\theta} \\ 1-\frac{1}{\bar{x}} &= 1-\frac{1}{\theta} \\ \theta &= \bar{x}\end{aligned}$$
Then I take the Second derivative of the log likelihood function and plug my critical value to see if its negative. $$\ell''_X(\theta)=-\frac{n\bar{x}-n}{(\theta-1)^2}+\frac{n\bar{x}}{\theta^2}$$
From here Im stuck as I cant tell by plugging this value whether the function will always be negative hence indicating a maximum. $$\ell''_X(\bar{x})=-\frac{\bar{x}-n}{(\bar{x}-1)^2}+\frac{n}{\bar{x}}$$
Maybe this is a sign that I'm way off here. Any pointers as to if and where I'm going wrong? Thanks!
$$\ell''_X(\theta) = -\frac{n \bar x - n}{(\theta - 1)^2} + \frac{n \bar x}{\theta^2} = n \left( \frac{-\theta^2(\bar x - 1) + \bar x(\theta - 1)^2}{(\theta - 1)^2 \theta^2} \right) = \frac{n(\theta^2 -2 \theta \bar x + \bar x)}{(\theta-1)^2 \theta^2}.$$ Note the denominator is never negative, and $n \ge 1$. So when $\theta = \bar x$, the remaining factor in the numerator is $$\bar x^2 - 2\bar x^2 + \bar x = \bar x - \bar x^2 = \bar x (1-\bar x).$$ Since $\bar x \ge 1$ because $X \ge 1$, it follows that $\bar x (1 - \bar x) \le 0$, with equality attained when $\bar x = 1$.