Geometric Distribution MLE - Second Order Condition I am trying to prove that $\dfrac {1}{\bar{x}}$ (as per the result in this youtube clip https://www.youtube.com/watch?v=0TSMugiWPc0) is certainly the MLE for geometric distribution.
When I try to confirm by then taking the second order condition. My calculation is that it is not less that zero! I am unsure how to write formulas on here to show what I have done, this is why I posted the clip, as it will show the first derivative.    
 A: Given data $X_1,...,X_n \sim \text{IID Geom}(\theta)$ (using the version of the geometric distribution where the value is the count of trials including the first success, so that $X_i \geqslant 1$) you have log-likelihood function:
$$\begin{equation} \begin{aligned}
\ell_\boldsymbol{x}(\theta) 
&= \sum_{i=1}^n \Big[ \ln (\theta) + (x_i-1) \ln (1-\theta) \Big] \\
&= n \ln \theta + \ln (1-\theta) \sum_{i=1}^n (x_i-1) \\[6pt]
&= n \ln \theta + n (\bar{x}-1) \ln (1-\theta) \quad \quad \text{for all } 0 \leqslant \theta \leqslant 1. \\[6pt]
\end{aligned} \end{equation}$$
For all $0 <\theta <1$ the first and second derivatives are:
$$\begin{equation} \begin{aligned}
\frac{d \ell_\boldsymbol{x}}{d\theta}(\theta) &= \frac{n}{\theta} - \frac{n(\bar{x}-1)}{1-\theta} = \frac{n(1-\bar{x}\theta)}{\theta (1-\theta)}, \\[10pt]
\frac{d^2 \ell_\boldsymbol{x}}{d\theta^2}(\theta) &= - \frac{n}{\theta^2} - \frac{n(\bar{x}-1)}{(1-\theta)^2} = - \frac{n(1-2\theta + \bar{x}\theta^2)}{\theta^2 (1-\theta)^2} <0.
\end{aligned} \end{equation}$$
We can see from the second derivative that the log-likelihood function is strictly concave.  Hence, the maximising value occurs at the unique critical point $\hat{\theta} = 1/ \bar{x}$.  Substitution of this value into the second derivative function yields the curvature:
$$\frac{d^2 \ell_\boldsymbol{x}}{d\theta^2}(\hat{\theta}) = - \frac{n \bar{x}^3}{ \bar{x}-1 } <0.$$
