Consider a random sample of size $n$ from a Pareto distribution, $X_i \sim PAR(\theta, \kappa =2)$. I have to compute the MLE, $\hat \theta$, to three decimale places.
So I started doing the following:
$$pdf: f(x;\theta,\kappa = 2) = \frac{\kappa}{\theta(1+\frac{x}{\theta})^{\kappa+1}}=\frac{2}{\theta(1+\frac{x}{\theta})^{3}}, 0<x,\theta$$
and calculated the Likelihood function:
$$L(\theta)=\prod_{i=1}^n f(x_i;\theta,2)=(\frac{2}{\theta})^n\prod_{i=1}^n(1+\frac{x_i}{\theta})^{-3}$$
Then, for more convenient calculations later, I take the natural logarithm from both sides:
$$ln(L(\theta))=n(ln(2)-ln(\theta))-3\sum_{i=1}^nln(1+\frac{x_i}{\theta})$$
Would like to calculate the maximum with respect to $\theta$, so:
$$\frac{d}{d\theta}ln(L(\theta))=-\frac{n}{\theta}+3\sum_{i=1}^n\frac{\frac{x_i}{\theta^2}}{1+\frac{x_i}{\theta}}=0$$
And finally get the following expression:
$$\sum_{i=1}^n \frac{x_i}{\theta + x_i}=\frac{n}{3}$$
From here, I get stuck. How should I go futher. I hope that I didn't do any mistake by the calculations and that the way of solving this is okay.
Thanks!!