I am generally having trouble understanding how to use product notation when calculating Maximum Likelihood Estimators. The example bellow is from a random sample $X_1,...,X_n$.
Find the MLE of $\theta$.
$f(x, x_0, \theta)=\theta x_{0}^{\theta} x^{-(1+\theta)}, \ for \ x>x_0, \ \theta > 1, \ \& \ x_0$ is known.
step one
$ L(x, \theta) = \prod_{i=1}^{n} \theta x_{0}^{\theta} x^{-(1+\theta)}$
step two
$ L(x, \theta) = \theta^{n} x_{0}^{n \theta} \prod_{i=1}^{n} x_{i}^{-(1+\theta)}$
step three
$ \ln [L(x, \theta)] = n \ln({\theta}) + n \theta \ln({x_{0}}) - (1+\theta)\sum_{i=1}^{n} \ln({x_i})$
step four
$ \dfrac{\partial}{\partial \theta} \ln [L(x, \theta)] = \dfrac{n}{\theta}+ n \ln{x_0} -\sum^{n}_{i=1} \ln{x_i}$
step five (set to 0, solve for $\theta$)
$\hat{\theta}= \dfrac {n}{\sum \ln {x_i}- n \ln{x_0}}$
The answer given is $\hat{\theta}=\dfrac{n}{\sum\ln(\frac{x_{i}}{x_0})}$
On step two, I am not sure if I correctly distributed the product, specifically $\prod_{i=1}^{n} x_{i}^{-(1+\theta)}$.
On step three, is $-(1+\theta)\sum_{i=1}^{n} \ln({x_i})$ the correct way to take the natural log of the product?
Finally, is my step five answer the same as the given answer given a little bit of algebra?
Thanks so much for your time and help!
Everything looks right except that at step three it should be $1+\theta$ instead of $1-\theta$.