Warning
The question is the second part of this question. The third and last part is found here.
Exercise
Let $X \thicksim Pa(\lambda,\theta)$ with density function: $ f(x; \theta, \lambda) = \frac{\lambda \theta^{\lambda}}{x^{\lambda+1}} $ where $x \geq \theta$ and $\theta > 0$.
From the first question I have found the CDF: $ F(x) = 1 - \left( \frac{\theta}{x} \right)^{\lambda} $
Let $\lambda $ known. Find the estimator of $\theta$ with MLE and establish if is unbiased.
Try
I find the log-likelihood function and I calculate the first two derivatives: $$ L(\theta) \propto \frac{\lambda^n\theta^{n\lambda}}{\prod_{i=0}^{n} x^{\lambda+1}_i} = \lambda^n\theta^{n\lambda} \left( \prod_{i=0}^{n} x_i \right)^{-(\lambda+1)} $$
$$ l(\theta) = \log L(\theta) = n\log(\lambda) + n\lambda \log(\theta) - (\lambda+1)\sum^{n}_{i=0} \log(x_i) $$
Since $\lambda$ is known, then $$ L(\theta) \propto \theta^{\lambda} 1(x_{(1)} \geq \theta) $$
where $1(\cdot)$ is the indicator function. For $\lambda > 0$, the likelihood function is monotone increasing on the interval $\theta \in (0, x_{(1)}]$. Hence, the MLE of $\theta$ is just the minimum $x_{(1)}$.
I verify if is unbiased:
$$ E(\hat{\theta}) = \int^{+\infty}_{\theta} \hat{\theta} f(\theta) dx = \int^{+\infty}_{\theta} x \frac{n\lambda \theta^{n\lambda}}{x^{n\lambda+1}} dx = n\lambda \theta^{n\lambda} \int^{+\infty}_{\theta} x^{-n\lambda} dx $$
$$ = n\lambda \theta^{n\lambda} \left[ \frac{x^{-(n\lambda-1)} }{-n\lambda+1} \right]^{+\infty}_{\theta} = n\lambda \theta^{n\lambda} \left(0 - \frac{\theta^{-(n\lambda-1)} }{-n\lambda+1} \right) = n\lambda \theta^{n\lambda} \frac{\theta^{-n\lambda}\theta}{n\lambda-1} $$
Finally: $$ E(\hat{\theta}) = \theta \left( \frac{n\lambda }{n\lambda-1} \right) $$
It's biased due to the constant $n\lambda/(n\lambda-1)$, but is necessarily itself because $P(X_{(1)} < \theta) = 0$ that's the sample minimum can never be less than $\theta$.
For the calculation of the MLE and the verify of unbiasedness, I have followed this question.
If I have missed anything else, please report it to me immediately.
