Warning
The question is the sequel to this question and it was divided into three parts.
The first is this, while the second 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 $x \geq \theta$, $\lambda > 0$ and $\theta > 0$.
The CDF is: $ F(x) = 1 - \left( \frac{\theta}{x} \right)^{\lambda} $
Let $\lambda$ known. Show if the statistic $S = \min\{X_1, ..., X_n\}$ is sufficient for $\theta$ and valuate if it's minimal.
Try
First, I determine the CDF and the density of $S=X_{(1)}$:
$$ F_S(x) = 1 - \left[1 - F(x)\right]^n = 1 - \left[ 1 - 1 + \left( \frac{\theta}{x} \right)^{\lambda} \right]^n = 1 - \left( \frac{\theta}{x} \right)^{n\lambda} $$
$$ f_S(x) = n\left[ 1 - F(x) \right]^{n-1} f(x) = \frac{n\lambda \theta^{\lambda}}{x^{\lambda+1}} \left( \frac{\theta}{x} \right)^{\lambda(n-1)} = \frac{n\lambda \theta^{n\lambda}}{x^{n\lambda+1}} $$
The statistic $S \thicksim Pa(n\lambda, \theta)$.
I apply the factorization theorem for sufficiency: $$ f(x;\theta) = \prod^{n}_{i=0} \frac{n\lambda \theta^{n\lambda}}{x^{n\lambda+1}_i} = (n\lambda)^n \theta^{n\lambda} \theta^{n^2 \lambda} \prod^{n}_{i=0} x^{-(n\lambda+1)} = g(x; \theta) \cdot h(x) $$
$$ g(x; \theta) = (n\lambda)^n \theta^{n\lambda} \theta^{n^2 \lambda} \qquad ; \qquad h(x) = \prod^{n}_{i=0} x^{-(n\lambda+1)} $$
It's sufficient. I verify the status of minimal using Lehmann-Scheffé theorem:
- $ f(x;\theta) = (n\lambda)^n \theta^{n\lambda} \theta^{n^2 \lambda} \prod^{n}_{i=0} x^{-(n\lambda+1)} $
- $f(y; \theta) = (n\lambda)^n \theta^{n\lambda} \theta^{n^2 \lambda} \prod^{n}_{i=0} y^{-(n\lambda+1)}$
Then: $$ \frac{L(\theta; x)}{L(\theta; y)} \propto \frac{(n\lambda)^n \theta^{n\lambda} \theta^{n^2 \lambda} \prod^{n}_{i=0} x^{-(n\lambda+1)}}{(n\lambda)^n \theta^{n\lambda} \theta^{n^2 \lambda} \prod^{n}_{i=0} y^{-(n\lambda+1)}} = \frac{\prod^{n}_{i=0} x^{-(n\lambda+1)}}{\prod^{n}_{i=0} y^{-(n\lambda+1)}} = c(X;Y) $$
It's also minimal.
If I have missed anything else, please report it to me immediately.