Distribution of $S = \min\{ X_1, ..., X_n\}$ and $c$ s.t. $E(c\hat{\theta}) = \theta$

Question

Warning

The question is the third and last part of this question.

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. Find the distribution of $S$ and establish if exists an unbiased estimator of $\theta$ in the class of estimators $c \cdot S$ where $c$ is an appropriate constant.

Try

The density and the CDF of $S = \min\{ X_1, ..., X_n \}$ are:

• $F_S(x) = 1 - \left( \frac{\theta}{x} \right)^{n\lambda}$
• $f_S(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 MLE estimator of the $X$ is the minimum: $\hat{\theta} = x_{(1)} = \min\{x_1, ..., x_n\}$.

The distribution of $S$ is the following: $$ F_S(x) = P(X_{(1)} \leq x) = 1 - P(X_{(1)} > x) $$

Since the $X_i$ are i.i.d., $$ \begin{align} F_S(x) & = 1 - P(min(X_1, ..., X_n) > x) \\ & = 1 - P(X_i > x, \qquad i = 1, ..., n) \\ & = 1 - P\left[ (X_1 > x) \cap (X_2 > x) \cap ... \cap (X_n > x) \right] \\ & = 1 - P\left[ (X_1 > x) \cdot (X_2 > x) \cdots (X_n > x) \right] \\ & = 1 - \left[ (1 - F_{X_1}(x)) \cdot (1 - F_{X_2}(x)) \cdots (1 - F_{X_n}(x)) \right] \\ & = 1 - \left[ (1 - F_X(x)) \cdot (1 - F_X(x)) \cdots (1 - F_X(x)) \right] \\ & = 1 - \left[ 1 - F_X(x) \right]^n \\ \end{align} $$

Since $F_X(x) = 1 - (\theta/x)^{\lambda}$, $$ F_S(x) = 1 - \left[1 - 1 + (\theta/x)^{\lambda} \right]^n = 1 - \left( \frac{\theta}{x} \right)^{n\lambda} $$

The statistic $S \thicksim Pa(n\lambda,\theta)$.

For the next point I have found a video that explains it. I set the following equation: $ E(c\hat{\theta}) = \theta $, hence: $$ \int^{+\infty}_{\theta} c\hat{\theta}f(\theta)dx $$

Let's solve the integral before: $$ \int^{+\infty}_{\theta} c\hat{\theta}f(\theta)dx = \int^{+\infty}_{\theta} cx\frac{n\lambda \theta^{n\lambda}}{x^{n\lambda+1}}dx = c\theta \left( \frac{n\lambda }{n\lambda-1} \right) $$

Finally: $$ c\theta \left( \frac{n\lambda }{n\lambda-1} \right) = \theta $$

$$ c = \left( \frac{n\lambda -1}{n\lambda} \right) $$

If I have missed anything else, please report it to me immediately.