# 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.

• Your algebra in the "Try" section is incorrect. Use the laws of probability to find the distribution of the minimum in terms of the distribution of the variables. See stats.stackexchange.com/questions/129145. It helps to work in terms of the survival function (aka complementary CDF). See stats.stackexchange.com/questions/102691 for examples. Search our site for many applications and generalizations.
– whuber
Jan 24 at 14:50
• @whuber I have tried to adjust the algebra. Is it right? Jan 31 at 12:02
• A quick glance indicates you have the ideas correct now.
– whuber
Jan 31 at 14:35
• @whuber Thank you. I'm glad to have resolved this issue too, refining the style a bit. Again, I thank you for your helpfulness. Jan 31 at 15:08