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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.

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    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Jan 24, 2023 at 14:50
  • $\begingroup$ @whuber I have tried to adjust the algebra. Is it right? $\endgroup$
    – iStats7238
    Commented Jan 31, 2023 at 12:02
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    $\begingroup$ A quick glance indicates you have the ideas correct now. $\endgroup$
    – whuber
    Commented Jan 31, 2023 at 14:35
  • $\begingroup$ @whuber Thank you. I'm glad to have resolved this issue too, refining the style a bit. Again, I thank you for your helpfulness. $\endgroup$
    – iStats7238
    Commented Jan 31, 2023 at 15:08

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