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Suppose you are sampling from a probability distribution which contains a reciprocal:

$$f(x; \alpha, \lambda) = \begin{cases}\frac{\lambda \alpha^\lambda}{x^{\lambda+1}}~~x\geq \alpha;\\ 0 ~~~~~~\text{otherwise};\end{cases}$$

You can't derive the MLE by minimising the likelihood. Instead you have to use a "trick" and realise that since the function is always decreasing, the likelihood is maximised at $Y = \min(X). $

Is there a trick to work out the expectation $E[Y]$?

I tried to calculate and integrate the PDF using an approach similar to here:

$$F_y(x) = \begin{cases} 0 ~~~~~~~~~~~x\lt\alpha; \\ \frac{\lambda^n \alpha^{n\lambda}}{x^{n(\lambda+1)}} ~~~~x\ge\alpha; \end{cases}$$

$$f_y(x) = \begin{cases} 0 ~~~~~~~~~~~x\lt\alpha; \\ \frac{-\lambda^n \alpha^{n\lambda}n(\lambda+1)} {x^{n(\lambda+1)+1}} ~~~~x\ge\alpha; \end{cases}$$

Integrating:

$$ E[Y] = \int_\alpha^\infty xf_y(x)dx = \frac{\lambda^n \alpha^{1-n}n(\lambda+1)} {n(\lambda+1)-1} $$

but I was expecting to get an unbiased estimate of $\alpha$ (or at least close to it).

SuMau
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  • $$F_y(x)=1-\frac{\alpha^{n\lambda}}{x^{n\lambda}}=\int_\alpha^xf_y(t)dt,,\Rightarrow,,f_y(t)=\frac{n\lambda\alpha^{n\lambda}}{t^{n\lambda+1}}$$ $$E[Y]=\int_\alpha^\infty f_y(t)tdt=n\lambda\alpha^{n\lambda}\int_\alpha^\infty\frac{dt}{t^{n\lambda}}=\frac{n\lambda\alpha^{n\lambda}}{n\lambda-1}\alpha^{1-n\lambda}=\frac{n\lambda\alpha}{n\lambda-1},\to,\alpha,,\text{at},,n\to\infty$$ – Svyatoslav Aug 25 '22 at 18:54
  • Thank you that solved it! Would you like to put it as an answer and for me to tick it? – SuMau Aug 25 '22 at 19:55
  • Thank you, this is fine :) – Svyatoslav Aug 25 '22 at 20:32

0 Answers0