3

A population density function is definded as $$ f(s)=\begin{cases} (W-1)s^{-W} & s\geq 1, \\ 0& \text{elsewhere}, \end{cases} $$

where $W>1$ is unknown. I just want to ask, how do i find the maximum likelihood estimator of $W$? Do I do the normal differentiation of this function first, and then do a 2nd order differentiation? Thanks.

Stefan Hansen
  • 25,582
  • 7
  • 59
  • 91
dorothy
  • 133

1 Answers1

1

This is a Pareto distribution.

You omit to say what sort of data you've got. I'm going to take a leap and guess you mean an i.i.d. sample, whose size let us call $n$. Then you've got $$ \ell(W)=\log L(W) = \log\Big((W-1)^n (s_1\cdots s_n)^{-W} \Big) = n\log(W-1)-W\log(s_1\cdots s_n). $$ So $$ \ell\,'(W) = \frac{n}{W-1} - \log(s_1\cdots s_n). $$ $$ = \frac{n-(W-1)\log(s_1\cdots s_n)}{W-1} \qquad \begin{cases} >0 & \text{if }1<W<\frac{n}{\log(s_1\cdots s_n)} +1, \\[12pt] =0 & \text{if }W=\frac{n}{\log(s_1\cdots s_n)} +1, \\[12pt] <0 & \text{if }W>\frac{n}{\log(s_1\cdots s_n)} +1 .\end{cases} $$ (Notice that to find these intervals, one needs to consider only the numerator, since the denominator is always positive.)

So $\widehat W = \dfrac{n}{\log(s_1\cdots s_n)}+1$.

Michael Hardy
  • 1
  • Fixed. ${{{{{{{{}}}}}}}}$ – Michael Hardy Jan 27 '13 at 16:27
  • One can also note that $\mathbb E_W(\log S)=W-1$ hence $\widehat W_n\to W$ almost surely when $n\to\infty$.. – Did Jan 27 '13 at 16:28
  • Perhaps. . . . . . But in practice $n$ will never approach $\infty$ and it may be of more interest to examine the probability distribution of $\widehat W$ for small $n$. – Michael Hardy Jan 27 '13 at 16:30
  • Not always--but anyway my point was rather to check that the result for $\widehat W_n$ was not too unreasonable. – Did Jan 27 '13 at 16:39
  • You could have an estimator that's very unreasonable for small $n$ but approaches the right limit as $n\to\infty$. – Michael Hardy Jan 27 '13 at 17:12
  • Yeeeess... I knoooow... Nevertheless, wondering what is the almost sure behaviour of an estimator when $n\to\infty$ may help to spot computation mistakes (for example this is how I instantly knew that your first answer was wrong, before determining the exact mistake you had made). – Did Jan 27 '13 at 17:21
  • I see. ${{{{{{}}}}}}$ – Michael Hardy Jan 27 '13 at 19:56
  • thanks for this. much appreciated. sorry for not mentioning that the sample data are random from the same population. – dorothy Jan 27 '13 at 23:44