Here is what the maximum likelihood method is:

Assume we have observed $x_1,x_2,...,x_n$ from the random independent variables $X_1,X_2,...X_n$

The joint distribution for $(X_1,X_2,...X_n)$ is given by

$f_{X_1,X_2...X_n}(x_1,x_2,...x_n) = \prod\limits_{k=1}^{n}f(x_k)$

We then would like give the the parameters values of $f(\cdot)$ so that we maximize:

$L = \prod\limits_{k=1}^{n}f(x_k)$.

(all $x_k$ are given)

Okey here is my question:

consider the random variable $Z := max(X_1-u,X_2-u,...X_N -u)$ , where $u$ is some threshold, $N,X_1,X_2$ are independent the $X_i$:s are all equally distributed, N has a poisson distribution with parameter $\lambda$ ,$(X_i-u)$ are ,for all $i$, pareto distributed . Also all $X_i$ exceeds $u$.

The distribution function for $Z$ - let us call it $G(x)$, is given,after some calculations, as :

$\mathbb{P}(Z \leq x) = G(x) = e^{-\lambda(1- F(x)) }$

Where $F(x) = \mathbb{P}(X_i -u \leq x)= 1 - (\dfrac{\alpha}{\alpha + x})^{\gamma}$, (the pareto distribution with parameters $\alpha,\gamma$)

if $x_1,x_2,..x_n$ are the realizations of $X_1,X_2,..$ then the Maximum likelihood estimate of the parameters$(\alpha,\lambda,\gamma)$ can be obtained by using $x_1 - u , x_2 - u , ...x_n-u$.(all this is from the book)

I have some difficulties of setting upp the maximum likelihood equation (in terms of the definition above)

I mean i would like to think that $G'(x)$ would take the place of $f(\cdot)$ in the definition above but then we would have only one single observation from $Z$ and that is $max(x_1 - u , x_2 - u , ...x_n-u)$ and that sound strange that (the book) would propose to estimate the parameters from one observation.

What Iam missing or what is missing here?


1 Answer 1


A few comments on the formulation the problem:

  1. If you observe all of the $X_i$ then you don't need $Z$ to estimate the parameters.

  2. Are you sure $Z$ is not the sum, and that $G$ is not the probability generating function?

  3. You can also denote $Y_i:=X_i-u$ and deal directly with $Y_i$ as you know their explicit distribution.

  4. Now you want to write down the joint distribution of $Y_1,\ldots,Y_N,N$. Note that the last $N$ is discrete and $Y_i$ are continuous.

  • $\begingroup$ thanks Liron for your answer. the answer for your first and second question is No. I think you are right about 3,4. @Liron The likelihood should look like? $\prod_i f(y_i)P(N=n) $ Where $y_i =$ is he realisation of $Y_i$ according to (3) in your comment $\endgroup$
    – Danny
    Feb 23, 2015 at 22:25

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