Expected value on subset of Pareto

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xag

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Hi -
I have a Pareto distribution X (xm=1, k known)*. High-value samples are filtered out and I want the expected value of the remaining. Namely: E[X|X<a].

*Wikipedia notations

Many thanks in advance.
Xavier
 
What is the definition of the conditional expectation you are referring to? How did you try to calculate it? Where did you get stuck?
 

xag

6
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Hi Pere -

What I'm looking for is a mean value.
I found out that the samples of interest belong to a larger set that roughly follow a Pareto distribution, and that among this larger set, the samples of interest are the ones whose values are below a certain number, 'a'.

I can estimate the parameters of the Pareto once for all, and calculate 'a' for each run. But I can't access the samples for each run, so if there was a way to predict it with 'k' (the Pareto shape) and 'a', that would be great.

My first guess is that the samples of interest also follow a Pareto, but I've not shown that. And in this case, could we express its shape K and scale Xm? Then the mean I'm looking for would be K Xm/(K-1).

Best regards,
Xavier
 

xag

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Ok. It seems I got all the way to the finish line and stopped just before :

I guess I just had to integrate the probability density function between 1 and a (it's defined between 1 and +oo).
It gives : (a ^(1 - k) - 1) * k / (1 - k)

Can anyone confirm?
 
I guess I just had to integrate the probability density function between 1 and a.
Can anyone confirm?
What you did gives you the probability of a sample with value less than a.

Try the following instead: First calculate the conditional distribution function[itex]\mathbb{P}(X\leqx| X \leq a)[/itex].
By definition this is equal to

[tex]
\frac{\mathbb{P}(X\leq x \wedge X \leq a)}{\mathbb{P}(X\leq a)}
[/tex]

(You already calculated the denominator.)

You can also easily calculate the numerator. For x>a the numerator is [itex]\mathbb{P}(X \leq a)[/itex] which cancels with denominator giving one. For x>a you calculate it in the same way as you did for [itex]\mathbb{P}(X \leq a)[/itex] with a replaced by x. Then you take the derivative with respect to x to find the conditional density function [itex]\rho_a(x)[/itex] (suported on [1,a]), which you then use to calculate the conditional expectation as
[tex]\mathbb{E}\left[X|X\leq a\right]=\int_1^a{x\rho_a(x)dx}[/tex]
A more direct though maybe less obvious formula would be
[tex]\mathbb{E}\left[X|X\leq a\right]=\frac{\int_1^a{x\rho(x)dx}}{\int_1^a{\rho(x)dx}}[/tex],
where [itex]\rho(x)[/itex] is the density function of your original Pareto distribution.
I suggest you try them both and check if they really are the same:smile:
-Pere
 
Last edited:

xag

6
0
Ok. It seems I got all the way to the finish line and stopped just before :

I guess I just had to integrate the probability density function between 1 and a (it's defined between 1 and +oo).
It gives : (a ^(1 - k) - 1) * k / (1 - k)

Can anyone confirm?
Forget this post: the thing I integrated was x * dF(x) and moreover the result I gave here is wrong.
 

xag

6
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Try the following instead: First calculate the conditional distribution function[itex]\mathbb{P}(X\leqx| X \leq a)[/itex].
[...]
I suggest you try them both and check if they really are the same
-Pere
Thank you so much.

In both cases I find [tex] \mathbb{E}\left[X|X\leq a\right]= \frac{k}{k - 1} \frac{a ^ k - a}{a ^ k - 1}[/tex]
 

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