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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

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- Thread starter xag
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- #1

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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

- #2

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- #3

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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

- #4

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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?

- #5

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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

-Pere

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- #6

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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.

- #7

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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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