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Expected value on subset of Pareto 
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#1
Mar1208, 12:37 PM

P: 6

Hi 
I have a Pareto distribution X (xm=1, k known)*. Highvalue samples are filtered out and I want the expected value of the remaining. Namely: E[XX<a]. *Wikipedia notations Many thanks in advance. Xavier 


#2
Mar1308, 09:30 AM

P: 588

What is the definition of the conditional expectation you are referring to? How did you try to calculate it? Where did you get stuck?



#3
Mar1308, 10:47 AM

P: 6

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/(K1). Best regards, Xavier 


#4
Mar1408, 02:07 PM

P: 6

Expected value on subset of Pareto
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? 


#5
Mar1508, 01:19 PM

P: 588

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[XX\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[XX\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 


#6
Mar1508, 06:10 PM

P: 6




#7
Mar1508, 06:17 PM

P: 6

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


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