sum of Paretos

A rather curious question on X validated about the evolution of

$\mathbb E^{U,V}\left[\sum_{i=1}^M U_i\Big/\sum_{i=1}^M U_i/V_i \right]\quad U_i,V_i\sim\mathcal U(0,1)$

when M increases. Actually, this expectation is asymptotically equivalent to

$\mathbb E^{V}\left[M\big/\sum_{i=1}^M 2U_i/V_i \right]\quad U_i,V_i\sim\mathcal U(0,1)$

or again

$\mathbb E^{V}\left[1\big/(1+2\overline R_{M/2})\right]$

where the average is made of Pareto (1,1), since one can invoke Slutsky many times. (And the above comparison of the integrated rv’s does not show a major difference.) Comparing several Monte Carlo sequences shows a lot of variability, though, which is not surprising given the lack of expectation of the Pareto (1,1) distribution. But over the time I spent on that puzzle last week end, I could not figure the limiting value, despite uncovering the asymptotic behaviour of the average.

This entry was posted on May 12, 2022 at 12:22 am and is filed under Books, Kids, R, Statistics with tags cross validated, inverse moment, Landau distribution, Pareto distribution, Slutsky's lemma, Solidarity with Ukraine, stable distribution. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

6 Responses to “sum of Paretos”

Gerard Letac Says:
May 12, 2022 at 5:32 pm

Et en appliquant $a^{-1}=\int_0^{\infty }e^{-as}ds$ à $a=\sum_{i=1}^M\frac{U_i}{V_i}$ on devrait y voir plus clair?

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- xi'an Says:
  May 13, 2022 at 11:56 am
  
  Pas vraiment…
  
  Reply
  - Gerard Letac Says:
    May 15, 2022 at 10:59 am
    
    Si
    $A(t)=(1-e^{-t})/t$
    alors
    $\mathbb E(e^{-sU_1/V_1}|V_1)=A(s/V_1)$
    et
    $\mathbb E(U_1e^{-sU_1/V_1}|V_1)=A'(s/V_1).$
    
    Consequence
    
    $\mathbb E[(\sum_{i=1}^M U_i)/(\sum_{i=1}^M U_i/V_i)]$
    $=ME[U_1 /(\sum_{i=1}^M U_i/V_i)]$
    $=M\mathbb E[U_1 \int_0^{\infty}\exp(-s\sum_{i=1}^M U_i/V_i)ds]$
    $=M\int_0^{\infty}\mathbb E[(A(s/V_1))^{M-1}A'(s/V_1)]ds$
    $=M\mathbb E(V_1)\int_0^{\infty}A(s)^{M-1}A'(s)ds=1/2.$
  - Letac Says:
    May 15, 2022 at 12:53 pm
    
    Correction A(t)=(1-e^{-t})/t evidemment. Si on remplace les lois uniformes de $U_i$ et $V_i$ par deux lois sur la demi doite quelconques la valeur doit etre encore jolie
  - xi'an Says:
    May 15, 2022 at 2:21 pm
    
    Les simulations ne semblent pas concorder avec cette valeur limite, puisqu’elles indiquent plutôt 1/8 ?!
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May 12, 2022 at 8:10 am

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