## 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.

### 6 Responses to “sum of Paretos”

1. Gerard Letac Says:

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?

• Pas vraiment…

• Gerard Letac Says:

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

$\mathbb E[(\sum_{i=1}^M U_i)/(\sum_{i=1}^M U_i/V_i)]$
$=ME[U_1 /(\sum_{i=1}^M U_i/V_i)]$
$=M\mathbb E[U_1 \int_0^{\infty}\exp(-s\sum_{i=1}^M U_i/V_i)ds]$
$=M\int_0^{\infty}\mathbb E[(A(s/V_1))^{M-1}A'(s/V_1)]ds$
$=M\mathbb E(V_1)\int_0^{\infty}A(s)^{M-1}A'(s)ds=1/2.$

• 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

• Les simulations ne semblent pas concorder avec cette valeur limite, puisqu’elles indiquent plutôt 1/8 ?!

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