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?

  2. […] article was first published on R – Xi'an's Og, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here) […]

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