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![\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)](https://s0.wp.com/latex.php?latex=%5Cmathbb+E%5E%7BU%2CV%7D%5Cleft%5B%5Csum_%7Bi%3D1%7D%5EM+U_i%5CBig%2F%5Csum_%7Bi%3D1%7D%5EM+U_i%2FV_i+%5Cright%5D%5Cquad+U_i%2CV_i%5Csim%5Cmathcal+U%280%2C1%29&bg=000000&%23038;fg=B0B0B0&%23038;s=0&%23038;c=20201002)
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)](https://s0.wp.com/latex.php?latex=%5Cmathbb+E%5E%7BV%7D%5Cleft%5BM%5Cbig%2F%5Csum_%7Bi%3D1%7D%5EM+2U_i%2FV_i+%5Cright%5D%5Cquad+U_i%2CV_i%5Csim%5Cmathcal+U%280%2C1%29&bg=000000&%23038;fg=B0B0B0&%23038;s=0&%23038;c=20201002)
or again
![\mathbb E^{V}\left[1\big/(1+2\overline R_{M/2})\right]](https://s0.wp.com/latex.php?latex=%5Cmathbb+E%5E%7BV%7D%5Cleft%5B1%5Cbig%2F%281%2B2%5Coverline+R_%7BM%2F2%7D%29%5Cright%5D&bg=000000&%23038;fg=B0B0B0&%23038;s=0&%23038;c=20201002)
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.
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