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Assume we have two independent, random variables, $X$ and $Y$ which follow the same Pareto distribution, i.e.:

$f(x) = \frac{1}{x^{2}}, x > 1$

Given that - what would be the joint probability $P(X + Y < a)$ of those variables? And generally speaking - how to obtain the CDF of $S = X + Y$?

macgivera
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    When they're independent, the relevant integral of the joint density reduces to a convolution of the densities. For the general continuous case see https://en.wikipedia.org/wiki/Convolution_of_probability_distributions immediately before the section labelled "Example Derivation": $f_Z(z) = \int_{-\infty}^\infty f_X(x), f_Y(z-x), dx$. In some cases properties of generating functions (pgfs, mgfs, characteristic functions, cumulant generating functions) might be used instead. – Glen_b Aug 25 '22 at 00:58
  • The case of two Paretos with different parameter is covered here: https://stats.stackexchange.com/questions/115291/what-distribution-results-in-adding-two-pareto-distributions (of course one can set $a=b$ there at least). – Glen_b Aug 25 '22 at 01:03
  • For identically-distributed Pareto variates see C. M. Ramsay (2006), "The Distribution of Sums of Certain I.I.D. Pareto Variates," Communication in Statistics- Theory and Methods 35(3):395-405 – Glen_b Aug 25 '22 at 01:09
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    I think it would be splendid for the OP to submit the general solution to this exercise, in terms of Hypergemetric2F1 functions, and see what the marker makes of it. – wolfies Aug 25 '22 at 04:15
  • @wolfies There's no need, because in this case the hypergeometric functions simplify: the integral has an elementary solution as a rational function of $a$ and $\log(a-1).$ Perhaps the easiest way to find it is to compute the convolution of the densities with partial fractions. The CDF is found in a similar manner. – whuber Aug 25 '22 at 14:09
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    Sure - I am aware of that, having solved it before posting my comment. That is precisely why I think it would be splendid to submit the general solution: above and beyond (like Star Trek). – wolfies Aug 25 '22 at 15:33

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