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Recursive integral equations for random weights averages: Exponential functions and Cauchy distribution

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  • Soltani, A.R.

Abstract

In this article, firstly, we prove that functions of the form ϕ(x)=ecxI(−∞,0)(x)+ebxI[0,+∞)(x), c,b constants, are the only solutions to the integral equation ϕ(x)=∫01ϕ(ux)ϕ((1−u)x)du. This indeed gives the result of Van Asshe (1987) who used the Schwartz distribution theory to prove that for i.i.d X and Y, UX+(1−U)Y=dX if and only if X has a Cauchy distribution. Secondly, by looking into certain recursive integral equations involving characteristic functions, we prove that if for an n≥2, the random weight mean U(1)X1+(U(2)−U(1))X2+⋯+(1−U(n−1))Xn has a Cauchy distribution, then X1 has a Cauchy distribution; random variables X1,…,Xn are i.i.d, the random weights are the cuts of (0,1) by a uniform sample. The multivariate analogue of this result is also provided.

Suggested Citation

  • Soltani, A.R., 2022. "Recursive integral equations for random weights averages: Exponential functions and Cauchy distribution," Statistics & Probability Letters, Elsevier, vol. 190(C).
  • Handle: RePEc:eee:stapro:v:190:y:2022:i:c:s016771522200147x
    DOI: 10.1016/j.spl.2022.109606
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    References listed on IDEAS

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    1. Soltani, Ahmad Reza & Roozegar, Rasool, 2012. "On distribution of randomly ordered uniform incremental weighted averages: Divided difference approach," Statistics & Probability Letters, Elsevier, vol. 82(5), pages 1012-1020.
    2. Soltani, A.R. & Homei, H., 2009. "Weighted averages with random proportions that are jointly uniformly distributed over the unit simplex," Statistics & Probability Letters, Elsevier, vol. 79(9), pages 1215-1218, May.
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