Return to Revisions

This does not answer the question, but reformulates it as an integration problem and addresses a question asked in the comments. Conditionally on $|z_1| = a_1$, ..., $|z_n| = a_n$, the probability that $x<c$ was computed in Kluyver, "A local probability problem" (1906) to be $$ \mathbb{P}\Big( \big|\sum_{i=1}^n z_i\big| < c \,\Big|\, |z_i|=a_i \Big) = c \int_0^\infty J_1(uc)\prod_{i=1}^n J_0(u a_i)\,\mathrm{d}u. $$ Using that $\int_0^1 2a\,J_0(u a) \mathrm{d}a = 2 J_1(u) /u$ and $\frac{\partial}{\partial c}( c J_1(u c)) = uc J_0(u c)$, this gives the density as $$ \rho(x) = \frac{\partial}{\partial x}\mathbb{P}\Big( \big|\sum_{i=1}^n z_i\big| < x\Big) = \int_0^\infty xu J_0(ux) \,\left(\tfrac{2}{u}J_1(u)\right)^n\mathrm{d}u. $$ (Note: there was a small mistake in the formula given in the comments.) By construction $\int_0^n\rho(x)\,\mathrm{d}x = 1$ as $\rho(x)$ vanishes for $x > n$. Hence \begin{align} \mathbb{E}\Big( \frac{1}{n}\big|\sum_{i=1}^n z_i\big|\Big) &= \frac{1}{n} \int_0^n \mathrm{d}x \int_0^\infty \mathrm{d}u\, x^2u\, J_0(ux) \left(\tfrac{2}{u}J_1(u)\right)^n \\ &= \frac{n^2}{3} \int_0^\infty u\,{_1F_2}(\tfrac{3}{2};1,\tfrac{5}{2}; -\tfrac14 n^2u^2) \left(\tfrac{2}{u}J_1(u)\right)^n, \end{align} where the second integral was kindly provided by mathematica.