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Does anyone know of any clever tricks that solve

$\large G_n(t) = \int_0^t dt_1 \int_0^{t_1} dt_2 \cdots \int_0^{t_{n-2}}dt_{n-1}\int_0^{t_{n-1}}dt_{n} e^{i\lambda(t_1-t_2+t_3-\cdots + t_{n-1}-t_n)}$

I've come up with a few recursion relations but I'm finding it hard to pin down the exact answer.

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    @fatbox May I ask where does such an integral arise?2012-03-08

1 Answers 1

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By rescaling $t_{k} = t u_{k}$ we get $ \begin{eqnarray} G_n(\lambda, t) &=& t^n n! \int_0^1 \mathrm{d} u_1 \frac{1}{n!} \int_0^{u_1} \mathrm{d} u_2 \cdots \int_0^{u_{n-1}} \mathrm{d} u_n \exp\left(i \lambda t \sum_{k=1}^n (-1)^{k-1} u_k \right) \\ &=& t^n n! \mathbb{E}\left( \exp\left(i \lambda t \sum_{k=1}^n (-1)^{k-1} U_{k:n} \right) \right) \end{eqnarray} $ where $U_{k:n}$ is the $k$ out of $n$ order statistics on a sample from uniform distribution. It is well known that the vector $(U_{1:n},U_{2:n},\ldots,U_{n:n})$ is equal in distribution to $ \left( \frac{X_1}{\sum_{k=1}^{n+1} X_k}, \frac{X_1+X_2}{\sum_{k=1}^{n+1} X_k}, \ldots, \frac{X_1+X_2 + \cdots + X_n}{\sum_{k=1}^{n+1} X_k} \right) $ where $X_i$ are i.i.d. exponential random variables with unit mean. Thus $G_n$ is related to the characteristic function of the beta distribution: $ \begin{eqnarray} \sum_{k=1}^n (-1)^{k-1} U_{k:n} &=& \frac{1}{\sum_{m=1}^{n+1} X_m} \sum_{k=1}^n (-1)^{k-1} \sum_{p=1}^{k} X_p = \frac{1}{\sum_{m=1}^{n+1} X_m} \sum_{k=1}^n (-1)^{k-1} \sum_{p=1}^{k} X_p \\ &=& \frac{1}{\sum_{m=1}^{n+1} X_m} (-1)^{n-1} \sum_{0\leqslant 2k < n} X_{n-2k} \\ &=& (-1)^{n-1} \frac{\sum_{0\leqslant 2k < n} X_{n-2k} }{ \sum_{0\leqslant 2k < n} X_{k} + \sum_{0\leqslant 2k < n+1} X_{n+1-2k} } = (-1)^{n-1} \frac{g_1}{g_1+g_2} \end{eqnarray} $ where $g_1$ follows $\Gamma\left( \left\lfloor \frac{n+1}{2} \right\rfloor\right)$ distribution and $g_2$ follows $\Gamma\left( \left\lceil \frac{n+1}{2} \right\rceil\right)$. The ratio $V = g_1/(g_1+g_2)$ is a $\operatorname{Beta}\left(\left\lfloor \frac{n+1}{2} \right\rfloor, \left\lceil \frac{n+1}{2} \right\rceil \right)$ random variable. Thus

$ G_n(t) = n! t^n \phi_V( (-1)^{n-1} \lambda t ) = n! t^{n} \cdot {}_1 F_1\left( \left\lfloor \frac{n+1}{2} \right\rfloor ; n+1; (-1)^{n-1} i \lambda t \right) $

Here is verification in Mathematica v8:

In[1]:= Table[   Hypergeometric1F1[Floor[(n + 1)/2], 1 + n, (-1)^(n - 1) I la t] -     Expectation[ Exp[I la t Sum[(-1)^(k - 1) x[k], {k, 1, n}]],      Array[x, n] \[Distributed]       OrderDistribution[{UniformDistribution[], n}, Range[n]],      Assumptions -> la > 0 && t > 0], {n, 2, 6}] // Expand  Out[1]= {0, 0, 0, 0, 0} 
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    Ver$y$ nice solution!2012-03-08