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Again from the Shepp and Lloyd paper "ordered cycle lengths in a random permutation", I found this puzzling equality. This one might require access to the paper itself since it's quite a mouthful:

In equation (15), they claimed it is straightforward that if there is an $F_r$ such that

$$\int_0^1 \exp(-y/\xi) dF_r(\xi) = \int_y^{\infty} \frac{E(x)^{r-1}}{(r-1)!} \frac{\exp(-E(x) -x)}{x}   dx $$

then $F_r$ will have moments $G_{r,m}$.

Here

$$G_{r,m} = \int_0^{\infty} \frac{x^{m-1}}{m!} \frac{E(x)^{r-1}}{(r-1)!} \exp(-E(x)-x)   dx$$

and

$$E(x) = \int_x^{\infty} \frac{e^{-y}}{y} dy$$

which is related to the thread Reference request for a "well-known identity" in a paper of Shepp and Lloyd

It looks to me like some sort of Laplace transform, but I can't manage to get the algebra to work, because of the inverse exponent $y/\xi$ with respect to $\xi$.

I will be happy enough if one can tell me why we are looking at the transform $\int_0^1 \exp(-y/\xi) dF_r(\xi)$ instead of the usual moment generating function $\int_0^1 \exp(-y \xi) dF_r(\xi)$, or maybe it's a typo?

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I fixed your LaTeX. We don't support those fancy environments around here! –  Qiaochu Yuan Mar 28 '10 at 19:08
    
Thanks. I wasn't aware. –  John Jiang Mar 28 '10 at 19:20

1 Answer 1

up vote 7 down vote accepted

This is a general fact: assume that $X$ is a positive random variable and that, for a given nonnegative function $g$, $\displaystyle E(\mathrm{e}^{-y/X})=\int_y^{\infty}g(x)\mathrm{d}x$ for every positive $y$. Then $\Gamma(s+1)E(X^s)=\displaystyle\int_0^{\infty}x^sg(x)\mathrm{d}x$ for every positive $s$.

To prove this, integrate the equality $\displaystyle\int_0^{\infty}\mathrm{e}^{-y/x}y^{s-1}\mathrm{d}y=\Gamma(s)x^s$ over $x>0$ with respect to the distribution of $X$ and change the order of integration in the LHS.

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C'est Chouette! Merci Beaucoup, Mon ami! –  John Jiang Mar 29 '10 at 16:05
    
De rien, tout le plaisir est pour moi. :-) –  Did Mar 29 '10 at 16:24

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