most recent 30 from http://mathoverflow.net 2013-05-24T02:07:15Z http://mathoverflow.net/feeds/question/19648 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19648/method-of-moments-and-laplace-transform-from-shepp-and-lloyd John Jiang 2010-03-28T18:58:19Z 2011-01-15T12:48:32Z

<p>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:</p> <p>In equation (15), they claimed it is straightforward that if there is an $F_r$ such that </p> <p>$$\int_0^1 \exp(-y/\xi) dF_r(\xi) = \int_y^{\infty} \frac{E(x)^{r-1}}{(r-1)!} \frac{\exp(-E(x) -x)}{x} &nbsp; dx $$</p> <p>then $F_r$ will have moments $G_{r,m}$. </p> <p>Here</p> <p>$$G_{r,m} = \int_0^{\infty} \frac{x^{m-1}}{m!} \frac{E(x)^{r-1}}{(r-1)!} \exp(-E(x)-x) &nbsp; dx$$</p> <p>and </p> <p>$$E(x) = \int_x^{\infty} \frac{e^{-y}}{y} dy$$</p> <p>which is related to the thread <a href="http://mathoverflow.net/questions/19526/reference-request-for-a-well-known-identity-in-a-paper-of-shepp-and-lloyd" rel="nofollow">http://mathoverflow.net/questions/19526/reference-request-for-a-well-known-identity-in-a-paper-of-shepp-and-lloyd</a></p> <p>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 $\xi$. </p> <p>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?</p>

http://mathoverflow.net/questions/19648/method-of-moments-and-laplace-transform-from-shepp-and-lloyd/19737#19737 Didier Piau 2010-03-29T15:40:25Z 2011-01-15T12:48:32Z

<p>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 $s$.</p> <p>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.</p>