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Right now I have a horribly-looking triple sum ($x,y,z$ are non-negative integers and $x+y+z=N$): $$ W_{12}(x,y)=\frac{x}{N}\sum_{l=0}^{x-1}\sum_{l'=0}^{y}\sum_{l''=0}^{z}{x-1 \choose l}p^{l}\left(1-p\right)^{x-1-l}{y \choose l'}q^{l'}\left(1-q\right)^{y-l'}{z \choose l''}q^{l''}\left(1-q\right)^{z-l''}\Theta(l'-l)\Theta(l'-l'')\\ =\frac{x}{N}\sum_{l=0}^{x-1}\sum_{l'=0}^{y}\sum_{l''=0}^{z}B_{x-1,p}(l)B_{y,q}(l')B_{z,q}(l'')\Theta(l'-l)\Theta(l'-l''), $$ where $\Theta(x)$ is the step function which is 1 whenever $x>0$. Because of the two step functions, it is very difficult to manipulate.

My question is: Is there a way to simplify the triple sum to something like $W_{12}(x,y)=\cdots\sum_{L=0}^{\cdots}\sum_{L'=0}^{\cdots}\cdots$?

Why I think this might be doable: a simpler version has been done in http://journals.aps.org/pre/pdf/10.1103/PhysRevE.79.046104.(see the Appendix) For those who are outside the paywall, please access here(this is a temporary link). The idea is to have a change of variable $L=l'-l$. But since I have two step functions, and $L=l'-l$ and $L'=l'-l''$ would be somehow correlated, I don't know how to apply the idea to $W_{12}(x,y)$.

Edit 1: Thanks David for pointing out the ambiguity in the wording of the question. First for convenience I just posted the relevant sections of the paper. The following simplifies $d_M=\dfrac{M}{N}\sum_{k=0}^{M-1}\sum_{k'=0}^{N-M}B_{M-1,p}(k)B_{N-M,q}(k')\Theta(k'-k)$ to $d_M=\dfrac{M}{N}\sum_{K=0}^{N-M-1}B_{N-1,p}(K)$. enter image description here enter image description here

So I guess $W_{12}(x,y)$ can be simplified as a sum of the product of two binomial coefficients like $W_{12}(x,y)=\cdots\sum_{L=0}^{\cdots}\sum_{L'=0}^{\cdots}\cdots$

Edit 2: The following figure shows the contour plot of $W_{12}$ when $N=900$ and $p=0.35$. Contours with reddish color is higher than contours with bluish color. One can see the obvious $x$ dependence of $W_{12}$ in some part of the region. One can also see the steep steps. enter image description here

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I assume, although you didn't actually say so, that $\Theta(x)$ is zero when $x\le0$. If so, then we have $\Theta(l'-l)=0$ for $l'\le l$, so we only need sum for $l'>l$, and we can then take $\Theta(l'-l)=1$. Likewise for $l''$, only it's the other way round: $\Theta(l'-l'')$ is zero if $l''\ge l'$. So $$W_{12}(x,y) =\frac{x}{N}\sum_{l=0}^{x-1}\sum_{l'=l+1}^{y}\sum_{l''=0}^{l'-1}B_{x-1,p}(l)B_{y,q}(l')B_{z,q}(l'')\ .$$ This still looks difficult, but at least we have removed the step functions, which is what you asked.

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  • $\begingroup$ I acknowledge that there is ambiguity in the wording of my question. Please see the edited question. $\endgroup$ – wdg May 28 '14 at 8:20

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