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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.