# Does the following function series converge?

Let $$f_n(x)=\frac{\frac{1}{(n-1)!}\sum_{k=0}^{\lfloor \alpha n-x\rfloor}C_{n-1}^{k}~(-1)^k(\alpha n-x-k)^{n-1}}{\frac{1}{n!}\sum_{k=0}^{\lfloor \alpha n\rfloor}C_{n}^{k}(-1)^k(\alpha n-k)^{n}},$$ where

• $$x\in[0,1]$$,
• $$C_{n}^{k}$$ is the binominal coefficient,
• and $$\alpha$$ is a constant such that $$0 \le \alpha \le 1$$.

Based on my intuition and numerical results, I guess the above series converges pointwise to a truncated exponential function $$g(x)=A\exp(-\lambda x)$$, where $$A$$ and $$\lambda$$ are parameters to be determined.

Question: How to prove or disprove this conjecture?

This question originates from my studies on the the marginal distribution of a uniform distribution defined over an $$n$$-dimensional simplex truncated by a unit cube, which can be defined as $$\mathscr{T}_n(t)=\bigg\{\vec{\mathbf{x}}:\sum_{i=1}^n x_i \le t, 0 \le x_i \le 1\bigg\}.$$ Considering a uniform distribution over the domain $$\mathscr{T}_n(\alpha n)$$, I obtained the density function of the marginal distribution in any dimension as $$p(x)=f_n(x)=\frac{\text{vol}\left(\mathscr{T}_{n-1}\left(\alpha n-x\right)\right)}{\text{vol}\left(\mathscr{T}_{n}\left(\alpha n\right)\right)},$$ where $$x\in[0,1]$$. It is known that the marginal distribution of the joint random vector uniformly distributed over a simplex with a finite and nonzero mean value will converge to an exponential distribution. For this reason, I guess the considered series converges to the density function of a truncated exponential distribution.

• In the denominator, do you mean $\sum_{k=0}^{\lfloor \alpha n\rfloor}$ or $\sum_{k=0}^{\lfloor \alpha n-x\rfloor}$? Jul 27 '20 at 17:33
• @Iosif Pinelis The latter one is right. Jul 28 '20 at 2:00

$$\newcommand{\si}{\sigma}$$ By the Irwin--Hall formula, your first displayed ratio is $$$$f_n(x)=\frac{P(S_{n-1}\le an-x)}{P(S_n\le an-x)}=\frac{P(S_{n-1}\le a(n-1)-(x-a))}{P(S_n\le an-x)},$$$$ where $$a:=\alpha\in[0,1]$$, $$x\ge0$$, $$S_n:=X_1+\dots+X_n$$, and $$X_1,\dots,X_n$$ are iid random variables each uniformly distributed on $$[0,1]$$.

If $$a=0$$ then $$P(S_n\le an-x)=0$$ for $$x\ge0$$, so that $$f_n(x)$$ is undefined. If $$a>1/2$$ then, by the law of large numbers, $$P(S_n\le an-y)\to1$$ (as $$n\to\infty$$) for any fixed real $$y$$, so that $$f_n(x)\to\frac11=1$$. If $$a=1/2$$ then, by the central limit theorem, $$P(S_n\le an-y)\to1/2$$ for any fixed real $$y$$, so that $$f_n(x)\to\frac{1/2}{1/2}=1$$.

It remains to consider the nontrivial case when $$a\in(0,1/2)$$. Since $$X_i$$ equals $$1-X_i$$ in distribution, we have $$$$f_n(x)=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_n\ge bn+x)},$$$$ where $$$$b:=1-a\in(1/2,1).$$$$ By Theorem 1 by Petrov, $$$$P(S_n\ge tn)\sim\frac{e^{nL_t(h_t)}}{h_t\si(h_t)\sqrt{2\pi n}} \tag{*}$$$$ uniformly in $$t$$ in any closed subinterval of the interval $$(1/2,1)$$, where $$$$L_t(h):=-ht+\ln R(h),\quad R(h):=Ee^{hX_1}=\frac{e^h-1}h,\quad\si(h):=m'(h),\quad m(h):=R'(h)/R(h)$$$$ for real $$h>0$$, and $$h_t\in(0,\infty)$$ is the only root of the equation $$$$m(h_t)=t.$$$$

The functions $$m$$ and $$\si$$ (on $$(0,\infty)$$) are smooth, and $$\si>0$$. So, $$m$$ is a smooth increasing function, and hence the function $$(1/2,1)\ni t\mapsto h_t$$ is smooth. So, if $$t\to t_0\in(1/2,1)$$, then $$$$h_t\si(h_t)\sim h_{t_0}\si(h_{t_0})$$$$ and $$$$\frac d{dt}L_t(h_t)=\frac{\partial L_t(h)}{\partial h}\Big|_{h=h_t}\;\frac{dh_t}{dt}-h_t =(-t+m(h_t))\;\frac{dh_t}{dt}-h_t=-h_t\sim-h_{t_0},$$$$ whence, by (*), $$$$\frac{P(S_n\ge tn)}{P(S_n\ge t_0n)}\sim \exp[-nh_{t_0}(t-t_0)(1+o(1))].$$$$ Using this with $$t_0=b$$ and $$t=b+x/n$$, we get $$$$\frac{P(S_n\ge bn+x)}{P(S_n\ge bn)}\sim e^{-h_b x}$$$$ for each real $$x$$. Hence, \begin{align} P(S_n\ge bn)&=\int_0^1 P(S_{n-1}\ge bn-z)\,dz \\ &=\int_0^1 P(S_{n-1}\ge b(n-1)+b-z)\,dz \\ &\sim P(S_{n-1}\ge b(n-1))\int_0^1 e^{-h_b(b-z)}\,dz \\ &=P(S_{n-1}\ge b(n-1))e^{-h_b b}R(h_b). \end{align}

We conclude that \begin{align} f_n(x)&=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_n\ge bn+x)} \\ &=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_{n-1}\ge b(n-1))} \frac{P(S_{n-1}\ge b(n-1))}{P(S_n\ge bn)} \frac{P(S_n\ge bn)}{P(S_n\ge bn+x)} \\ &\sim e^{-h_b(x-a)}\frac{e^{h_b b}}{R(h_b)}\,e^{h_b x} =\frac{e^{h_b}}{R(h_b)} \end{align} for each real $$x$$.

For an illustration, here are the graphs $$\{(x,f_n(x)/\frac{e^{h_b}}{R(h_b)})\colon|x|<5\}$$ with $$a=0.25$$ for $$n=100$$ (left) and $$n=500$$ (right):

• Very helpful infinitesimal analysis based on the LDT. I have corrected a critical mistake in the previous post, but your method is also applicable for the edited question. Based on my own undestanding of your method, I have verified the edited function series indeed converges to the truncated exponential density function. Jul 28 '20 at 2:11