The mode-one conjecture for the largest exponential m-spacing

A strict majority bound for the maximizing index.

Let $Y_k$, $k\geq 1$, be independent exponential random variables with rates $k$, and, for an integer $m\geq 2$, put

$$Z_r=Y_r+\cdots+Y_{r+m-1},\qquad R=\operatorname*{\arg\,\max}_{r\geq 1} Z_r.$$

We prove the stronger inequality

$$\mathbb P(R=1)>\frac12.$$

Consequently $1$ is the unique mode of $R$, proving Henze's Conjecture 1.

The argument separates short windows from long ones. For long windows, simple comparisons with the first window suffice. The remaining three lengths are handled by an explicit sufficient event whose probability has an elementary polynomial lower bound.

First, $Y_k\to 0$ almost surely: for every $\varepsilon>0$, the series $\sum_k\mathbb P(Y_k>\varepsilon)=\sum_k e^{-k\varepsilon}$ converges. Thus $Z_r\to0$ almost surely. Since $Z_1>0$, a maximum is attained at a finite index. Any two distinct windows differ by a nontrivial linear combination of independent continuous random variables, so ties have probability zero. This also establishes directly the existence and uniqueness used in the definition of $R$.

1. Exact pairwise comparison probabilities

Write $(x)_a=x(x+1)\cdots(x+a-1)$. A sum of independent exponential variables with consecutive rates $b,b+1,\ldots,b+a-1$ has the property that its negative exponential is beta distributed with parameters $(b,a)$. Indeed, its Laplace transform is

$$\prod_{j=b}^{b+a-1}\frac{j}{j+s} =\frac{\mathrm B(b+s,a)}{\mathrm B(b,a)},\qquad s\geq0,$$

where $\mathrm B$ denotes the beta integral. In particular, a sum with rates $1,\ldots,a$ has distribution function $(1-e^{-t})^a$ for $t\geq0$.

For $1\leq a\leq m-1$, cancellation of the overlapping terms gives

$$\begin{aligned} \mathbb P(Z_{a+1}>Z_1) &=\mathbb P(Y_{m+1}+\cdots+Y_{m+a}>Y_1+\cdots+Y_a)\\ &=\frac{(a)_a}{(m+a+1)_a}. \end{aligned}$$

The two sums on the first line are independent. The beta identity proves the second line by taking the expectation of $(1-e^{-t})^a$ at the first sum.

For $r\geq m+1$, the two entire windows are independent, and the same calculation gives

$$\mathbb P(Z_r>Z_1)=\frac{(m)_m}{(r+m)_m}.$$

The total of these nonoverlapping comparison probabilities is

$$\begin{aligned} T_m&:=\sum_{r=m+1}^{\infty}\mathbb P(Z_r>Z_1) =\frac{(m)_m}{(m-1)(2m+1)_{m-1}}\\ &=\frac{(2m-1)!(2m)!}{(m-1)!(m-1)(3m-1)!}. \end{aligned}$$

Here the sum telescopes using

$$\frac1{(x)_m} =\frac1{m-1}\left(\frac1{(x)_{m-1}}-\frac1{(x+1)_{m-1}}\right).$$

For $m=5,6$, the union bound therefore yields

$$\mathbb P(R\ne1) \leq\sum_{a=1}^{m-1}\frac{(a)_a}{(m+a+1)_a}+T_m =\begin{cases} \dfrac{5923}{12012},&m=5,\\[4pt] \dfrac{28481}{72930},&m=6. \end{cases}$$

Both fractions are strictly smaller than $1/2$.

2. All lengths at least seven

Consider the event

$$E_m=\{Y_1>Y_{m+1}+\cdots+Y_{2m-1}\}.$$

On this event, the first window beats every overlapping competitor: for $1\leq a\leq m-1$,

$$Y_1+\cdots+Y_a>Y_{m+1}+\cdots+Y_{m+a}.$$

Independence and the exponential distribution of $Y_1$ give

$$\mathbb P(E_m)=\prod_{k=m+1}^{2m-1}\frac{k}{k+1} =\frac{m+1}{2m}.$$

Subtracting the probability of any nonoverlapping competitor winning, we obtain

$$\mathbb P(R=1)\geq\frac{m+1}{2m}-T_m.$$

For $m\geq2$, the factorial formula gives

$$\frac{T_{m+1}}{T_m} =\frac{4(m^2-1)}{3m^2} \frac{2m+1}{3m+1}\frac{2m+1}{3m+2} <\frac{40}{63}<\frac23\leq\frac{m}{m+1}.$$

We used $(2m+1)/(3m+1)\leq5/7$ and $(2m+1)/(3m+2)<2/3$. Hence $mT_m$ decreases. Since

$$7T_7=\frac{7007}{19380}<\frac12,$$

we have $T_m<1/(2m)$ for every $m\geq7$. The preceding lower bound is therefore strictly greater than $1/2$.

3. The remaining lengths two, three and four

Define the smallest initial average

$$C_m=\min_{1\leq a\leq m}\frac{Y_1+\cdots+Y_a}{a}.$$

If every $Y_k$ with $k\geq m+1$ is smaller than $C_m$, then the first window wins. For an overlapping competitor, the removed $a$ initial terms sum to at least $aC_m$, while the replacement terms sum to less than $aC_m$. A nonoverlapping window sums to less than $mC_m\leq Z_1$.

Let $U_m=e^{-C_m}$, and write $f_m$ for its density on $(0,1)$. Conditioning on the first $m$ variables gives

$$\mathbb P(R=1)\geq \int_0^1 f_m(u)\prod_{k=m+1}^{\infty}(1-u^k)\,du.$$

For the three lengths in question the distribution functions of $U_m$ are

$$\begin{aligned} \mathbb P(U_2\leq u)&=2u^2-u^3,\\ \mathbb P(U_3\leq u)&=3u^3-4u^5+2u^6,\\ \mathbb P(U_4\leq u)&=4u^4-7u^7-4u^8+14u^9-6u^{10}. \end{aligned}$$

For completeness, these formulas can be obtained without any distributional approximation. Set $A_j=Y_1+\cdots+Y_j$. Their joint density on $0<a_1<\cdots<a_m$ is

$$m!\exp(a_1+\cdots+a_{m-1}-ma_m).$$

For $c\geq0$, define $F_0(x,c)=1$ and successively

$$F_j(x,c)=\int_{jc}^{x} e^tF_{j-1}(t,c)\,dt \qquad (x\geq jc).$$

Integrating the joint density over $a_j\geq jc$ gives

$$\mathbb P(C_m\geq c) =m!\int_{mc}^{\infty}e^{-mt}F_{m-1}(t,c)\,dt.$$

Evaluating this formula for $m=2,3,4$ and substituting $u=e^{-c}$ gives exactly the three polynomials above. In particular their derivatives are

$$\begin{aligned} f_2(u)&=4u-3u^2,\\ f_3(u)&=9u^2-20u^4+12u^5,\\ f_4(u)&=16u^3-49u^6-32u^7+126u^8-60u^9. \end{aligned}$$

It remains to lower-bound the infinite product. For $0<u<1$ and $K\geq m+1$, the union bound for the tail factors gives

$$\prod_{k=m+1}^{\infty}(1-u^k) \geq\prod_{k=m+1}^{K}(1-u^k) \left(1-\frac{u^{K+1}}{1-u}\right).$$

The right-hand side is allowed to be negative; the inequality remains valid. Because $f_m$ is a density, multiplication and integration preserve the inequality. Put

$$L_{m,K}=\int_0^1 f_m(u)\prod_{k=m+1}^{K}(1-u^k) \left(1-\frac{u^{K+1}}{1-u}\right)\,du.$$

These are integrals of explicit polynomials: the denominator $1-u$ cancels against $1-u^{m+1}$. Expanding the finite products and using $\int_0^1u^d\,du=1/(d+1)$ gives

$$\begin{aligned} L_{2,7}&=\frac{4534722274987}{9025222055850}>\frac12,\\[3pt] L_{3,7}&=\frac{14501251235773}{28880710578720}>\frac12,\\[3pt] L_{4,8}&=\frac{690798675487591181}{1345655451257488800}>\frac12. \end{aligned}$$

Thus $\mathbb P(R=1)>1/2$ also for $m=2,3,4$. Together with the preceding cases, this proves the bound for every $m\geq2$. For every $r\geq2$,

$$\mathbb P(R=r)\leq1-\mathbb P(R=1)<\mathbb P(R=1),$$

which proves the conjecture, with a unique mode at $1$.