Block size in Geometric(p)-biased permutations

Fix a probability distribution $\mathbf p = (p_1, p_2, \cdots)$ on the positive integers. The first block in a $\mathbf p$-biased permutation can be visualized in terms of raindrops that land at each positive integer $j$ with probability $p_j$. It is the first point $K$ so that all sites in $[1,K]$ are wet and all sites in $(K,\infty)$ are dry. For the geometric distribution $p_j= p(1-p)^{j-1}$ we show that $p \log K$ converges in probability to an explicit constant as $p$ tends to 0. Additionally, we prove that if $\mathbf p$ has a stretch exponential distribution, then $K$ is infinite with positive probability.


Introduction
One way to visualize a general rainstick process is to partition the unit interval [0, 1] into sets I i = [a i , a i+1 ) with (a i ) i≥1 strictly increasing and a i → 1 as i → ∞. Raindrops fall, one after the other, on points chosen uniformly at random from the unit interval. A segment becomes wet after a raindrop falls on it. We can think of X j as the index of the segment hit by the jth raindrop. Thus, the rainstick process can be thought of as occurring on [0, 1], or on N + = {1, 2, . . .}. Let N be the first time that for some K, [1, K] is wet and [K, ∞) is dry.
To state our first three results, we define the constant b = log 2 − For nonnegative random variables X and Y we write X Y if P[X ≥ z] ≤ P[Y ≥ z] for all z ≥ 0. Theorem 1. K Geo(pe −b/p ). Thus, EK ≤ (1/p)e b/p and for all ǫ > 0 P[K > e (b+ǫ)/p ] → 0, as p → 0.
The proof begins by extending the rainstick process to all of the integers, with rain falling on j ∈ Z in continuous time at rate (1 − p) j−1 p. Let M t be rightmost wet point at time t. To obtain our upper bound we make the drastic modification that after the maximum M t increases, all of the sites in (−∞, M t ) are reset to being dry. Let G be the event that all sites j < M t become wet before the boundary moves again. We then show that P[G] ≥ e −b/p . Thus the number of tries we need to have the event G occur is stochastically dominated by a Geo(e −b/p ) random variable. Since the boundary increases by a Geo(p) distributed amount each time and the sum of a Geo(q) number of independent Geo(p) is Geo(pq) we obtain the stochastic upper bound on K. Given the drastic simplification in the last paragraph, it is surprising that the constant b is sharp.
To prove Theorem 2 we get an upper bound on P[K = k] by estimating the probability of the event G k,t = {all sites ≤ k are wet and all sites > k are dry at time t}.
One can write an explicit formula for P[G k,t ], see (7). Bounding ∞ 0 P[G k,t ] dt leads to the desired result.
Since the maximum of n geometrics grows logarithmically in n it follows easily from these two theorems that N grows as a double exponential: When p = 0.1, exp(e b/p ) = e 101,113 = 1.21 × 10 49313 so it seems unlikely that one could find our results by simulation.
Since the time grows doubly exponentially fast for the geometric as p → 0, it should not be surprising that when the tail of the distribution is stretched exponential there is positive probability that the process never terminates.
Theorem 4. Fix α ∈ (0, 1) and let K α be the size of the first block in a p-biased permutation where P[X 1 ≥ k] = C α e −k α with C α a normalizing constant.
To accomplish this, they needed an upper bound for the expected size of the first block. To find this they used a Markov chain M n defined by M n = max{M n−1 , Z n } − 1 where the Z i are i.i.d. geoemtric(p). To relate this to the rainstick construction, we use more colorful language to create a process that we call the paintstick process. Imagine that on the ith step, a paintball falls on Z i . All of the integers left of Z i are painted red to indicate that they must all be used before the first block is formed. The integer at Z i is removed, and the values to its right shift to the left by 1. Let M ′ n be the number of red sites at time n. The first block is created at K ′ = min{k : M ′ k = 0}. Unlike the rainstick process, every arrival changes the process, so the time to form a block and the block size are both equal to K ′ .
Taking the log of both sides of (1), then using that (1 − p) 1/p → e −1 we see that 6 . Our next result shows that the mean gives the typical block size not just an upper bound.
The analysis of the paintstick process is much easier than that of the rainstick since the state of the paintstick is always an interval while in the rainstick one has to understand the structure of the empty sites behind the front. Our proof of Theorem 7 does more than just confirm that the mean is accurate. It shows that the process returns to J ≈ (1/p) log(1/p) many times before a large deviation takes the chain to 0.
The results for the rainstick are proved in Section 2 in the order that they were stated. Our result for the paintstick is proved in Section 3.

Asymptotics for K and N for the rainstick
We adopt a continuous time perspective for how the X j are revealed. The advantage is that, by Poisson thinning, points arrive at each integer as independent Poisson processes. Let M t denote the maximum point in the process at time t with M 0 = X 1 . Keeping with the rainstick metaphor, we will refer to integers left of M t yet to be covered as dry sites, and covered integers as wet sites. Accordingly, let H t be the number of dry sites in [ 2.1. Proof of Theorem 1. Given M t = m, we may rescale time so that raindrops arrive at m at rate 1, and raindrops arrive at site m + ℓ at rate After this rescaling, a new maximal wet site arrives at rate If we allow for dry sites all the way to −∞, then we can extend the rate in (2) to all ℓ ∈ Z. Observe that by independence we have M t is unchanged when we allow for rain to arrive at the non-positive integers as well. To bound H t from above, we consider a "forgetful" version of this process: whenever M t increases we reset all of the sites in (−∞, M t ) to be dry. This modification has no effect on M t . LetĤ t be the number of dry sites in (−∞, M t ) in this forgetful process. The natural coupling ensures that H t ≤Ĥ t , so it takes longer forĤ t to reach 0 than its counterpart H t . Thus,η := inf{t :Ĥ t = 0} η. It follows that We now show that the probability that all of the sites behind M t become wet before M t increases is at least e −b/p .
Proof. Because of the time-scaling described at (2) and (3), the arrival time t 1 is exponential with rate p −1 (1 − p). Conditioning on t 1 gives Let α = log 2. The right side may be bounded below by restricting the domain of integration Taking the log of the infinite product and changing variables l = x/p we have The last sum is a Riemann sum that evaluates the function at the right endpoint of each interval. Since the integrand is increasing, the sum is larger than Therefore, which establishes the lemma.
We can now quickly deduce Theorem 1.
Proof of Theorem 1. Recall that (4) gives K Mη. By Lemma 8 we have all of the dry sites become wet with probability at least e −b/p . However, if the maximum increases, it does so by a Geo(p) distributed amount. When this occurs we reset all of the sites behind the new maximum to be dry, thus restarting the dynamics. It follows that The terms are independent because the amount the maximum increases by (Geo(p)) does not depend on how long it takes to cover all of the dry sites behind it. This is because we reset all sites (−∞, M n ) to be dry each time the maximum increases. In light of (4) we then have the claimed dominance K Mη Geo(pe −b/p ).

2.2.
Proof of Theorem 2. We need to show that for any ǫ > 0, if p is sufficiently small. Therefore, if α ∈ (0, 1), the probability that this occurs at least p −α times in succession is at least Since the maximum moves forward by at least p −1 at each occurrence, it follows that K ≥ p −α p −1 in this case. That is, P[K < p −(α+1) ] 6p 1−α as p → 0. The lemma now follows with, say, α = 3/4.
Our main estimate is an exponential bound for larger values of K.
Proposition 10. For any ǫ > 0, we have if p is sufficiently small.
With these two estimates it is elementary to prove Theorem 2.
Proof of Theorem 2. Start by decomposing P[K ≤ e (b−ǫ)/p ] as The first term vanishes as p → 0, by Lemma 9. The second term is bounded by which also vanishes, by Proposition 10 with ǫ ′ = ǫ/2.
It remains to prove Proposition 10. During the proof we will state two lemmas, which we establish immediately after.

Proof of Proposition 10.
In what follows we write u p as shorthand for exp(−b/p + o(1/p)). The constants implicit in o(1/p) may change from line to line, but in every instance this quantity is independent of k ≥ 2/p 3/2 . We will run the process at rate 1/ Let G j,t be the event that at time t (in this new time scale) all sites less than or equal to j are wet and all larger than j are dry. Call such a formation a j-block. We claim that With this new time scaling, drops fall on site j at rate (1 − p) j−k , and drops fall in the region [k + 1, ∞) at rate Therefore, given that a block forms at k, the block will have an expected lifetime of p 1−p before a drop falls in [k + 1, ∞). The bound (6) follows from this: To bound the right-hand side of (6), we note that by Poisson thinning At any time t there is an integer j(t) that maximizes P[G j,t ] over all j ≥ 0. Note that j(t) implicitly depends on k. We describe how P[G j(t),t ] behaves in the following two lemmas.
Lemma 12. Let t 0 = 3 log 2. For any n > 1, holds if p is sufficiently small (depending on n, but not on k).
The combination of Lemma 11 and Lemma 12 allows us to control the integral in (6) from t 1 to ∞: To control the integral over small times t < t 1 < ǫ, we use the bound for ǫ is small enough. Since b < 2 and since t 1 < ǫ for p small, it follows that Proof of Lemma 11. For fixed t, define the real number j * = j * (t) by t(1 − p) j * −k = log 2. From (7), we see that the ratio is greater than 1 if j < j * and it is less than 1 if j > j * . If j * ≥ 1, this shows that P[G j,t ] is maximized at j(t) = max{j ∈ Z : j ≤ j * }. Otherwise, P[G j,t ] is maximized at j = 0. Observe that j * (t) = j(t) = k when t = log 2. The fact that t(1 − p) j * −k = log 2 implies p(j(t) − k) = log(t/ log 2) + O(p) as p → 0. Now we estimate P[G j(t),t ], assuming t ≥ t 1 . Recall that k ≥ 2p −3/2 is assumed in Proposition 10. This and the assumption that t ≥ t 1 guarantees that j(t) ≥ p −3/2 . By (7) we have The last term is equal to −(1 − p) 1+j(t)−j * (t) log 2 which converges to − log 2 as p → 0, uniformly in k, since −1 ≤ j(t) − j * (t) ≤ 0. The first sum is a Riemann sum approximating an integral. Let us write the sum as where the index set is Since p(j(t) − k) = log(t/ log 2) + O(p), the upper index in the sum is This converges to − log log 2 as p → 0, uniformly over t ≥ t 1 and k ≥ 2p −3/2 . The lower index in I t is which converges to −∞ as p → 0, also uniformly over t ≥ t 1 and k. Consequently, and the convergence is uniform over t ≥ t 1 and k ≥ 2p −3/2 . The final equality gives the claimed bound on P[G j(t),t ].

Proof of Theorem 3.
Proof. Let β < α. If n = exp(e β/p ) the probability that a site beyond e α/p becomes wet before time n is as p → 0. Therefore, if a < a ′ < b, we must have Since P[K ≥ e a ′ /p ] → 1 by Theorem 2, this implies that P[N ≤ exp(e a/p )]) → 0.
If β > α, the probability in (12) goes to 1, as p → 0. This means that if c > c ′ > b, we must have P M n ≥ e c ′ /p , n = exp(e c/p ) → 1, as p → 0.
On the other hand, the event {K ≥ e c ′ /p } contains the event {M n ≥ e c ′ /p , n = exp(e c/p )}∩ {N ≥ e c/p }. Since P[K ≥ e c ′ /p ] → 0, by Theorem 1, we conclude that P[N ≥ e c/p ] → 0.
To deduce that EN = ∞ we observe that Indeed, given X 1 > 1, then the number of draws from (X j ) j≥2 to obtain the point Weighting these conditional expectations by P[X 1 = n] and summing over n ≥ 2 gives EN = ∞.

Proof of Theorem 4.
Proof. Scaling time to eliminate the normalizing constant C α , we can assume that rain lands on k at rate e −k α − e −(k+1) α for k = 1, 2, . . .. When the maximal wet site is at M t , let H t be the number of dry sites in [M t − ℓ(t), M t )) where ℓ(t) = M 1−α t /2. We divide by 2 so that if at time t there is a jump of more than k 1−α , which will increase the size of the viewing window, all of the sites within ℓ(t) of the boundary are vacant. The first block size K is finite only ifH t hits zero in finite time.
As before to simplify the arithmetic we run time at rate exp(M α t ). When we do this, jumps of M t larger than 2ℓ(t) occur at rate e −(Mt+2ℓ(t)) α e M α t ≈ e −α since when j ≪ k we have which is approximately α when j = k 1−α and k is large. Such jumps in M resetH to its maximal value of M 1−α /2.
Using the same reasoning as in the proof of Theorem 2, we see that H t jumps from i to i − 1 at a rate no more than r i = e α i/ℓ(t) + i/ℓ(t). If ǫ = e −α /(e α + 1) then when i/ℓ(t) < ǫ we have r i < e −α . So, given that the maximal wet site is at M t = k andH t = ℓ(t) = k 1−α /2 (its maximum possible value), the probability thatH becomes 0 before resetting to its maximum value is ≤ σ k = (1/2) ǫℓ(t) = (1/2) ǫk 1−α /2 . Since k σ k < ∞, it follows that with positive probabilityH will reset to its maximum value infinitely many times before hitting zero. Thus, P[K = ∞] > 0.

Asymptotics for K ′ for the paintstick
In this section we study paintstick process. Our main goal is to prove Theorem 7. To begin, we define the process. At any time the process consists of an interval 1, . . . , M ′ n of red sites. Red paint balls fall on i ≥ 1 with probability p(1 − p) i−1 . When a ball lands on i, the site i explodes, [i + 1, ∞) shifts to the left by 1, and sites [1, i − 1] are painted red. Given M ′ n = k, the process jumps [1, k]), k → m > k with probability p(1 − p) m (the ball must land at m + 1).
A ball landing on k + 1 does nothing. So, the probability that M ′ n increases when M ′ n = k is (1 − p) k+1 .
Thus, we have proved the desired result under the assumption that the chain starts at J rather than 0. To complete the proof we will show that (15) P 0 [T J < T 0 ] → 1.