Lecture 41 — Poisson Convergence Under Dependence, Derangements, Empty Boxes, and Compound Poisson Limits
This lecture extends Poisson convergence beyond independent Bernoullis to certain dependent structures such as random permutations and occupancy problems.
We conclude with the finite-dimensional Poisson limit theorem, which leads to compound Poisson distributions.
1. Reminder: Poisson Limit for Triangular Arrays of Bernoulli’s
(Page 1.)
Let $X_{n,k}\in{0,1,2,\dots}$.
Assume:
- Rare events: \(\sum_{k=1}^n P(X_{n,k}=1)\to\lambda.\)
- No dominant term: \(\max_{1\le k\le n} P(X_{n,k}=1)\to 0.\)
- Negligible “large counts”: \(\sum_{k=1}^n P(X_{n,k}\ge 2)\to 0.\)
Then:
\[S_n=\sum_{k=1}^n X_{n,k} \quad\Rightarrow\quad \mathrm{Poisson}(\lambda).\]Define:
\[Y_{n,k}=X_{n,k}\mathbf{1}_{\{X_{n,k}\le 1\}},\qquad T_n=\sum_{k=1}^n Y_{n,k}.\]From (3):
\[P(S_n\neq T_n) \le \sum_{k=1}^n P(X_{n,k}\ge 2)\to 0.\]So $S_n-T_n\to 0$ in probability, and since $T_n\Rightarrow\mathrm{Pois}(\lambda)$,
Slutsky gives $S_n\Rightarrow\mathrm{Pois}(\lambda)$.
2. Poisson Approximation for Dependent Events
Example: Fixed Points of a Random Permutation
(Pages 1–2.)
Let $\pi$ be a uniform random permutation of ${1,\dots,n}$.
Define indicator events:
Let:
\[S_n=\sum_{k=1}^n \mathbf{1}_{A_k}=\#\{\text{fixed points of }\pi\}.\]These events are not independent, but the dependence becomes negligible as $n\to\infty$.
Compute probabilities
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For a single fixed point: \(P(A_k)=\frac{(n-1)!}{n!}=\frac{1}{n}.\)
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For two fixed points: \(P(A_k\cap A_\ell)=\frac{(n-2)!}{n!}=\frac{1}{n(n-1)}.\)
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For $j$ fixed points: \(P(A_{i_1}\cap\cdots\cap A_{i_j}) =\frac{(n-j)!}{n!}.\)
Derive $P(S_n=0)$
By inclusion–exclusion (page 2):
\[\begin{aligned} P(S_n=0) &= 1 - \binom{n}{1}\frac{1}{n} + \binom{n}{2}\frac{1}{n(n-1)} - \binom{n}{3}\frac{1}{n(n-1)(n-2)} + \cdots \\ &= \sum_{k=0}^n \frac{(-1)^k}{k!} \ \xrightarrow{n\to\infty}\ e^{-1}. \end{aligned}\]Thus:
\[P(S_n=0)\to e^{-1}.\]Show full Poisson(1) convergence
We want:
\[P(S_n = k) \to \frac{e^{-1}}{k!},\qquad k=0,1,2,\dots\]Use the combinatorial identity (page 2):
\[P(S_n=k) = \frac{P(S_{n-k}=0)}{k!}.\]Since $P(S_{n-k}=0)\to e^{-1}$, we get:
\[P(S_n=k)\to \frac{e^{-1}}{k!}.\]Therefore:
\[S_n \Rightarrow \mathrm{Poisson}(1) \qquad\text{even though the events are dependent}.\]This is the classical result that the number of fixed points in a random permutation converges to Poisson(1).
3. Example: Number of Empty Boxes
(“Balls in Boxes” Occupancy Problem)
(Page 2–3.)
Place $n$ balls uniformly at random into $n$ boxes.
Let:
= the number of empty boxes.
Compute:
\[EY_k=P(\text{box }k\text{ empty}) =(1-1/n)^n\to e^{-1}.\]Thus:
\[ES_n = nEY_k \to \lambda = 1/e.\]Heuristic principle on page 3:
If $ES_n \to \lambda$, and dependence is weak, then $S_n\Rightarrow\mathrm{Pois}(\lambda)$.
Indeed, one can verify the triangular-array Poisson conditions by analyzing occupancy indicators.
Conclusion:
\[S_n = \#\{\text{empty boxes}\} \;\Rightarrow\; \mathrm{Poisson}(1/e).\]4. Important Extension: Removing Condition (3)
(Page 3.)
We now allow $X_{n,k}$ to take more than two values, still iid across $k$ for each fixed $n$, but not merely 0–1.
Let:
\[P(X_{n,k}=a_i)=p^{(n)}_i, \qquad i=1,\dots,I, \qquad 1\le k\le n.\]Assume that the limits:
\[n\,p^{(n)}_i \to \lambda_i,\qquad 1\le i\le I\]exist (finite).
If some $n p^{(n)}_i$ does not converge or does not go to 0 or finite limit, we cannot obtain a Poisson limit.
The limit law
Define independent Poisson variables:
\[Y_i \sim \mathrm{Poisson}(\lambda_i),\qquad 1\le i\le I.\]Then:
\[\sum_{k=1}^n X_{n,k} \;\Rightarrow\; \sum_{i=1}^I a_i Y_i.\]This is a compound Poisson distribution (page 3–4).
Interpretation:
- $Y_i$ counts how many times the value $a_i$ appears,
- Total number of summands is $T=\sum_i Y_i\sim\mathrm{Poisson}(\lambda)$,
where $\lambda = \sum_{i=1}^I \lambda_i$.
Thus:
\[\boxed{ \sum_{k=1}^n X_{n,k} \ \Rightarrow\ \sum_{i=1}^I a_i Y_i, }\]a fundamental finite-dimensional Poisson limit.
5. Compound Poisson Interpretation
(Page 4.)
Let ${Z_k}_{k\ge1}$ be iid with:
\[P(Z_k=a_i)=\lambda_i/\lambda, \quad \lambda=\sum_{i=1}^I \lambda_i.\]Let $T\sim \mathrm{Poisson}(\lambda)$, independent of ${Z_k}$.
Define the random sum:
\[\sum_{k=1}^T Z_k.\]Then (page 4):
\[P(Z_k=a_i) = \frac{\lambda_i}{\lambda},\]and the total sum has the same limit law as the previous Poisson decomposition:
\[\sum_{k=1}^T Z_k\ \overset{d}{=}\ \sum_{i=1}^I a_i Y_i.\]This is the standard representation of a compound Poisson random variable.
6. Exam Notes (Page 4)
The lecture ends with exam pointers:
- Be able to state the major limit theorems,
- Know how to prove convergence in distribution, Slutsky, etc.,
- CLT with truncation (Steps A, B, C),
- Lindeberg’s condition for triangular arrays,
- Poisson convergence,
- Characteristic function calculations.
Cheat–Sheet Summary — Lecture 41
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A general triangular array with rare events and negligible “≥2 jumps” satisfies: \(\sum_k X_{n,k}\Rightarrow\mathrm{Poisson}(\lambda).\)
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Fixed points in random permutations:
\(S_n \Rightarrow \mathrm{Poisson}(1),\) despite strong dependence. -
Empty boxes in occupancy with $n$ balls, $n$ boxes:
\(S_n \Rightarrow \mathrm{Poisson}(1/e).\) -
Multivariate Poisson limits:
If $n p_i^{(n)}\to \lambda_i$, then
\(\sum_{k=1}^n X_{n,k} \Rightarrow \sum_{i=1}^I a_i Y_i, \quad Y_i\sim\mathrm{Poisson}(\lambda_i).\) -
Representation as a compound Poisson: \(\sum_{i=1}^I a_i Y_i \stackrel{d}{=} \sum_{k=1}^T Z_k, \qquad T\sim\mathrm{Poisson}(\lambda), \ Z_k\sim (\lambda_i/\lambda).\)
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