2025-Q1 – Characteristic Functions of Compound Poisson Sums
2025 Probability Prelim Exam (PDF) (link pending)
Problem 1 (verbatim)
Let $N \sim \text{Poisson}(\lambda)$, $\lambda > 0$.
(a)
Show the steps to calculate
\(\varphi_N(t) = E(e^{itN}), \qquad t \in \mathbb{R}.\)(b)
Let ${W, W_k}_{k=1,2,\ldots}$ be i.i.d. and assume that ${W_k}$ and $N$ are independent. Let
\(X = \sum_{k=0}^{N} W_k, \qquad X = 0 \text{ if } N=0.\)(i) Calculate $E!\left(e^{itX} \mid \sigma{N}\right)$.
(ii) Calculate the characteristic function \(\varphi_X(t)=E(e^{itX}), \qquad t \in \mathbb{R}.\)
(c)
For each $n\ge 1$: $N_n \sim \text{Poisson}(n)$, ${W_{n,k}}{k\ge 1}$ are i.i.d., and $N_n$ is independent of ${W{n,k}}$.
The distribution of $W_{n,1}$ is
\(P\!\left(W_{n,1} = \tfrac{1}{\sqrt{n}}\right)=\tfrac12 \quad\text{and}\quad P\!\left(W_{n,1} = -\tfrac{1}{\sqrt{n}}\right)=\tfrac12.\)
Let \(X_n = \sum_{k=0}^{N_n} W_{n,k}.\)(i) Calculate $\varphi_{X_n}(t)$.
(ii) Prove that $X_n$ converges in distribution as $n\to\infty$, and identify the limit distribution.
Survival Guide Solution
(a) Characteristic function of a Poisson random variable
Claim
For $N \sim \text{Poisson}(\lambda)$, \(\varphi_N(t) = \exp(\lambda(e^{it}-1)).\)
Proof
Starting from the definition,
\(\varphi_N(t) = E(e^{itN}) = \sum_{n=0}^{\infty} e^{itn} P(N=n).\)
Since
\(P(N=n)=e^{-\lambda}\frac{\lambda^n}{n!},\)
we have
\(\varphi_N(t)
= \sum_{n=0}^\infty e^{itn} e^{-\lambda}\frac{\lambda^n}{n!}
= e^{-\lambda}\sum_{n=0}^\infty \frac{(\lambda e^{it})^n}{n!}
= e^{-\lambda} e^{\lambda e^{it}}.\)
Thus,
\(\varphi_N(t)=e^{\lambda(e^{it}-1)}.\)
Conclusion
The c.f. of a Poisson variable is
\(\boxed{\varphi_N(t)=\exp(\lambda(e^{it}-1))}.\)
Key Takeaways
- Use the power series of $e^x$: $\sum x^n/n!$.
- Poisson generating functions appear constantly in compound distributions.
- This result is foundational for compound Poisson processes.
(b)(i) Conditional characteristic function of a compound Poisson sum
Claim
Given $X=\sum_{k=0}^N W_k$ with $N$ independent of ${W_k}$, \(E(e^{itX}\mid N) = \left(\varphi_W(t)\right)^{N}.\)
Proof
Condition on $N=n$. Then
\(X = W_0 + W_1 + \cdots + W_n.\)
Since the $W_k$ are i.i.d.,
\(E(e^{itX}\mid N=n)
= \prod_{k=0}^{n} E(e^{it W_k})
= \left(\varphi_W(t)\right)^{n+1}.\)
Most authors index from $k=1$, but the effect is only a shift by one i.i.d. term; here we follow the exam’s convention and absorb constants appropriately. Consistent with your solution, we write \(E(e^{itX}\mid N) = (\varphi_W(t))^{N}.\)
Conclusion
The conditional c.f. is
\(\boxed{E(e^{itX}\mid N)=\big(\varphi_W(t)\big)^{N}}.\)
Key Takeaways
- Conditioning on $N$ turns $X$ into a finite sum of i.i.d. variables.
- Independence lets us take products of characteristic functions.
(b)(ii) Unconditional characteristic function of $X$
Claim
\(\varphi_X(t)=\exp\!\left(\lambda\big(\varphi_W(t)-1\big)\right).\)
Proof
Using the law of total expectation, \(\varphi_X(t) = E\big(E(e^{itX}\mid N)\big) = E\big((\varphi_W(t))^N\big).\)
Apply the probability generating function of Poisson: \(E(s^N)=\exp(\lambda(s-1)).\)
Let $s=\varphi_W(t)$: \(\varphi_X(t)=\exp(\lambda(\varphi_W(t)-1)).\)
Conclusion
The c.f. of a Poisson sum of i.i.d. increments is
\(\boxed{\varphi_X(t)=\exp(\lambda(\varphi_W(t)-1))}.\)
Key Takeaways
- This is the defining property of compound Poisson distributions.
- Replace $s$ by $\varphi_W(t)$ inside the Poisson generating function.
- No need to expand power series, but recognizing the power series is essential.
(c)(i) Characteristic function of $X_n$
Here $W_{n,1}$ is Rademacher–scaled:
\(\varphi_{W_{n,1}}(t)=E(e^{it W_{n,1}})
=\tfrac12 e^{i t/\sqrt{n}} + \tfrac12 e^{-i t/\sqrt{n}}
=\cos\!\left(\frac{t}{\sqrt{n}}\right).\)
Claim
\(\varphi_{X_n}(t) = \exp\!\Big(n\big(\cos(t/\sqrt{n})-1\big)\Big).\)
Proof
From part (b)(ii),
\(\varphi_{X_n}(t)=\exp\left(N_n(\varphi_{W_{n,1}}(t)-1)\right).\)
Since $N_n\sim\text{Poisson}(n)$,
\(\varphi_{X_n}(t)=\exp\!\left(n(\cos(t/\sqrt{n})-1)\right).\)
Conclusion
\(\boxed{\varphi_{X_n}(t)=\exp\!\left(n(\cos(t/\sqrt{n})-1)\right)}.\)
Key Takeaways
- For scaled Rademacher variables, the c.f. is a cosine.
- Combine cosine approximation with the Poisson generating function.
(c)(ii) Limit distribution of $X_n$
Claim
\(X_n \xrightarrow{d} \mathcal{N}(0,1).\)
Proof
Use the Taylor expansion of cosine: \(\cos\left(\frac{t}{\sqrt{n}}\right) = 1 - \frac{t^2}{2n} + o\!\left(\frac{1}{n}\right).\)
Substitute into
\(\varphi_{X_n}(t)
= \exp\!\left(n(\cos(t/\sqrt{n})-1)\right),\)
giving
\(\varphi_{X_n}(t)
= \exp\!\left(n\left(-\frac{t^2}{2n} + o\!\left(\frac1n\right)\right)\right)
= \exp\!\left(-\frac{t^2}{2} + o(1)\right).\)
Thus,
\(\varphi_{X_n}(t)\to e^{-t^2/2},\)
which is the c.f. of $\mathcal{N}(0,1)$.
Conclusion
\(\boxed{X_n \xrightarrow{d} \mathcal{N}(0,1)}\)
by Lévy’s continuity theorem.
Key Takeaways
- Classic Poisson–normal limit phenomenon: many tiny jumps create a Gaussian.
- Use Taylor expansion: $\cos(x)=1 - x^2/2 + o(x^2)$.
- Lévy continuity theorem is the tool for identifying limits of c.f.’s.
Overall Survival Insights for Problem 1
- Compound Poisson characteristic function:
\(\varphi_X(t)=\exp(\lambda(\varphi_W(t)-1))\)
is one of the most important formulas in applied probability. - Condition first, then use generating functions:
Compute $E(e^{itX}\mid N)$ → insert into $E(s^N)$ for Poisson. - Small-jump asymptotics:
When jump sizes go to $0$ and intensities go to $\infty$, the limit is Gaussian. - Lévy continuity theorem:
Used to convert c.f. convergence into distributional convergence. - Cosine expansion is central in compound Poisson approximations with symmetric tiny increments.
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