MathIsimple

Probability Theory – Problem 2: Prove that follows a Poisson distribution with parameter

Question

A counting process {N(t),t0}\{N(t),t\geqslant0\} is defined to be a nonhomogeneous Poisson process with intensity function λ(t),t0\lambda\left(t\right),t\geqslant0 if:

(i) N(0)=0(i)\ N(0)=0

(ii) N(t),t0N(t),t\geqslant0 has independent increments

(iii) P{N(t+h)N(t)=1}=λ(t)h+o(h)(i i i)\ P\{N(t+h)-N(t)=1\}=\lambda\left(t\right)h+o\left(h\right)

(iv) P{N(t+h)N(t)2}=o(h).(i v)\ P\{N(t+h)-N(t)\geqslant2\}=o\left(h\right).

Prove that N(t+s)N(t)N(t+s)-N(t) follows a Poisson distribution with parameter tt+sλ(u)du\int_{t}^{t+s}\lambda\left(u\right)\,d u.

Step-by-step solution

Step 1. Partition the interval [t,t+s][t,t+s] into nn equal subintervals and set tk=t+ks/nt_k=t+k\,s/n. By the independent increments property, N(t+s)N(t)=k=0n1(N(tk+1)N(tk)).N(t+s)-N(t)=\sum_{k=0}^{n-1}\big(N(t_{k+1})-N(t_k)\big). Step 2. By definition, for each subinterval, P{N(tk+1)N(tk)=1}=λ(tk)sn+o ⁣(sn),P{N(tk+1)N(tk)2}=o ⁣(sn),P\{N(t_{k+1})-N(t_k)=1\}=\lambda(t_k)\frac{s}{n}+o\!\left(\frac{s}{n}\right), \qquad P\{N(t_{k+1})-N(t_k)\ge2\}=o\!\left(\frac{s}{n}\right), so each subinterval produces at most one jump with high probability. Step 3. Summing these probabilities yields k=0n1(λ(tk)sn+o ⁣(sn))tt+sλ(u)du,\sum_{k=0}^{n-1}\left(\lambda(t_k)\frac{s}{n}+o\!\left(\frac{s}{n}\right)\right) \to \int_t^{t+s}\lambda(u)\,du, which is a Riemann sum converging to the definite integral. Step 4. Since the subintervals are independent and each admits at most one jump with small probability, N(t+s)N(t)N(t+s)-N(t) is a sum of independent Bernoulli random variables. Letting nn\to\infty and applying the Poisson limit theorem, P{N(t+s)N(t)=k}=exp ⁣(tt+sλ(u)du)(tt+sλ(u)du)kk!.P\{N(t+s)-N(t)=k\} =\exp\!\left(-\int_t^{t+s}\lambda(u)\,du\right) \frac{\big(\int_t^{t+s}\lambda(u)\,du\big)^k}{k!}.

Final answer

QED.

Marking scheme

Here is the grading rubric for the problem.

1. Checkpoints (max 7 pts total)

Score exactly one chain | take the maximum subtotal among chains; do not add points across chains.

Chain A: Partition and Limit Method [Official Approach]

  • [1 pt] Partitioning: Divide the time interval [t,t+s][t, t+s] into nn equal (or arbitrary) subintervals and explicitly express the total increment N(t+s)N(t)N(t+s)-N(t) as the sum of the increments over each subinterval.
  • [1 pt] Local Probability Analysis: Correctly invoke conditions (iii) and (iv) to show that in the kk-th subinterval the probability of exactly one jump is approximately λ(tk)Δt\lambda(t_k)\Delta t (or λ(tk)sn\lambda(t_k)\frac{s}{n}), and the probability of two or more jumps is o(Δt)o(\Delta t).
  • [2 pts] Riemann Sum Limit: Sum the jump probabilities over all subintervals and correctly identify the limit as the definite integral tt+sλ(u)du\int_{t}^{t+s}\lambda(u)\,du.
  • *Note: If only the summation is written without passing to the integral limit, award 1 pt.*
  • [3 pts] Distribution Conclusion: Using the independent increments property (ii), invoke the Poisson Limit Theorem (or the limit distribution of a Bernoulli trial sequence) to conclude that the sum follows a Poisson distribution with the above integral as its parameter.
  • *Note: If only the expectation is shown to equal the integral without justifying why the distribution is Poisson, no credit is awarded for this checkpoint.*

Chain B: ODE / Probability Generating Function Method

  • [2 pts] Establishing the Difference Relation: Define Pn(τ)=P(N(t+τ)N(t)=n)P_n(\tau) = P(N(t+\tau)-N(t)=n) or the characteristic/generating function, and use the law of total probability together with conditions (iii)-(iv) to establish a relation between times τ\tau and τ+h\tau+h (e.g., Pn(τ+h)Pn(τ)(1λh)+Pn1(τ)λhP_n(\tau+h) \approx P_n(\tau)(1-\lambda h) + P_{n-1}(\tau)\lambda h).
  • [2 pts] Derivation of the ODE: Let h0h \to 0 and correctly derive the Kolmogorov forward equations Pn(τ)=λ(t+τ)Pn(τ)+λ(t+τ)Pn1(τ)P_n'(\tau) = -\lambda(t+\tau)P_n(\tau) + \lambda(t+\tau)P_{n-1}(\tau) (or the corresponding PDE for the generating function).
  • *Key point: The intensity function must be time-dependent λ(t+τ)\lambda(t+\tau), not a constant.*
  • [1 pt] Solving the Zeroth-Order Term: Correctly solve P0(τ)=exp(tt+τλ(u)du)P_0(\tau) = \exp\left(-\int_t^{t+\tau}\lambda(u)\,du\right), or obtain the logarithmic part of the characteristic function.
  • [2 pts] Induction or Inversion: Solve for the general term Pn(τ)P_n(\tau) by mathematical induction, or invert/Taylor-expand the characteristic function, to arrive at the complete Poisson distribution formula.

Total (max 7)


2. Zero-credit items

  • Transcription Only: Merely copying the definitions (i)-(iv) from the problem statement with no substantive derivation.
  • Circular Reasoning: Directly assuming N(t)N(t) is a nonhomogeneous Poisson process and invoking its distributional properties as a conclusion, without deriving them from the microscopic conditions (i)-(iv).
  • Unsupported Assertion: Writing down the final formula P(N(t+s)N(t)=k)=P(N(t+s)-N(t)=k) = \dots with no intermediate partition-and-sum or ODE derivation.

3. Deductions

  • [Cap at 3/7] Constant Intensity Fallacy: Assuming λ(t)λ\lambda(t) \equiv \lambda (constant) throughout the proof. This drastically simplifies the core difficulty of the problem (handling the variable-limit integral), so even if the logic is otherwise sound, the total score is capped at 3.
  • [-1] Missing o(h)o(h) Terms: Completely ignoring or mishandling the o(h)o(h) terms when deriving the ODE or taking limits (e.g., writing =0=0 instead of \approx or o(h)o(h)), thereby compromising mathematical rigor.
  • [-1] Incorrect Integration Limits: Writing the wrong limits of integration in the final result or derivation (e.g., 0sλ(u)du\int_0^s \lambda(u)du without accounting for the time shift tt; the correct form is tt+sλ(u)du\int_t^{t+s}\lambda(u)du or equivalently 0sλ(t+u)du\int_0^s \lambda(t+u)du).
  • [-1] Logic Gap: In Chain A, after computing the probability sum, directly asserting the Poisson distribution without mentioning that the subinterval increments are independent or invoking the rare-event principle.
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