Math Problem Statement

Let N(t) be a Poisson process with parameter λ. Define a new process X(t) = N(t) - λt. Find E[X(t)] and E[X(t) | X(s) = x] where s < t.

Solution

We are tasked with analyzing the process X(t)=N(t)λtX(t) = N(t) - \lambda t, where N(t)N(t) is a Poisson process with rate λ\lambda. Let's solve for the required expectations.

1. Finding E[X(t)]\mathbb{E}[X(t)]

Since N(t)N(t) is a Poisson process with parameter λt\lambda t, we know: E[N(t)]=λt.\mathbb{E}[N(t)] = \lambda t. Thus, for X(t)=N(t)λtX(t) = N(t) - \lambda t: [ \mathbb{E}[X(t)] = \mathbb{E}[N(t)] - \lambda t = \lambda t - \lambda t = 0. ]

2. Finding E[X(t)X(s)=x]\mathbb{E}[X(t) \mid X(s) = x], where s<ts < t

We need the conditional expectation E[X(t)X(s)=x]\mathbb{E}[X(t) \mid X(s) = x]. To analyze this, note that: X(t)=N(t)λt,X(s)=N(s)λs.X(t) = N(t) - \lambda t, \quad X(s) = N(s) - \lambda s. This implies: N(t)=X(t)+λt,N(s)=X(s)+λs.N(t) = X(t) + \lambda t, \quad N(s) = X(s) + \lambda s. Since N(t)N(t) is a Poisson process, the increments N(t)N(s)N(t) - N(s) are independent of N(s)N(s) and follow a Poisson distribution with parameter λ(ts)\lambda (t-s). Therefore: N(t)N(s)Poisson(λ(ts)).N(t) - N(s) \sim \text{Poisson}(\lambda (t-s)). Using this, we write: X(t)=N(t)λt=(N(s)+(N(t)N(s)))λt.X(t) = N(t) - \lambda t = \big(N(s) + (N(t) - N(s))\big) - \lambda t. Substituting N(s)=X(s)+λsN(s) = X(s) + \lambda s, we get: X(t)=(X(s)+λs)+(N(t)N(s))λt.X(t) = (X(s) + \lambda s) + (N(t) - N(s)) - \lambda t. Simplify: X(t)=X(s)+(N(t)N(s))λ(ts).X(t) = X(s) + (N(t) - N(s)) - \lambda (t-s). Now take the expectation conditional on X(s)=xX(s) = x. Since N(t)N(s)N(t) - N(s) is independent of N(s)N(s), and its mean is λ(ts)\lambda (t-s): [ \mathbb{E}[X(t) \mid X(s) = x] = x + \mathbb{E}[N(t) - N(s)] - \lambda (t-s). ] Substitute E[N(t)N(s)]=λ(ts)\mathbb{E}[N(t) - N(s)] = \lambda (t-s): E[X(t)X(s)=x]=x+λ(ts)λ(ts)=x.\mathbb{E}[X(t) \mid X(s) = x] = x + \lambda (t-s) - \lambda (t-s) = x.

Final Results

  1. E[X(t)]=0\mathbb{E}[X(t)] = 0.
  2. E[X(t)X(s)=x]=x\mathbb{E}[X(t) \mid X(s) = x] = x.

Let me know if you want further clarifications!

Relative Questions

  1. How does the independence of Poisson increments influence the conditional expectation?
  2. What would change if the Poisson process had a time-varying rate λ(t)\lambda(t)?
  3. Can we generalize this result for a different transformation of N(t)N(t)?
  4. How does this relate to martingale properties of X(t)X(t)?
  5. What are practical scenarios where such processes might be modeled?

Tip

Always consider the independence and distributional properties of increments when analyzing Poisson processes.

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Math Problem Analysis

Mathematical Concepts

Poisson Process
Conditional Expectation
Stochastic Processes

Formulas

E[N(t)] = λt for Poisson process
E[X(t)] = E[N(t)] - λt
X(t) = X(s) + (N(t) - N(s)) - λ(t-s)

Theorems

Independence of Poisson Increments
Conditional Expectation Properties

Suitable Grade Level

University (Advanced Probability/Statistics)