To solve the problem, let’s analyze each part:

Problem Summary:

1. $N(t)$ is a Poisson process with rate parameter $\lambda$, meaning $N(t) \sim \text{Poisson}(\lambda t)$.
2. $X(t) = N(t) - \lambda t$ defines a new process.

We are tasked to compute:

• $\mathbb{E}[X(t)]$,
• $\mathbb{E}[X(t) | X(s) = x]$ for $s < t$.

Step 1: Find $\mathbb{E}[X(t)]$

From the definition of $X(t)$: $X(t) = N(t) - \lambda t.$

The expectation of $N(t)$ is: $\mathbb{E}[N(t)] = \lambda t.$

Using linearity of expectation: [ \mathbb{E}[X(t)] = \mathbb{E}[N(t)] - \lambda t = \lambda t - \lambda t = 0. ]

Thus: $\mathbb{E}[X(t)] = 0.$

Step 2: Find $\mathbb{E}[X(t) | X(s) = x]$, $s < t$

Key Property:

The increments of a Poisson process are independent. For $s < t$, the process $N(t) - N(s)$ follows a Poisson distribution with parameter $\lambda (t - s)$.

Rewrite $X(t)$ and $X(s)$:

$X(s) = N(s) - \lambda s,$ $X(t) = N(t) - \lambda t.$

Conditional on $X(s) = x$, we can express $N(s)$ as: $N(s) = x + \lambda s.$

For $t > s$, the increment $N(t) - N(s)$ is independent of $N(s)$ and has distribution: $N(t) - N(s) \sim \text{Poisson}(\lambda (t - s)).$

Thus: $N(t) = N(s) + (N(t) - N(s)).$

Substitute $N(t)$ into $X(t)$: $X(t) = N(t) - \lambda t = \big(N(s) + (N(t) - N(s))\big) - \lambda t.$

Using $N(s) = x + \lambda s$: $X(t) = \big((x + \lambda s) + (N(t) - N(s))\big) - \lambda t.$

Simplify: $X(t) = x + (N(t) - N(s)) - \lambda (t - s).$

Conditional Expectation:

Given $X(s) = x$, the term $N(t) - N(s)$ has expectation: $\mathbb{E}[N(t) - N(s) | X(s) = x] = \lambda (t - s).$

Substitute this into $X(t)$: [ \mathbb{E}[X(t) | X(s) = x] = x + \mathbb{E}[N(t) - N(s) | X(s) = x] - \lambda (t - s). ]

$\mathbb{E}[X(t) | X(s) = x] = x + \lambda (t - s) - \lambda (t - s).$

Simplify: $\mathbb{E}[X(t) | X(s) = x] = x.$

Final Results:

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

Would you like further clarification or additional examples?

Related Questions:

1. What is the variance of $X(t)$?
2. How does $X(t)$ behave as $t \to \infty$?
3. What is the distribution of $X(t)$?
4. Can we generalize the result for $\mathbb{E}[X(t) | N(s) = n]$?
5. What happens if $\lambda$ is time-dependent?

Tip:

Always use the independence property of Poisson increments when working with conditional expectations in Poisson processes.