- #1

docnet

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- Homework Statement
- Please see the red step, I don't think it's a correct step because the parts don't add up right.

- Relevant Equations
- .

Let ##N_t## be the Poisson point process with the probability of the random variable ##N_t## being equal to ##x## is given by $$\frac{(\lambda t)^xe^{-\lambda t}}{x!}.$$ ##N_t## has stationary and independent increments, so for any ##\alpha\geq 0, t\geq 0,## the distribution of ##X_t = N_{t+\alpha} -N_t## is independent of ##t.##

I'm trying to show this explicitly.

$$

\begin{align*}

X_{t}&= N_{t+\alpha}-N_{t} \\

&= N_{t} + N_\alpha -N_{t} \\

&= N_\alpha \\

&=\frac{(\lambda \alpha)^xe^{-\lambda \alpha}}{x!} .

\end{align*}$$

I don't think it's correct because

$$\begin{align*}

N_{t} + N_\alpha &= \frac{(\lambda t)^xe^-{\lambda t}}{x!} +\frac{(\lambda \alpha)^xe^{-\lambda \alpha}}{x!} \\

&\neq \frac{(\lambda (t+\alpha))^xe^{-\lambda (t+\alpha)}}{x!}= N_{t+\alpha} .

\end{align*}$$

I'm trying to show this explicitly.

$$

\begin{align*}

X_{t}&= N_{t+\alpha}-N_{t} \\

&= N_{t} + N_\alpha -N_{t} \\

&= N_\alpha \\

&=\frac{(\lambda \alpha)^xe^{-\lambda \alpha}}{x!} .

\end{align*}$$

I don't think it's correct because

$$\begin{align*}

N_{t} + N_\alpha &= \frac{(\lambda t)^xe^-{\lambda t}}{x!} +\frac{(\lambda \alpha)^xe^{-\lambda \alpha}}{x!} \\

&\neq \frac{(\lambda (t+\alpha))^xe^{-\lambda (t+\alpha)}}{x!}= N_{t+\alpha} .

\end{align*}$$

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