Poisson process is identical on equal intervals?

[docnet]

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

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

 

[docnet]

The latex formatting issues are fixed.

 

[Orodruin]

You are not trying to show the correct thing. What your last equation would imply is that the probability of $N_t$ and $N_\alpha$ being $x$ is equal to the probability of $N_{\alpha+t}$ being $x$.

What you should be showing is that the probability of $N_{\alpha+t}$ being $x$ is the same as the sum of the probabilities of $N_t$ being $y$ at the same time as $N_\alpha$ is $x-y$ for all $y$.

A random variable is not equal to the probability of it being $x$.

 

[docnet]

You are not trying to show the correct thing. What your last equation would imply is that the probability of $N_t$ and $N_\alpha$ being $x$ is equal to the probability of $N_{\alpha+t}$ being $x$.

You're so right! I knew it didn't quite make sense but I didn't have the knowledge to pinpoint the reason. Thank you.

What you should be showing is that the probability of $N_{\alpha+t}$ being $x$ is the same as the sum of the probabilities of $N_t$ being $y$ at the same time as $N_\alpha$ is $x-y$ for all $y$.

Would it be better, then, to say
$$
\begin{align*}
N_{t+\alpha} -N_t&=\Big(\mathbb{E}\Big[ \frac{(\lambda t)^ye^{-\lambda t}}{y!} \Big]+\frac{(\lambda (t+\alpha))^xe^{-\lambda (\alpha)}}{x!}\Big)-\mathbb{E}\Big[ \frac{(\lambda t)^ye^{-\lambda t}}{y!} \Big]\\
&= \frac{(\lambda \alpha)^xe^{-\lambda (\alpha)}}{x!}= N_{\alpha} ?
\end{align*}$$

A random variable is not equal to the probability of it being $x$.

You're right! So instead the probability function of the random variable $N_t$ gives the probabilities of it being $x$, given that $x=0$ with probability $1$ at $t=0$?

 

[Orodruin]

Would it be better, then, to say
$$
\begin{align*}
N_{t+\alpha} -N_t&=\Big(\mathbb{E}\Big[ \frac{(\lambda t)^ye^{-\lambda t}}{y!} \Big]+\frac{(\lambda (t+\alpha))^xe^{-\lambda (\alpha)}}{x!}\Big)-\mathbb{E}\Big[ \frac{(\lambda t)^ye^{-\lambda t}}{y!} \Big]\\
&= \frac{(\lambda \alpha)^xe^{-\lambda (\alpha)}}{x!}= N_{\alpha} ?
\end{align*}$$

Not really. You are still mixing up the probabilities with the random variable itself.

You're right! So instead the probability function of the random variable $N_t$ gives the probabilities of it being $x$, given that $x=0$ with probability $1$ at $t=0$?

Yes, which is definitely not the same as $N_t$ itself. The probsbilities are any numbers between 0 and 1 and $N_t$ is a discrete RV which takes imteger values.

 

[docnet]

So I looked at my notes again, and it turns out one way to show (maybe it's wrong) that distribution of $N_{\alpha+t}-N_t$ is independent of $t$ is by using the definition of the counting process: for any $\alpha,t\geq 0$, $N_{\alpha+t}-N_t$ equals the number of events that occur in the interval $(\alpha,t]$, then use the fact that $N_t$ has stationary increments to say $N_{\alpha+t}-N_t=N_\alpha$ for all $\alpha,t\geq 0$.

Is it ok to say that the independence of the increments of $N_t$ is not required because the stationary property? I thank you for your effort and patience in helping me learn Stochastic processes.