Poisson process is identical on equal intervals?

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

The latex formatting issues are fixed.

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##.

Orodruin said:

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.

Orodruin said:

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

Orodruin said:

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##?

docnet said:

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.

docnet said:

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.

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.