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It's pretty straightforward, thanks to the fact that the interarrival times of a Poisson process are exponentially distributed and the exponential distribution has the memoryless property. (As an asi …

1) Think of each of those $k+m$ events as having occurred either before $t$ or after $t$, and you're trying to calculate the probability that exactly $k$ of them occurred before $t$, which leaves exac …

Your mistake lies in the transition between this statement: $$\mathbb{E}[S_4 | N(1) = 2] = \mathbb{E}[S_2 + T_3 + T_4 | N(1) = 2]$$ and this statement (the next line): $$\mathbb{E}[S_4 | N(1) = 2] …

Let us assume we have a Poisson process with an arrival rate of $\lambda$. After some time $t$, $N_t$ unobserved arrivals have occurred. After some more time, say $\tau$, we observe that $N_{t+\tau} …

Unfortunately, the maximum likelihood estimate of the rate parameter for a Poisson process that's sampled over a predetermined interval $T$ does not have a finite mean (or higher moments). This is be …

You've sort of mixed up two distinct approaches to this problem, hence your understandable confusion. One approach is to ignore the time dimension and simply generate a sample of $Z_i, i \in \{1, \do …

Each Poisson arrival will see the system either in state $E$ or not-$E$ ($\bar{E}$). One might suspect that we can construct confidence intervals for the long run fraction of the time in state $E$ ( …

You can find the areas of the counties in California from https://en.wikipedia.org/wiki/List_of_counties_in_California. This will enable you to generate a count $n_i$ from the appropriate Poisson dis …

We can use the law of total expectations to simplify the problem as follows: $$\mathbb{E} N(\min(t,T^*)) = \lambda \mathbb{E}\min(t,T^*)$$ Finding $\mathbb{E} \min(t,T^*)$ is straightforward: $$\mathb …

You have made a slight, but important, mistake in your likelihood function. Let's try constructing it differently, breaking the $T=13,500$-length time period into $13,500$ time periods of length $1$ …

It's a mixture of three distributions, and can be found pretty easily by brute force, if one allows oneself to handwave over some important details (e.g., "is $\lambda < \mu$"). Let $n$ be the numb …

1) Yes, the mean of the service time distribution is just the mean of the Gamma(2,$\lambda)$ distribution. 2) The traffic intensity of the system is the arrival rate / the service rate, in this case: …

Here's another approach that uses a common trick with characteristic functions to avoid having to work out the sums / integrals. I'll set $\lambda = 1$ without loss of generality, it simplifies nota …