Confused about when to treat probability calculation as small interval and use $\lambda$ in Poisson process I am doing MITx 6.431x and I am really confused about Poisson process, in particular about when to treat the calculation as a small interval and when not.
Suppose we are merging two Poisson processes, say Red and Green. Then P(the kth arrival is red)=$\frac{\lambda}{\lambda_{Red}+\lambda_{Green}}$.
Why? This is as if we are looking at a small interval here. Precisely speaking, we are taught to use the following when $\tau$ as an interval is 'small'.
$P(k,\tau) \approx
\begin{cases}
\ 1-\lambda\tau,  & \text{if $k$ =0} \\[2ex]
\lambda\tau, & \text{if $k$ =1} \\[2ex]
\ 0, & \text{if $k$ >1}
\end{cases}$
$\frac{\lambda}{\lambda_{Red}+\lambda_{Green}}$ is different from $\lambda\tau$ obviously, but in the lecture the former is derived from the latter, which seems to suggest we are still assuming a small interval. Furthermore, why shouldn't we use, say, $p_{N_\tau}(k)=P(k,\tau)=\frac{(\lambda \tau)^ke^{-\lambda \tau}}{k!}$ to calculate, where $p_{N_\tau}(k)$ means the probability of having $k$ arrivals in interval $\tau$?
In the lecture, $\lambda$ is used to calculate probability when it is explicitly said that we are only concerning ourselves with small intervals. But it seems really vague when exactly do we use $\lambda$, consider the following:

All ships travel at the same speed through a wide canal. Each ship
takes $t$ days to traverse the length of the canal. Eastbound ships
(i.e., ships traveling east) arrive as a Poisson process with an
arrival rate of $λ_E$ ships per day. Westbound ships (i.e., ships
traveling west) arrive as an independent Poisson process with an
arrival rate of $λ_W$ ships per day. A pointer at some location in the
canal is always pointing in the direction of travel of the most recent
ship to pass it.
What is the probability that the next ship to pass will be westbound?

The solution here tells me that P(next ship is westbound)=$\frac{\lambda_W}{\lambda_{E}+\lambda_{W}}$. But nothing in the question suggests that we are considering a 'small' interval. Why shouldn't we use $p_{N_\tau}(k)=P(k,\tau)=\frac{(\lambda \tau)^ke^{-\lambda \tau}}{k!}$?
 A: So, the small interval thing is actually not particularly relevant here. You can take the small interval thing to be somewhat of a heuristic derivation of how a Poisson process is derived. (If you sum up n i.i.d. Bernoulli random variables with parameter $1/p$, you get a random variable $X_n$ whose distribution converges to Poisson(1)). You can argue loads of properties through it heuristically (and rigorously too, though at that point, arguing the properties through actual definitions of Poisson processes is simply easier.)
Under the hood, in this problem, you are actually using the splitting property of Poisson processes, if you are interested, their proofs you can find in Durrett EOSP (https://services.math.duke.edu/~rtd/EOSP/EOSP2E.pdf; Theorem 2.11). I will elaborate further on your ship example to show you the power of splitting.
Intuitively, for the ship problem, consider a big ship company that sends ships according to a Poisson process with rate $\lambda_W + \lambda_E$. Now, you have a conductor that tells each ship independently to be westbound with probability $\frac{\lambda_W}{\lambda_W + \lambda_E}$ and eastbound otherwise. What is remarkable is that now, both your eastbound and westbound ships are also going to be independent, and they will be Poisson processes with rates $\lambda_E$ and $\lambda_W$ respectively. So, conversely, whenever you are given two independent Poisson processes, you can imagine that they are actually coupled somewhere in the background together, and this magical splitting makes two independent streams. Hopefully from this view, it is clear as to why the probability of the first ship being westbound after any time $t$ will be $\frac{\lambda_W}{\lambda_W + \lambda_E}$.
Let me address your actual question: Here, it is kind of hard to work with the Poisson distribution itself, since when we are looking at the first ship, we are kind of subtly conditioning on it, and thus, the distributions kind of change. There is no fixed $\tau$; it is more like, at the random time $T$ when the next ship passes, what is the probability that it is westbound?
A: This may help: let $T_{r},T_{g}$ be the first arrival times for the "red" and "green" processes, denoted $N_r(t)$ and $N_g(t)$.  Then, $P(T_{r}>t)=P(N_{r}(t)=0)=e^{-\lambda_{r}t}$, so $T_{r}$ has an exponential distribution with rate $\lambda_{r}$; similarly $T_{g}$ is exponential with rate $\lambda_{g}$.
So, to see the probability "red" arrives first, assuming the "red" and "green" processes are independent, we can compute $$P(T_{r}<T_{g})=\iint_{\{(s,t):0<s<t\}} \lambda_{g}e^{-t\lambda_g} \lambda_r e^{-s\lambda_r}dsdt
$$
$$
\int_0^\infty\int_0^t\lambda_{g}e^{-t\lambda_g} \lambda_r e^{-s\lambda_r}dsdt = \int_0^\infty \lambda_ge^{-\lambda_g t}(1-e^{-\lambda_rt})dt
$$
$$
=1-\frac{\lambda_g}{\lambda_g+\lambda_r} = \frac{\lambda_r}{\lambda_r+\lambda_g}.
$$
