The Poisson process is a type of [[Stochastic Process]] that models the occurrence of random, [[Independence of Events|independent events]] over time or space. The number of events occurring in any fixed time interval follows a [[Poisson Distribution]]. The Poisson process is the continuous-time counterpart of the [[Bernoulli Process]], therefore sharing similar assumptions. ## Assumptions - *Independence:* The number of successes in any two disjoint time intervals are independent of each other. I.e. the number of successes during one section does not tell me anything about the number of successes in another section. ![[poisson-process-independence.png|center|400]] - *Homogeneity:* The probability of $k$ successes during a fixed interval length $\tau$ is always the same, no matter where we put the interval. Therefore $\mathbf P(k, \tau)$ is fully defined without any reference to the starting time $t$. ![[poisson-process-homogeneity.png|center|400]] - *Small interval probabilities:* During a very small interval $\delta$ there cannot be gt;1$ successes. The probability of success is the success rate per unit of time $\lambda$, multiplied by the length of the time interval $\tau$. $ \mathbf P(k, \tau) \approx \begin{cases} 1 - \lambda \tau &\text{if } k=0\\ \lambda \tau &\text{if } k=1\\ 0 &\text{if } k>1 \end{cases} $ ## Time Until First Success We want to find the [[Probability Density Function|PDF]] of the time until the first success in a Poisson process. We begin by establishing the [[Cumulative Density Function|CDF]] until the first success $F_{T-1}(t)$. ![[poisson-process-first-success.png|center|400]] $ F_{T_1}(t)=\mathbf P(T_1\le t)= 1 - \mathbf P(T_1>t) = 1- \mathbf P(0,t) $ The probability that the time of the first success $T_1$ happens before $t$, is the same as $1$ minus the probability that $T_1$ happens after $t$. The latter implies that there have been $0$ successes until $t$, which we can plug-into our Poisson formula. $ \begin{align} \mathbf P(k, \tau)&=\frac{(\lambda \tau)^k}{k!}*e^{-\lambda \tau} \\[8pt] \implies \mathbf P(0,t)&=\frac{(\lambda t)^0 * e^{-\lambda t}}{0!} = e^{-\lambda t} \end{align} $ Now we can define the CDF, and differentiate to get to the PDF $f_{T_1}(t)$, which turns out to follow an [[Exponential Distribution]]. $ \begin{align} F_{T_1}(t)&=1-\mathbf P(0,t)= 1- e^{-\lambda t} \\[6pt] f_{T_1}(t)&=\int_{-\infty}^\infty 1- e^{-\lambda t}=\lambda e^{-\lambda t} \end{align} $ **Time until k-th success:** We denote the time until the $k$-th success in a Poisson process as $Y_k$, which follows an [[Erlang Distribution]]: $ f_{Y_k}(y) \approx\frac{\lambda^{k} y^{k-1} * e^{- \lambda y}}{(k-1)!}$ ## Merged Poisson Process A Poisson process is fully defined by the parameter $\lambda \tau$, which is the product of the success rate $\lambda$ and the interval length $\tau$. When two independent processes $X_1 \sim \text{Pois}(\lambda_1)$ and $X_2 \sim \text{Pois}(\lambda_2)$ are merged (i.e. occur simultaneously), the result is also a Poisson process $X_3$: $ X_3 \sim \mathrm{Pois}(\lambda_1+\lambda_2)$ ![[poisson-process-merged.png|center|400]] To prove that the merged process $X_3$ is also a Poisson process, it needs to meet assumptions. **Independence:** Since both individual processes are Poisson, this means that they are already independent of time. Also independence is given between the individual processes. Therefore the merged process is also independent. **Homogeneity:** In a Poisson process the success rate has to be constant over time. Wrapping up the matrix above, gives following probabilities. By neglecting all $\delta^2$ terms for $X_3$, we can prove that the merged process has a success rate $\lambda_3 = \lambda_1+ \lambda_2$. $ X_3 = \begin{cases} 0: & 1- (\lambda_1+ \lambda_2)*\delta \\ 1: & (\lambda_1+\lambda_2)*\delta \end{cases} $ **Small Interval Probabilities:** When $\delta$ is small the probability of $\ge 2$ successes during that interval $\approx 0$. During a small interval of length $\delta$ the following combinations out of the two processes are possible: | | $X_1=0$ | $X_1=1$ | $X_1=2$ | | ------- | ------------------------------------------- | --------------------------------------- | ------------------------------------ | | $X_2=0$ | $(1-\lambda_1 \delta)*(1-\lambda_2 \delta)$ | $\lambda_1 \delta*(1-\lambda_2 \delta)$ | $O(\delta^2) * (1-\lambda_2 \delta)$ | | $X_2=1$ | $(1-\lambda_1 \delta)* \lambda_2\delta$ | $\lambda_1 \delta * \lambda_2 \delta$ | $O(\delta^2) * \lambda_2 \delta$ | | $X_2=2$ | $(1- \lambda_1 \delta)*O(\delta^2)$ | $\lambda_1 \delta * O(\delta^2)$ | $O(\delta^2)*O(\delta^2)$ | Since $\delta$ is a very small number, every term that includes $\delta^2 \approx 0$ and can be neglected, which cancels out all scenarios with successes $\ge 2$. | | $X_1=0$ | $X_1=1$ | $X_1=2$ | | ------- | ------------------------------------------- | --------------------------------------- | ------- | | $X_2=0$ | $(1-\lambda_1 \delta)*(1-\lambda_2 \delta)$ | $\lambda_1 \delta*(1-\lambda_2 \delta)$ | 0 | | $X_2=1$ | $(1-\lambda_1 \delta)* \lambda_2\delta$ | 0 | 0 | | $X_2=2$ | 0 | 0 | 0 | Rewrite the remaining terms. | | $X_1=0$ | $X_1=1$ | $X_1=2$ | | ------- | ------------------------------------------------------------------ | ---------------------------------------------- | ------- | | $X_2=0$ | $1-\lambda_2 \delta-\lambda_1 \delta - \lambda_1\lambda_2\delta^2$ | $\lambda_1 \delta- \lambda_1\lambda_2\delta^2$ | 0 | | $X_2=1$ | $\lambda_2\delta - \lambda_1\lambda_2\delta^2$ | 0 | 0 | | $X_2=2$ | 0 | 0 | 0 | Again, the newly identified terms with a factor $\delta^2 \approx 0$. This results in a final table of probabilities for each outcome. | | $X_1=0$ | $X_1=1$ | $X_1=2$ | | ------- | ------------------------------------- | ------------------ | ------- | | $X_2=0$ | $1-\lambda_2 \delta-\lambda_1 \delta$ | $\lambda_1 \delta$ | 0 | | $X_2=1$ | $\lambda_2\delta$ | 0 | 0 | | $X_2=2$ | 0 | 0 | 0 | ## Attributing Successes in the Merged Process In a merged process $X_3 \sim \mathrm{Pois}(\lambda_1+\lambda_2)$, we want to determine from which original process $(X_1$ or $X_2)$ a given success originated. **Attribution rule:** The conditional probability that a success in $X_3$ at time $t$ comes one of the original processes (e.g. $X_1$), given that a success occurred at time $t$ in general: $ \mathbf P(X_1 \vert \text{success at} \: t)= \frac{\lambda_1}{\lambda_1+ \lambda_2}$ **Derivation:** Using [[Bayes Rule]] for conditional probabilities. $ \mathbf P(A \vert B) = \frac{\mathbf P(A \cap B)}{\mathbf P(B)} = \frac{\lambda_1 \delta}{(\lambda_1+\lambda_2)\delta}=\frac{\lambda_1}{(\lambda_1+\lambda_2)} $ where: - Intersection $(A \cap B)$: A success happens at time $t$ and originated from $X_1$. - Condition $(B)$: A success happens at time $t$ from either process. ## Split Poisson Process A Poisson process $X$ can be split into two (or more) new Poisson processes $\{X_1,X_2\}$, by assigning each success from $X$ to one of the new processes with probabilities $\{q,\, 1-q\}$. * $q$: Probability of assigning a success to $X_1$ * $(1-q)$: Probability of assigning a success to $X_2$ ![[poisson-process-split.png|center|400]] The resulting new processes $\{X_1,X_2\}$ are both Poisson processes, meeting the assumptions. - *Independence:* In the original process $X$ there is independent between all $X_i$. Also the allocation parameter $q$ is independent of $X$. - *Homogeneity:* This is given since the probability of success is linear to $\delta$, so relative to the length of the time window, no matter at which point of the process. - *Small interval probabilities:* In the original process the probability of $\ge 2$ arrivals at the same time is already negligible as it is $O(\delta^2)$. In any of the split processes it is even smaller, since these double arrivals still have to be allocated to the same sub-process. $ X_1: \begin{cases} \ge 2 & O(\delta^2) \\ =1 & \lambda\delta q\\ =0 & \text{otherwise} \end{cases} $ $ X_2: \begin{cases} \ge 2 & O(\delta^2) \\ =1 & \lambda\delta (1-q)\\ =0 & \text{otherwise} \end{cases} $ >[!note:] >In contrast to Bernoulli process, the split processes are independent of each other. Knowing that a success happened in $X_1$ at a specific point in continuous time does not provide insights on $X_2$. This is because that specific point in time is infinitely small.