Poisson and renewal processes

Idea: Arrivals with no memory

A Poisson process of rate $\lambda$ has independent counts in separate windows.

• Count in time $t$: mean and variance both $\lambda t$
• Blocks at $9$ an hour, window of $20$ minutes: $9\times \frac{20}{60}=3$
• Answer with $\lambda t$ first, then compute

Formula: Poisson count

$P(N(t)=k)=e^{-\lambda t}\frac{(\lambda t{)}^k}{k!}$

The chance of exactly k arrivals in a window of length t.

Idea: Gaps forget the past

Gaps are independent exponentials with mean $1/\lambda$. Nothing is ever due.

• $P(T>s+u\mid T>s)=P(T>u)$
• After a quiet half hour, the next block is no closer
• Empty window with $\lambda t=3$: $e^{-3}\approx 0.0498$

Formula: Merging streams

$P(\text{from}\text{stream}a)=\frac{a}{a+b}$

Merged Poisson streams add their rates; each arrival's source is a coin with these odds.

Quick check: Is the next order a buy?

Buys arrive at $6$ a minute and sells at $3$, both Poisson. What is the chance the next order is a buy?

Example: Where did the blocks fall?

$4$ blocks came in $1$ hour. What is the chance that exactly $1$ fell in the first $15$ minutes?

1. Given the count, the times are uniform: $\mathrm{Binomial}\left(4,\frac{1}{4}\right)$
2. $\left(\genfrac{}{}{0ex}{}{4}{1}\right)\times \left(\frac{1}{4}\right)\times {\left(\frac{3}{4}\right)}^3=\frac{27}{64}$

Idea: Renewal: gaps of any shape

Independent gaps with one distribution and mean $\mu$ give arrivals at rate $1/\mu$.

• Reward per unit time: $\frac{E[R]}{E[X]}$, a ratio of means
• Gaps of $1$ or $19$ seconds, chances $0.9$ and $0.1$: mean $0.9\times 1+0.1\times 19=2.8$

Example: The inspection paradox

Gaps are $1$ second with chance $0.9$, $19$ with chance $0.1$. You arrive at a random moment. Expected length of your gap?

1. $E[X]=0.9\times 1+0.1\times 19=2.8$
2. $E[X^2]=0.9\times 1+0.1\times 361=37$
3. Your gap: $\frac{37}{2.8}=\frac{185}{14}$

Common mistake: The average gap is not your gap

Long gaps cover more of the clock, so a random moment lands in them more often.

Instead Use $\frac{E[X^2]}{E[X]}$. For Poisson, your gap has mean $2/\lambda$, not $1/\lambda$.

Idea: Say it out loud

Name the property you use: independent counts, no memory, merging or splitting.

• Real trades come in bursts: a constant rate is a first model
• Count variance above the mean: a Poisson error bar is too narrow