Bernoulli/Poisson processes

Bernoulli Processes

Flip $n$ coins, each of which comes up heads with probability $p$: this is a Bernoulli process (so named because each coin is just a Bernoulli random variable).

Regarding this process, we can ask several interesting questions:

1. What’s the chance that we get exactly $k$ heads?
2. What’s the chance that the index of the first coin which comes up heads is $i$ (one-indexed)?
3. If $n$ is infinite, what is the expected value of $i$ (first index w/ heads)?

Answers

1. $\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$
2. $(1-p{)}^{i-1}p$
3. $\frac{1}{p}$. One simple way to obtain this result is to notice that $\mathbb{E}[i]=1+(1-p)\mathbb{E}[i]$.

Currently, events are discrete, since they are in chunks of one coin flip at a time. How can we make them continuous instead?

Poisson Processes

Instead of a process parameterized by $p$, we now use $\lambda$ to denote the expected # of successes (heads) within each one-second interval. In other words, the chance of a success in an interval of size $\mathrm{d}t$ is $\lambda \mathrm{d}t$.

First off, what is $P(0,\tau )$, the chance that there are no successes at all in a $\tau$-second interval? Note that

$$P(0,\tau +\mathrm{d}t)=P(0,\tau )(1-\lambda \mathrm{d}t)=P(0,\tau )-\lambda P(0,\tau )\mathrm{d}t$$

This differential equation is precisely the definition of the exponential function! That is:

$$P(0,\tau )=e^{-\lambda \tau }$$

What about $P(1,\tau )$ the chance of one success in a $\tau$-second interval? Integrating over all possible times for this single success, we get that:

$$P(1,\tau )=e^{-\lambda \tau }{\int }_0^{\tau }\frac{\lambda \mathrm{d}t}{1-(\lambda \mathrm{d}t)}=e^{-\lambda \tau }{\int }_0^{\tau }\lambda \mathrm{d}t=e^{-\lambda \tau }\lambda \tau$$

Essentially, we start assuming that all trials failed, then choose one failure to “toggle” to a success. Notice that the denominator vanishes because $1-\lambda \mathrm{d}t=1$.

Now, in general, what is $P(k,\tau )$, the chance of $k$ successes in a $\tau$-second interval? Applying the same logic, we get that

$$P(k,\tau )=e^{-\lambda \tau }\frac{(\lambda \tau {)}^k}{k!}$$

Another derived random variable to consider is $T$, the time of first success. Note that $1-\text{cdf}(T)=P[T\ge \tau ]=P(0,\tau )=e^{-\lambda \tau }$. From this, we can derive the PDF:

$$\text{pdf}(T)=-\frac{\mathrm{d}}{\mathrm{d}\tau }P[T\ge \tau ]=\lambda e^{-\lambda \tau }$$

Note that $\mathbb{E}[T]$ is still $\frac{1}{\lambda }$, since $\mathbb{E}[T]=\mathrm{d}t+(1-\lambda \mathrm{d}t)\mathbb{E}[T]$.

A more interesting problem: what’s $\mathbb{E}[T^n]$? We can use a similar argument as we did for expected value (taking advantage of the fact that the Poisson process is memoryless) to get that

$$\mathbb{E}[T^n]=(1-\lambda \mathrm{d}t)(\mathbb{E}[T^n]+\mathrm{d}t\mathbb{E}[T^{n-1}])$$

Rearranging, we get:

$$\lambda \mathrm{d}t\mathbb{E}[T^n]=\mathrm{d}t\mathbb{E}[T^{n-1}]$$

Here, we use the fact that $(1-\lambda \mathrm{d}t)\mathrm{d}t=\mathrm{d}t$ since $(\mathrm{d}t{)}^2=0$. From here, we can show that

$$E[T^n]=\frac{n!}{{\lambda }^n}$$

As a direct consequence, $\text{Var}(T)=\frac{1}{{\lambda }^2}$.

Now, let $N$ be the number of success in a $\tau$-second interval. Then, we can show that $E[N]=\text{Var}(N)=\lambda \tau$. The fact that $E[N]=\text{Var}(N)$ is actually very special, since typically mean and variance have different dimensions and thus this equality is highly unlikely. However, $N$ is a “counting” variable, which means it’s dimensionless and thus the above expression makes perfect sense.