9-probability - Processes

The sample space for bernoulli process is the set of all binary sequences.

For a given number of trials, how many successes do we have? call this random variable $S_n$ (number of successes in $n$ trails)
What is $p_{S_n}$? well for any $k=0\to n$,
$p_{S_n}(k)=nCk{p}^k(1-p{)}^{n-k}$
$E[S_n]=np$, $var(S_n)=np(1-p)$
Next, for a given number of successes, how many trials do we have? call is $T_n$ (number of trails for $n$ successes)
What is $p_{T_n}(k)$ for any $k\in \mathbb{N}$, that is, what is the probability that I spend $k$ trials to get $n$ sucesses?
well if $k<n$, then the probability is zero.
if $k=n$, then the probability is simply $p_{T_n}(n)=p^n$.
what is $p_{T_n}(n+r)$? well, the last $n+r$th trial has to be a success, else if we have $n$ successes before the last trial, we have used less than $n+r$ trails to get $n$ successes, and if we have less than $n-1$ successes before the last trial, we we never get $n$ successes including the last trail.
Therefore $p_{T_n}(n+r)=p\cdot p_{S_{n+r-1}}(n-1)$
That is, the last trial has to be a success, and independently, given the previous $n+r-1$ trials, we need a total of $(n-1)$ successes. Hence, $$p_{T_{n}}(n+r) =(n+r-1) \ C \ (n-1) \ p^n(1-p)^r$$ (pascal PMF)

Now what is $E[T_n]$? Well $E[T_n]=\sum _{r=0}^{\mathrm{\infty }}{p}_{T_n}(n+r)\ast (n+r)$ which seems harder to calculate.
Well the memoryless property (which comes from independence) is an absolute UNIT here.
Let us say your cousin, saw $n-1$ successes so far, where $T_{n-1}$ models the Random variable, which depicts the numbs of trails needed to see $n-1$ successes. And now you walk into the room. How many more trials do you need to see another success? that is modeled by $T_1$ of course. So $E[T_n]=E[T_{n-1}]+E[T_1]$. Doing this repeatedly,
we have $E[T_n]=nE[T_1]=\frac{n}{p}$.
Similarly, $var(T_n)=n\cdot var(T_1)=\frac{n(1-p)}{p^2}$

Poisson process

He we talk about continuous time, and use the idea of time homogeneity. In any time interval $\tau$, the probability of $k$ arrivals in that time is modeled by $p(k,\tau )$. Equal time intervals have equal probability for any number of arrivals.
and disjoint time intervals and independent.
for any small time interval $\delta$, we have $p(k,\delta )\approx 1-\lambda \delta \text{if}k=0,\lambda \delta \text{if}k=1,0\text{if}k>1$.
Discretize the interval $[0,t]$ as $[0,\delta ],(\delta ,2\delta ],\dots (n-1\delta ,n\delta ]$. whenre $n\delta =t$.
Then each discretized interval acts as a Bernoulli trial.

Hence we ask the question again:
For a length of interval, discretized as $t=n\delta$, Let the random variable $A_t$ denote the number of arrivals in the interval $[0,t]$. Then, the probability mass, approximated by discretization looks like a bernoulli process $p_{A_t}(k)=nCk(\lambda \delta {)}^k\cdot (1-\lambda \delta {)}^{n-k}$

now, $p_{A_t}(k)=\underset{n\to \mathrm{\infty }}{lim}nCk{\left(\frac{\lambda t}{n}\right)}^k\cdot (1-\frac{\lambda t}{n}{)}^{n-k}$.
Without doing any calculus, because we don't want to, we write the result
$p_{A_t}(k)=\frac{(\lambda t{)}^k{e}^{-\lambda t}}{k!}$. (poission distribution)
$E[A_t]=\lambda t=var(A_t)$. this is pretty cool! so the standard deviation of the poission is the square root of the mean.

Now, for a number of arrivals $k$, let the random variable $T_k$ denote the length of time interval needed to see $k$ arrivals.
using the anologous arugment, if we want to see $k$ the time $t$, we have to see exactly one arrival in the interval $(t-\delta ,t]$ and exactly have $k-1$ arrivals in the time $[0,t-\delta ]$.
Since the two intervals are disjoint, and since $T_k$ is a continuous random variable
we have $f_{T_k}(t)\delta \approx \underset{\delta \to 0}{lim}{p}_{A_{t-\delta }}(k-1)\cdot \lambda \delta$ Do some rigorous cancellation of the deltas, and some plug and chug

$T_1=\lambda e^{-\lambda }$.
And due to the fact that $T_k$ is memoryless,
$E[T_k]=kE[T_1]=k\lambda e^{-\lambda }$.

Merging and splitting:

Poisson processes:

Remember that the arrival rate $\lambda$ is enough to determine a poisson process. Imagine we have two poison signals ${\lambda }_1,{\lambda }_2$. And a machine that records an arrival, if in some small time interval, any one or both of the poison signals record an arrival.
let us use some notation and call $Q({\lambda }_1)$ and $Q({\lambda }_2)$ for the two Poisson signals, and $Q\ast$ for the merged signal.
Then, for $Q\ast$ to record an arrival in the small interval $\delta$, we have three mutually exclusive cases: only $Q({\lambda }_1)$ records an arrival, with probability ${\lambda }_1\delta (1-{\lambda }_2\delta )$, only $Q({\lambda }_2)$ records an arrival, with probability $(1-{\lambda }_1\delta ){\lambda }_2\delta$, or they both do, with prob ${\lambda }_1{\lambda }_2{\delta }^2$.
And so the probability that $Q\ast$ records a signal in the small interval $\delta$ is the sum of these three possibilities which is $({\lambda }_1+{\lambda }_2)\delta$. Therefore $Q\ast$ is eqivalent to the poisson process with arrival rate ${\lambda }_1+{\lambda }_2$ !!!!!!!! (we have to ingore higher order terms like ${\delta }^2$).

In the small interval $\delta$, given that we record in arrival in $Q\ast$, what is the probability that $Q({\lambda }_1)$ also recorded a signal in that interval, using conditioning. we get (ignoring ${\delta }^2$ terms) that given $Q\ast$ recorded an arrival, the probability that $Q({\lambda }_1)$ recorded that signal is $\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2}$.
Analogously we can merge to bernoulli processes, $B(p_1),B(p_2)$ and the probability that in any single trail, the merged $B\ast$ succeeds is $p1(1-p2)+p2(1-p1)+p1p2$ hence $B\ast$ is the bernoulli process with each trail succeeded with probability $p1+p2-p1p2\approx p_1+p_2$.

Splitting poison

Given a poisson signal $Q\ast$ we split it into two signals $Q_1,Q_2$, such that (indepedently) in any time interval of length $\delta$, if $Q\ast$ records a signal, with probability $p$ I sent it to $Q_1$ and with probability $1-p$ I send it to $Q_2$.
It is not difficult to see that $Q_1$ is a possion process with arrival rate $\lambda p$ and $Q_2$ is one with arrival rate $\lambda (1-p)$, where $\lambda$ is the arrival rate of $Q\ast$ itself.

We can analogously split the one Bernoulli process into two this way as well. :)