lec19.typ

== Lecture 19

The Poisson Process

*Def*
$(N_t : 0 <= t#footnote[continuously] < oo)$ is a rate $lambda$ Poisson Process — $"PP"(lambda)$ — if

1. $N_0 = 0$
2. $N_(t+s) - N_s =^d "Poisson"(t lambda)$, *for any* $t,s >= 0$
3. Indep. increments:
$
  N_(t_1) - N_(t_0), N_(t_3) - N_(t_2), dots, N_(t_n) - N_(t_(n-1))#footnote[sequence of disjoint intervals]
$
indep. for *any* $t_0 < t_1 < dots < t_n$.

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#text(size: 9pt)[
  $
    N_(t_1) - N_(t_0) & = "# points in" (t_0, t_1] \
    "(2.)" & =^d "Poisson"(lambda (t_1 - t_0)#footnote[length of $(t_0, t_1]$])
  $

  $
    "(3.)" "*Indep.* of" N_(t_2) - N_(t_1) & = "# points in" (t_1, t_2] \
    & =^d "Poisson"(lambda (t_2 - t_1)#footnote[length of $(t_1, t_2]$])
  $

  #figure[
    #image(
      "./figs/p19_06m.png",
      width: 50%,
    )
  ]

  (1.) $N_0 = 0$ because no points yet at time 0.
]

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In particular, all

$
  N_t#footnote[$=N_t - N_0$] & = "# points in" (0, t] \
                             & = "# points by time" t \
                             & =^d "Poisson"(lambda t).
$

This is why it is called a Poisson Process.

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*Constructing a rate $lambda$ Poisson Process:*

$tau_1, tau_2, dots " IID Exp"(lambda)$

Let $T_n = sum_(i=1)^n tau_i =^d "Gamma"(n, lambda)$.

If we think of $tau_i =$ lifetime of $i$th lightbulb, then $T_n$ is the time at which we have gone through $n$ lightbulbs in total.

#figure()[
  #image(
    "./figs/p19_12m.png",
    width: 60%,
  )
]

$T_0 = 0, quad T_1 = tau_1, quad T_2 = tau_1 + tau_2$

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Let $N_t = max {n: T_n <= t} = "# of * to left of" t$

#figure[
  #image(
    "./figs/p19_15m.png",
    width: 85%,
  )
]

Poisson is discrete in values but continuous in time.

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We will show that $(N_t)_(t>=0)$ is a $"PP"(lambda)$.

Clearly $N_0 = max{n>=0: T_n <= 0} = 0$.

So we need to check (2.) & (3.) in definition above.

First step:

*Lemma.* $N_t ~ "Poisson"(lambda t)$

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#text(size: 10pt)[
  *Proof.* ${ N_t = n } = { T_n <= t < T_(n+1) }.$

  I.e., to have exactly $n$ points by time $t$, we need $n^"th"$ point to arrive by time $t$ & $(n+1)^"th"$ point to arrive at some later time $>t$.

  $
    therefore P(N_t = n)#footnote[$T_n$, $T_(n-1)$ indep, why?] = integral_0^t f_(T_n)(s)#footnote[PDF for arrival time of $n^"th"$ point, $T_n = sum_(i=1)^n tau_i$] P(tau_(n+1) > t-s) d s
  $

  #figure[
    #image(
      "./figs/p19_21m.png",
      width: 60%,
    )
  ]
]

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$
  & T_n ~ "Gamma"(n, lambda), quad f_(T_n)(s) = lambda e^(-lambda s) ((lambda s)^(n-1)) / ((n-1)!) \
  & tau_(n+1) ~ "Exp"(lambda), quad P(tau_(n+1) > t-s) = e^(-lambda(t-s))
$

$
  therefore P(N_t = n) & = integral_0^t lambda e^(-lambda s) ((lambda s)^(n-1))/((n-1)!) e^(-lambda(t-s)) d s \
  & = e^(-lambda t) lambda^n/((n-1)!) integral_0^t s^(n-1) d s = e^(-lambda t) ((lambda t)^n)/(n!) \
  & = P("Poisson"(lambda t) = n).
$

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We now check parts (2.) & (3.) in definition of $"PP"(lambda)$.

The key is LoM#footnote[lack of memory] & Lemma.

Proof of (2): $N_(t+s) - N_s ~ "Poisson"(lambda t).$

#figure[
  #image(
    "./figs/p19_30m.png",
    width: 60%,
  )
]

By LoM the lightbulbs burning at time $s$ starts afresh.

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So if we ignore $*$ to left of $s$, & start counting $*$ starting at time $s$, we have another IID sequence of $"Exp"(lambda)$ inter-arrival times.

So by Lemma, the \# of $*$ between $s$ & $s+t$ is $"Poisson"(lambda t)$.
Note the \# of such $*$
is $N_(s+t)#footnote[\# of $*$ to left of $t+s$] - N_s#footnote[\# of $*$ to left of $s$]$.

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Moreover, observe that

$
  (N_(t+s) - N_s)_(t>=0) quad \& quad (N_r)_(r<=s)
$

are independent.#footnote[This follows by 1. LoM, 2. $tau_i$ are IID.]

This is because the length of time until a \* after $s$ is indep. of whatever happened before time $s$ (again by LoM).

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#text(size: 11pt)[
  Finally, we check (3.)

  Proof of indep. increments:

  This follows by induction.
  The previous slide shows $(N_(t+t_(n-1)) - N_(t_(n-1)))_(t>=0)$ & $(N_r)_(r<=t_(n-1))$ indep.

  $therefore$ In particular,

  $
    underbrace(N_(t_n) - N_(t_(n-1)), #text[depends on \ $(N_(t+t_(n-1)) - N_(t_(n-1)))_(t>=0)$]), underbrace(#text[$N_(t_(n-2)) - N_(t_(n-3)), dots, N_(t_1) - N_(t_0)$], #text[depends on \ $(N_r)_(r<=t_(n-1))$])
  $

  are indep.
]

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We'll skip §2.2.2 on "more realistic models" for now.
$"PP"(lambda)$ pts arrive at rate $lambda$, independent of time $t$.

The non-homogeneous#footnote[$lambda_t$ depends on $t$] Poisson Process on p.105 is interesting.
We may put something on HW \#3 about this -- take a look yourself.

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§2.3 -- Compound Poisson Processes.

It is often useful to add one more layer of randomness.

*Eg* $(N_t)_(t>=0)$ is a $"PP"(lambda)$ modelling occurrences of earthquakes over time on Hayward fault.

Moreover, suppose each earthquake

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Has an IID magnitude $Y_i$ which is also indep. of $(N_t)$.

Then

$
  S_t = sum_(i=1)^(N_t) Y_i#footnote[Compound Poisson Process.]
$

is the total magnitude felt along fault by time $t$.

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Note: For a regular $"PP"(lambda)$, all $Y_i = 1$.

- More examples of compound Poisson Processes in §2.2.

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*Theorem.* $(N_t)$ a $"PP"(lambda)$.
$(Y_i)$ an IID sequence indep. of $(N_t)$.

Let

$
  S_t = sum_(i=1)^(N_t) Y_i
$

denote the Compound Poisson Process.

Then

$
    "E"S_t & = lambda t "E"Y \
  "Var"S_t & = lambda t "E"(Y^2).
$

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In a similar way, we can calculate 2nd moment:

$
  "E"(S_t^2) &= sum_(n=0)^oo P(N_t = n) "E"(S_t^2 | N_t = n) \
  &= sum_(n=0)^oo P(N_t = n) [n "Var"Y + (n "E"Y)^2#footnote[$"Var"X = "E"(X^2) - ("E"X)^2$]] \
  &= "Var"Y "E"(N_t) + ("E"Y)^2 "E"(N_t^2) \
  &= lambda t ( "Var"Y + ("E"Y)^2 ) \
  &= lambda t "E"(Y^2) quad square
$

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#text(size: 10pt)[
  Proof

  $
    "E"(S_t) &= sum_(n=0)^oo P(N_t = n) "E"(S_t | N_t = n) \
    &= sum_(n=0)^oo P(N_t = n) n "E"Y#footnote[$S_t | (N_t=n) = sum_(i=1)^n Y_i$] \
    &= "E"Y sum_(n=0)^oo n P(N_t = n) \
    &= "E"Y dot "E"(N_t) \
    &= lambda t "E"Y#footnote[$N_t ~ "Poisson"(lambda t)$]
  $
]

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*Eg* Customers arrive at a store according to a rate $lambda = 81$ Poisson Process.

Suppose each customer spends an IID amount of money with mean \$8 and SD \$6. Find the mean revenue after 1 day, i.e. at time $t=1$. Also find its SD.

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$
  R = sum_(i=1)^(N_1) Y_i, quad "E"Y = 8, quad "Var"Y = 36, quad N_1 ~ "Poisson"(81)
$

By theorem, $"E"R = 81#footnote[$lambda t = lambda dot 1 = 81$] dot 8#footnote[$"E"Y$] = \$648$

$
  \& "Var"R & = 81 "E"(Y^2) \
            & = 81 ("Var"Y + ("E"Y)^2) \
            & = 81 (36 + 64) = 8,100
$

$therefore "SD"(R) = sqrt(8100) = \$90$