lec16.typ

== Lecture 16

Start with 1 particle $X_0 = 1$.

Suppose that each particle in any generation gives birth to an IID number of particles before it dies, according to some *distribution*#footnote["offspring distribution" often denoted by $xi$.] with mean $mu$.

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Let $X_n$ = \# particles in $n$th generation.
($X_n$) is a MC.

$
  & X_0 = 1 \
  & X_n | (X_(n-1) = m) = sum_(i=1)^m xi_i^(n)
$

where $xi_i^(n)$ iid, $E xi = mu$.

*Q: Will MC ever visit (the absorbing state) 0 -- i.e. will the population die out eventually?*

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We will ignore the trivial case $P(xi = 1) = 1$. In this case each particle gives birth to 1 particle, $mu=1$, and the population survives.

#figure[
  #image(
    "./figs/p15_39m.png",
    height: 60%,
  )
]

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In all other cases, survival only depends on whether $mu > 1$ or $mu <= 1$.

Theorem
$(X_n)$ a BP with offspring distribution $xi$, $mu = E xi$. Suppose $P(xi=1) != 1$.
Then

$
  rho = P("extinct") & = P(X_n=0 " eventually") \
                     & = cases(
                         1 & quad mu <= 1,
                         <1 & quad mu > 1
                       )
$

Moreover, $rho = P("extinct")$ is smallest positive solution to $x = sum_k P(xi=k)x^k$.

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- The borderline case $mu=1$ is the most surprising.
- Also very useful that this result also gives us a formula for $P("survive") = 1 - rho$.

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Recall

$
  & X_n | (X_(n-1)=m) = sum_(i=1)^m xi_i \
  & => E(X_n | X_(n-1)=m) = m mu \
  & => E(X_n | X_(n-1)) = mu X_(n-1) \
  & => E(X_n) = mu E(X_(n-1))
$

Since $X_0 = 1$, this implies

$
  E(X_n) = mu^n quad "for all" n >= 0
$

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Hence

$
  E(X_n) ->^(n->oo) cases(
    0 & quad mu < 1,
    1 & quad mu = 1,
    oo & quad mu > 1
  )
$

By Markov's Inequality,

$
  & P(X_n > 0) <= E(X_n) -> 0 "if" mu < 1 \
  & therefore P("extinct") -> 1 "for" mu <= 1
$

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For $mu > 1$, we need to be more careful:

Recall, $rho = P("extinct")$. By FSA,
$
  rho = sum_(k=0)^oo P(xi=k) rho^k
$

Why?

#figure[
  #image(
    "./figs/p16_10m.png",
    width: 60%,
  )
]

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$
  phi(t) = sum_(k=0)^oo P(xi=k) t^k
$

is called the *generating function* of the RV $xi$.

From the previous slide,
$
  rho = phi(rho)
$

I.e. $rho$ is a root of $phi$.
Recall that the theorem claims that $rho$ is the *smallest* pos. root.

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Also note that $theta=1$ is always a root:

$
  phi(theta) = sum_(k=0)^oo P(xi=k) theta^k
$

Set $theta=1$:

$
  phi(1) = sum_(k=0)^oo P(xi=k) = 1
$

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*Lemma*. $rho$ is smallest pos. solution of $phi(theta) = theta$, $theta in [0,1]$.

*Proof*.

$
  underbrace(P(X_n=0), "Die out by time" n) = sum_(k=0)^oo P(xi=k) underbrace(P(X_(n-1)=0)^k, #footnote[All family trees of all k children of 1st particle must die out (independently) by time $n-1$.])
$

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Let $rho_n = P(X_n=0)$ \
By previous slide $rho_n = phi(rho_(n-1))$.

Note $rho_0 <= rho_1 <= rho_2 <= dots$ \
This is because $X_(n-1)=0 => X_n=0$ for any $n$. \
($A supset B => P(A) >= P(B)$)

All $rho_n <= 1$. So by calculus (Monotone Convergence Theorem) the sequence converges:

$
  lim_(n->oo) rho_n = rho_oo (= sup_n rho_n)
$

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$
  rho_n#footnote[$-> rho_oo$] = phi(rho_(n-1)#footnote[$-> rho_oo$]) => rho_oo = phi(rho_oo)
$

So $rho_oo$ is a solution to $theta = phi(theta)$.

To finish proof of lemma, we show $rho_oo$ = smallest sol. in $[0,1]$.

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Let $p$ = smallest pos. sol. to $phi(theta) = theta$ in $[0,1]$.

We show $p = rho_oo$.

Note: $phi(theta) = sum_(k=0)^oo P(xi=k) theta^k$ is increasing in $theta$, since all $P(xi=k) >= 0$.

$rho_0 = P(X_0=0) = 0 <= p$, since $X_0=1$.

Since $phi$ increasing, $phi(rho_0) <= phi(rho)$ \
$=> rho_1 <= rho$ since $phi(rho_0) = rho_1$, and $phi(rho)=rho$.

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Repeating argument,
all $rho_n <= rho$.

Take $n->oo$, $rho_oo <= rho quad square$

Using lemma, we can now study cases $mu>1$ and $mu=1$:

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#text(size: 11pt)[
  $mu > 1$
  If $P(xi = 0) = 0$, then clearly $rho = 0$. Also, $phi(theta) = sum_(k=1)^oo P(xi=k) theta^k$ (sum starts at $k=1$), so $phi(0) = 0$.

  If $P(xi=0) > 0$, then

  $
     phi(theta) & = sum_(k=0)^oo P(xi=k) theta^k \
    phi'(theta) & = sum_(k=1)^oo P(xi=k) dot k theta^(k-1) \
        phi'(1) & = sum_(k=1)^oo P(xi=k) dot k = mu.
  $

  If $mu>1$, slope of $phi$ at $theta=1$ is larger than 1 = slope of diagonal
]

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#figure[
  #image(
    "./figs/p16_32m.png",
    width: 80%,
  )
]

Blue: generating function $phi$ \
Black: diagonal $y = theta$ \
$phi(0) = P(xi = 0)$

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#text(size: 11pt)[
  $mu = 1$
  Recall, we exclude trivial case where $P(xi = 1) = 1$. Then $mu = 1$ and $rho = 0$.

  Suppose $mu=1$ & $P(xi=1)<1$. We show $phi$ has no root $theta < 1$.

  Note

  $
    phi'(theta) & = sum_(k=1)^oo P(xi=k) k theta^(k-1) \
                & < sum_(k=1)^oo P(xi=k) k = mu = 1
  $

  if $theta < 1$.
]

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$therefore$ If $theta < 1$ then

$
  integral_theta^1 phi'(u) d u & = phi(1) - phi(theta) \
                               & = 1 - phi(theta) \
                               & < integral_theta^1 1 d u = 1 - theta
$

$=> phi(theta) > 1 - (1-theta) = theta$.

$therefore phi(theta) > theta$ for $theta < 1$.

$therefore theta=1$ only root in $[0,1] quad square$

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Eg Binary Branching:

#figure[
  #image(
    "./figs/p16_43m.png",
    width: 100%,
  )
]

$
  & phi(theta) = sum_(k=0)^oo P(xi=k) theta^k = 1 - alpha + alpha theta^2 \
  & phi(theta) = theta => 0 = (theta-1)(alpha theta - (1-alpha)).
$

Roots are $theta=1$ & $theta = (1-alpha)/alpha < 1$ for $alpha > 1/2$ when $mu > 1$.

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We turn now to:
[D] §1.9 & 1.10 on *Exit Distributions & Times*. (Basically just FSA).

- We have already seen some of this in homework & workshops.
- Basis for this is FSA (First Step Analysis)

Eg: Gambler's ruin, $P_x ("Jackpot")$ & $P_x ("Ruin")$ is the exit distrib. from ${1, 2, dots, N-1}$ started at $x$.