Heavy Watal

Wakeley輪読会 2章2節

Book
Coalescent Theory — An Introduction
Author
John Wakeley
Publisher
Roberts & Company
Errata
PDF
輪読担当
岩嵜航
日程
2015-03-20

2. Probability Theory

2.1 Fundamentals

2.1.1 Events, Probabilities, and Random Variables

2.1.2 Probability Distributions

2.2 Poisson Processes

The backbone of the neutral coalescent:

Poisson distribution
the number of events that occur over a fixed period of time
Exponential distribution
waiting time until a first event occurs

The poisson process is a counting process.

$$\begin{split} P[K(t) = 1] &= \lambda t + o(t) \\ P[K(t) \ge 2] &= o(t) \end{split}$$
where
$K(t)$ is the number of observed events before $t$. $K(0) = 0$
$\lambda$ is the rate of occurrence per unit time
$o(t)$ goes to faster than $t$
These implies, within a sufficiently short period of time, $\delta t$,
$P[K(\delta t) = 1] \approx \lambda \delta t$
$P[K(\delta t) \ge 2]$ is negligible (i.e., two events don’t occur at the same time)

The number of events over time 0 to $t$ (or from arbitrary starting time $s$ to $s + t$) is Poisson distributed, for $k = 0, 1, 2, …$,

$$\begin{split} P[K(t) = k] \;&=\; \frac {(\lambda t)^k} {k!} e^{-\lambda t} \\ P[K(t+s) - K(s) = k] \;&=\; P[K(t) = k] \end{split}$$

(Eq. 2.53) Waiting time to the first event is exponentially distiributed,

$$\begin{split} f_T(t) = \lambda e^{-\lambda t} \end{split}$$
memoryless (無記憶性)
The number of coin-tosses required to observe the next heads is independent of previous results.
The waiting times between successive events are i.i.d. (independent and identically distributed).

(Eq. 2.54) The waiting time $W$ until the $n$ th event (= the sum of $n$ i.i.d. wating times) can be derived by $n - 1$ successive convolutions:

$$\begin{split} f_{W,1}(t) &= f_T(t) = \lambda e^{-\lambda t} \\ f_{W,2}(t) &= \int _0^t \lambda e^{-\lambda (t-t')} \; \lambda e^{-\lambda t'} dt' \\ &= \lambda ^2 e^{-\lambda t} \Big[1\Big]_0^t \\ &= t\lambda^2 e^{-\lambda t} \\ f_{W,3}(t) &= \int _0^t \lambda e^{-\lambda (t-t')} \; f_2(t') dt' \\ &= \frac {t^2 \lambda^3 e^{-\lambda t}} {2!} \\ &\;\vdots \\ f_{W,n}(t) &= \lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}} {(n - 1)!} \end{split}$$

This is the gamma distribution.

(Eq. 2.55) The mean and the variance are

$$\begin{split} \mathrm{E}[W] &= \sum^n \mathrm{E}[T] \\ &= \sum^n \frac 1 \lambda \\ &= \frac n \lambda \\ \mathrm{Var}[W] &= \sum^n \mathrm{Var}[T] \\ &= \sum^n \frac 1 {\lambda^2} \\ &= \frac n {\lambda^2} \end{split}$$

These hold even when $n$ is not an integer if we replace the factorial $(n - 1)!$ with gamma function, $\Gamma(n) = \int^\infty_0 x^{n-1} e^{-x} dx$. (Eq. 2.57)


The coalescent considers the events that have a very small probability of occurring in any single generation:

They will each form a Poisson process.

2.2.1 Poisson Process Results for the Coalescent

The Sum of Independent Poissons
Two independent Poisson random variables:
$X_1$ with occurrence rate $\lambda _1$
$X_2$ with occurrence rate $\lambda _2$

(Eq. 2.58) The distribution of $Y = X_1 + X_2$ can be obtained by convolution:

$$\begin{split} P[Y=k] \;&=\; \sum _{i=0}^k P[X_1=i] \; P[X_2=k-i] \\ &=\; \sum _{i=0}^k \frac {\lambda _1^i} {i!} e^{-\lambda _1} \frac {\lambda _2^{k-i}} {(k-i)!} e^{-\lambda _2} \\ &=\; e^{-\lambda _1} e^{-\lambda _2} \sum _{i=0}^k \frac {\lambda _1^i} {i!} \frac {\lambda _2^{k-i}} {(k-i)!} \frac {k!}{k!} \\ &=\; \frac {e^{-(\lambda _1 + \lambda _2)}} {k!} \sum _{i=0}^k {k \choose i} \lambda _1^i \lambda _2^{k-i} \\ &=\; \frac {e^{-(\lambda _1 + \lambda _2)}} {k!} (\lambda _1 + \lambda _2)^k \\ &=\; \text{Poisson distribution with occurrence rate } \lambda _1 + \lambda _2 \end{split}$$

The sum of independent Poisson processes is another Poisson process.

The Probability that the First Event Is of a Particular Type

(Eq 2.60) The probability that $X_1$ is observed before $X_2$ is given simply by the relative rate of the event (i.e., as a fraction of the total rate):

$$\begin{split} P[T_1 < T_2] &= \int _0^\infty P[T_2>t]\; f_{T_1}(t) dt \\ &= \int _0^\infty \left(e^{-\lambda _2 t} \right)_\text{Eq. 2.59}\; \lambda _1 e^{-\lambda _1 t} dt \\ &= \lambda _1 \int _0^\infty e^{-(\lambda _1 + \lambda _2) t} dt \\ &= \lambda _1 \left[-\frac {e^{-(\lambda _1 + \lambda _2)t}} {\lambda _1 + \lambda _2} \right]_0^\infty \\ &= \frac {\lambda _1} {\lambda _1 + \lambda _2}, \end{split}$$

using (Eq 2.59)

$$\begin{split} P[T>t] \;&=\; \int _t^\infty \lambda e^{-\lambda t} dt \\ &=\; \left[-e^{-\lambda t} \right]_t^\infty \\ &=\; e^{-\lambda t}. \end{split}$$
The Time to the First Event among Independent Poissons

(Eq. 2.61) The distribution of $T = \min(T_1, T_2)$

$$\begin{split} P[T>t] \;&=\; P[\min(T_1, T_2) > t] \\ &=\; P[T_1 > t \;\cap\; T_2 > t] \\ &=\; P[T_1 > t]\; P[T_2 > t] \\ &=\; e^{-\lambda _1 t} e^{-\lambda _2 t} \\ &=\; e^{-(\lambda _1 + \lambda _2) t} \end{split}$$

Therefore, $f_ {\min(T_1, T_2)}(t) = (\lambda _1 + \lambda _2) e^{-(\lambda _2 + \lambda _1)t}$

There is a one-to-one correspondence between cumulative distribution $P[T \le t]$ and probability densities $f_T(t)$

The Number of Events Required to See a Particular Outcome

$X_2, X_2, X_2, …, X_2, \boldsymbol{X_1}$, …

(Eq. 2.62) How many $X_2$ (e.g., mutation events) occur before $X_1$ (e.g., common ancector event)?

$$\begin{split} P[K=k] \;&=\; P[\text{First }X_1\text{ occurs at }K\text{th trial}] \\ &=\; P[X_2\text{ occurs }k - 1\text{ times first, then }X_1\text{ occurs}] \\ &=\; \left(\frac {\lambda _2} {\lambda _1 + \lambda _2} \right)^{k-1} \frac {\lambda _1} {\lambda _1 + \lambda _2} \end{split}$$

This is the geometric distribution with the rate $p = \frac {\lambda _1} {\lambda _1 + \lambda _2}$.

The process is just like a series of Bernoulli trials with probability of success $p = \frac {\lambda _1} {\lambda _1 + \lambda _2}$.

Tying All This Together: A Filtered Poisson Process

Reinterpret $f_T(t)$ with the sum rule and product rule (see Eq. 2.7, 2.8).

$$\begin{split} f_T(t)\; &=\; \sum _{k=1}^\infty f_T(t \mid K=k)\; P[K=k] \\ &=\; \sum _{k=1}^\infty (\text{Eq. }2.54) (\text{Eq. }2.62) \\ &=\; \sum _{k=1}^\infty (\lambda _1 + \lambda _2) e^{-(\lambda _1 + \lambda _2)t} \frac {\{(\lambda _1 + \lambda _2)t\}^{k-1}} {(k-1)!}\; \left(\frac {\lambda _2} {\lambda _1 + \lambda _2} \right)^{k-1} \frac {\lambda _1} {\lambda _1 + \lambda _2} \\ &=\; \lambda _1 e^{-(\lambda _1 + \lambda _2)t}\; \sum _{k=1}^\infty \frac {(\lambda _2 t)^{k-1}} {(k-1)!} \\ &=\; \lambda _1 e^{-(\lambda _1 + \lambda _2)t}\; \sum _{k=0}^\infty \frac {(\lambda _2 t)^k} {k!} \\ &=\; \lambda _1 e^{-(\lambda _1 + \lambda _2)t}\; e^{\lambda _2 t} \\ &=\; \lambda _1 e^{-\lambda _1 t} \end{split}$$

Taylor series of $e^x$

$$\begin{split} e^x \;&=\; 1 + \frac x 1 + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots \\ &=\; \sum _{k=0}^\infty \frac {x^k} {k!} \end{split}$$
Filtered Poisson process = Poisson process with rate $\lambda p\; (=\lambda _1)$
- total occurrence rate $\lambda = \lambda _1 + \lambda _2$
  • acceptance rate (proportion of the focal event) $p = \frac {\lambda _1} {\lambda _1 + \lambda _2}$

2.2.2 Convolutions of Exponential Distributions

The sum of the waiting times of $n$ events:

If the rate does not change with each event ($\lambda _i = \lambda \text{ for all } i$),
→ gamma-distributed (Eq. 2.54)
If the rate changes with each event ($\lambda _i \neq \lambda _j \text{ for } i \neq j$),
(e.g., in the study of genealogies, the rate of coalescence changes every time a coalescent event occurs)
→ obtained by successive convolution of exponential distribution (Eq. 2.63, 2.64)
$$\begin{split} f_{T_1 + T_2}(t) &= \int _0^t f_{T_1}(s)\; f_{T_2}(t-s)ds \\ &= \int _0^t \lambda _1 e^{-\lambda _1 s}\; \lambda _2 e^{-\lambda _2 (t-s)}ds \\ &= \lambda _1 \lambda _2 e^{-\lambda _2 t} \int _0^t e^{-(\lambda _1 -\lambda _2)s}ds \\ &= \lambda _1 \lambda _2 e^{-\lambda _2 t} \left[-\frac 1{\lambda _1 - \lambda _2} e^{-(\lambda _1 -\lambda _2)s} \right]_0^t \\ &= \frac {\lambda _1} {\lambda _1 - \lambda _2} \lambda _2 e^{-\lambda _2 t} \left(1 - e^{-(\lambda _1 -\lambda _2)t} \right) \\ &= \frac {\lambda _1} {\lambda _1 - \lambda _2} \lambda _2 e^{-\lambda _2 t} + \frac {\lambda _2} {\lambda _2 - \lambda _1} \lambda _1 e^{-\lambda _1 t} \\ &= \frac {\lambda _1} {\lambda _1 - \lambda _2} f_{T_2}(t) + \frac {\lambda _2} {\lambda _2 - \lambda _1} f_{T_1}(t) \\ (&= \text{weighted sum of the original distributions})\\[1ex] f_{T_1 + T_2 + T_3}(t) &= \int _0^t f_{T_1 + T_2}(s)\; f_{T_3}(t-s)ds \\ &= \frac {\lambda _2} {\lambda _2 - \lambda _1} \frac {\lambda _3} {\lambda _3 - \lambda _1} \lambda _1 e^{-\lambda _1 t} + \frac {\lambda _3} {\lambda _3 - \lambda _2} \frac {\lambda _1} {\lambda _1 - \lambda _2} \lambda _2 e^{-\lambda _2 t} + \frac {\lambda _1} {\lambda _1 - \lambda _3} \frac {\lambda _2} {\lambda _2 - \lambda _3} \lambda _3 e^{-\lambda _3 t} \\[1ex] f_{T_1 + T_2 + T_3 + T_4}(t) &= \cdots \\ &\;\vdots \\ f_{\sum _{i=1}^n T_i}(t) &= \sum_{i=1}^n \lambda _i e^{-\lambda _i t} \prod _{j=1,\;j \neq i} \frac {\lambda _j} {\lambda _j - \lambda _i} \end{split}$$

We will use this to obtain the distribution of the total waiting time to the MRCA of the entire samples (Eq. 3.27)