Probabilistic Models for Mapping

Recall the following defintion:  

• A non decreasing function $N: R_{0}^{+} \rightarrow N$ where N(t) = number of events until time t
• N(0) = 0
• The number of events in disjoint intervals are independent
  

  $$P(N(t+s) - N(s) = n) = e^{-\lambda t} \frac {({\lambda t})^{n}} { n! } \mbox{, for } n = 0,1,\ldots \mbox { and } s \geq 0$$ (10.2)

  
  

• Distribution of the number of events in an interval is stationary, i.e. depends only on the length of the interval. The expected number of events in an interval of length t is given by $E(N(t)) = \lambda t$.

Denote:

• T_n = time between event n-1 and event n
• S₀ = 0
• $S_{\imath} = \sum_{i=1}^{i} T_{\imath}$

Recall that inter-event times in a Poisson process are i.i.d. random variables, exponentially distributed with parameter $\lambda$, i.e.,

$$Pr(T_{i} > t) = e^{- \lambda t}$$ (10.3)

If it is known that $n \geq 1$ events occurred in a Poisson process until time t, then the inter-arrival times $\{S_{1},S_{2},...,S_{n} \}$ are distributed uniformly and independently in [0,t]. Assume clone length L, genome length G, and choose N clones at random. What is the expected fraction of the genome covered by clones? For a random point b, and an arbitrary clone C the probability of the point b being included in the clone c is given by:

$$Pr(b \in c) = \frac {L} {G}$$ (10.4)

and therefore, the probability of b being out of all the clones is given by:

$$Pr(\forall c : b \notin c) = (1 - \frac {L} {G})^{N} = (1 - \frac {L} {G})^{G \frac {N} {G}} \sim e^{-\frac {NL} {G}}$$ (10.5)

with the last approximation being valid when $L \ll G$ and $N \ll G$.  

$$R=\frac{NL}{G}$$

is said to be the redundancy of the clone set.  

is given by

$$E(\mbox{fraction not covered}) = e^{-R}$$ (10.6)

where R is the redundancy of the clone set Table 10.2 shows that using redundancy factor of 2 to 5 gives a good coverage of the genome segment considered.

 

Table 10.1: Coverage of genome segment depending on redundancy factor

 

R	Coverage
1	0.63
2	0.865
3	0.95
4	0.98
5	0.993

Define the length by setting clone length = 1, and denote N = number of clones, R = redundancy factor, and assume that the clone starting positions follow a Poisson process with rate $\lambda$. We define a minimal overlap factor $\theta$ between clones to identify overlap. A set of clones covering a continuous segment of the genome, together with their physical distances is called a contig. Contigs are sometimes referred to as islands.

Theorem 10.5   Lander-Waterman 1988 [4]:

1.

The expected number of apparent islands is given by

$$Ne^{-R(1-\theta)}$$ (10.7)

2.

The expected number of apparent islands with exactly $j \geq 1$ clones is given by

$$N e^{-2R(1-\theta)} {(1-e^{-R(1-\theta)})}^{(j-1)}$$ (10.8)

3.

The expected number of clones in an apparent island is given by

$$e^{-R(1-\theta)}$$ (10.9)

4.

The expected length of an apparent island is

$$\frac {e^{-R(1-\theta)} - 1} {R} + \theta$$ (10.10)

Proof:We will prove the first item of the theorem. In order to prove the formula for expected number of apparent islands, we define J(x) as follows:

$$\begin{eqnarray*}{} \centering J(x) &= & Pr(\mbox{two points } a, b = a + X ... ...r(\mbox{there are no left-end points in the interval } [b-1,a]) \end{eqnarray*}$$

Since [b-1,a] is of length 1-x, J(x) can be computed from the redundancy factor R as follows:

$$J(x) = \begin{cases} e^{-R(1-x)} & 0 \leq x \leq 1, \\ 1 & \text{otherwise}. \end{cases}$$ (10.11)

The number of islands is the number of times leaving a clone without detecting an overlap. Let E_c denote the event of a clone c being the right-hand clone of an island. If the right-hand side of the island is at a point t, we require that t and $t - \theta$ are not covered by a common clone (other than c) . The probability of such an event E_c is given by:

$$P(E) = J(\theta)$$ (10.12)

and the expected number of apparent islands is therefore given by:

$$Exp(\mbox{number of apparent islands}) = N \cdot J(\theta) = Ne^{-R(1-\theta)}$$ (10.13)