当代种群DNA样本分支突变概率的条件概率与全概率求解

开发者阿泽

2026-5-19

Alright, let's tackle this coalescent theory probability problem step by step. I'll walk through solving both the conditional probability and the total probability you've asked about, using standard results from population genetics and probability theory.

1. Solving the Conditional Probability $\mathbb{P}(M>0 | T_2,...,T_n)$

First, let's ground ourselves in the setup:

$T_i$ are independent exponential random variables with rate $\binom{i}{2}$ — this is standard for coalescent waiting times, where $T_i$ represents the time spent in the phase with $i$ active lineages.
$M$ counts the total mutations on a specific "solid branch" in the coalescent tree.

For any branch of fixed length $L$, the number of mutations $M$ follows a Poisson distribution with parameter $\lambda L$, where $\lambda$ is the mutation rate per unit time (a core assumption in molecular evolution models).

To simplify the conditional probability, we use the complement rule (calculating the chance of zero mutations and subtracting from 1):
$$
\mathbb{P}(M>0 | T_2,...,T_n) = 1 - \mathbb{P}(M=0 | T_2,...,T_n)
$$

For a Poisson random variable, the probability of zero mutations is $e^{-\lambda L}$, where $L$ is the length of the solid branch (fully determined by the observed $T_2,...,T_n$). If the solid branch corresponds to a single sample's lineage from the present back to the most recent common ancestor (MRCA), its length is the sum of all coalescent waiting times:
$$
L = T_2 + T_3 + ... + T_n
$$

Substituting this in, we get the final conditional probability:
$$
\mathbb{P}(M>0 | T_2,...,T_n) = 1 - e^{-\lambda \sum_{i=2}^n T_i}
$$

If the solid branch refers to a different segment of the tree, just replace $L$ with the appropriate sum of $T_i$ terms — the core logic stays identical.

2. Solving the Total Probability $\mathbb{P}(M>0)$

To find the total (unconditional) probability, we use the law of total expectation, which averages the conditional probability over all possible values of $T_2,...,T_n$:
$$
\mathbb{P}(M>0) = \mathbb{E}\left[\mathbb{P}(M>0 | T_2,...,T_n)\right]
$$

Substituting the conditional probability we derived earlier:
$$
\mathbb{P}(M>0) = \mathbb{E}\left[1 - e^{-\lambda L}\right] = 1 - \mathbb{E}\left[e^{-\lambda L}\right]
$$

Since the $T_i$ are independent, the expectation of the exponential of their sum is the product of their individual expectations. For $L = \sum_{i=2}^n T_i$, this means:
$$
\mathbb{E}\left[e^{-\lambda L}\right] = \prod_{i=2}^n \mathbb{E}\left[e^{-\lambda T_i}\right]
$$

Next, we calculate $\mathbb{E}\left[e^{-\lambda T_i}\right]$ for an exponential random variable $T_i$ with rate $\binom{i}{2}$. Using the moment-generating function for exponential variables, we get:
$$
\mathbb{E}\left[e^{-\lambda T_i}\right] = \frac{\binom{i}{2}}{\binom{i}{2} + \lambda}
$$

Putting it all together, the total probability simplifies to:
$$
\mathbb{P}(M>0) = 1 - \prod_{i=2}^n \frac{\binom{i}{2}}{\binom{i}{2} + \lambda}
$$

If you prefer an expanded form, substitute $\binom{i}{2} = \frac{i(i-1)}{2}$:
$$
\mathbb{P}(M>0) = 1 - \prod_{i=2}^n \frac{i(i-1)/2}{i(i-1)/2 + \lambda}
$$

内容的提问来源于stack exchange，提问作者Btzzzz