Clarification on posterior computation for genomic data in bayesian framework

Question

I'm reading Trifinov et al. (2013) and I am trying to understand the designed framework. I list some excerpts to focus on the passage that is not clear to me:

A lesion $M$ is a result of a random process with a given distribution $P^m(M)$, which specifies the probability of occurrence of the lesion $M $ due to $m$utation.

[...] To model selection we assume that the genotype resulting from a disjoint set of such lesions $S = \{M_1, ⋯, M_t \}$ expresses the phenotype when a lesion in the sequence contains a driver gene. We consider two possibilities for the posterior probability that a given gene $D$ is a driver of the phenotype, given that the genotype of the set $S$ expresses the phenotype: a global posterior and a local posterior. To obtain the two posteriors we define first the likelihood that a lesion $M$ is a driver lesion given that $D$ is a driver gene as

$$L(M|D) \quad \underline{deff} \quad P^m(M)·δ(D ∈ M)$$

where $δ(D ∈ M)$ is $1$ if the lesion $M$ includes the gene $D$ and $0$ otherwise, with corresponding posterior

$$P(D|M)= \frac{δ(D∈M)·P^d(D)}{∑_{G∈M}P^d(G)}$$

where $P^d(D)$ is the prior probability that $D$ is a $d$river of the phenotype.

I am trying to understand how the posterior has beed computed, i.e. how to pass from the first to the second equation.

I have little experience on bayesian statistics, except for the basic understanding required by the Wikipedia pages (1, 2). For instance, following from the Wikipedia's likelihood definition, first definition translates to:

$$L(M|D) = P(D|M) = P^m(M)·δ(D ∈ M)$$

So, the probability that $D$ is a driver gene given that $M$ is a driver lesion is equal to the probability of lesion $M$ due to mutation if $D$ is in $M$, $0$ otherwise. Is it correct?

Otherwise, could it be that $L(M|D)$ and $P(D|M)$ are both probabilities, in particular the first defining the probability of $M$ given $D$, and the second defining the posterior distribution of $D$ given $M$?

If this is the case, it unfolds to:

$$P(D|M)=\frac{P(M|D)·P(D)}{P(M)}$$ $$=\frac{P^m(M)·δ(D ∈ M)·P^d(D)}{P^m(M)}$$ $$=δ(D ∈ M)·P^d(D)$$

But in this case, the sum at the denominator of the second equation is missing.

How can I reduce the calculation of the posterior? What can I read to have a solid understanding of it?

Edit

I am now pretty sure that the derivation of the second equation should follow simply using the Bayes theorem, i.e.:

$$posterior = \frac{likelihood·prior}{marginal\ likelihood}$$

with

$likelihood=P^m(M)·\delta(D \in M)$

$prior=P^d(D)$

but I am unable to compute the marginal likelihood correctly. How can it be done?

• Trifonov, Vladimir, et al. "MutComFocal: an integrative approach to identifying recurrent and focal genomic alterations in tumor samples." BMC systems biology 7.1 (2013): 25.

Answer 1

You have the right idea in your second attempt, using Bayes' theorem. However, keep in mind that $P(D \mid M)$ is the probability that $D$ is a driver gene given that $M$ expresses the phenotype. Therefore, the denominator must be the marginal probability that $M$ expresses the phenotype, which we can write as $$P(M) = \sum_G P(M \mid G) P(G).$$ This is different from $P^m(M),$ which is the probability that lesion $M$ occurs due to mutation.

Substitute in the above equation to the denominator, and keep in mind that $\sum_G \delta(G \in M) f(G) = \sum_{G \in M} f(G).$ Then you should get the expression given in the passage.

EDIT: While the above assumes that $L(M \mid D) = P(M \mid D),$ which is how I interpret their expression for the "likelihood", it's possible that they simply mean that $L(M \mid D) \propto P(M \mid D)$ where the proportionality constant is independent of $D.$ If that's the case the math still works out, as you can simply cancel the proportionality constant from the numerator and denominator. I should clarify that I'm simply showing how to get to the result purely from a mathematical perspective. I can't speak for a conceptual understanding here because I don't know this field, nor did I read the paper.