Python function for a hacky approximation of the F_Tt factors in the Bryant et al SNAPP method as specialized for use by a homework assignment in the Kelly and Holder "Likelihood methods in Biology" class

gistfile1.py

def calc_F_Tt(r_Tb, n_Tt, r_Tt, param_vec):
    if r_Tt > n_Tt:
        return 0.0
    Ne = param_vec[ParamOrder.PopSize].value
    mu = param_vec[ParamOrder.MutRate].value
    tau = param_vec[ParamOrder.DivTime].value

    # The probability of coalescence along the Turkish branch goes to 1 as tau gets big or Ne gets small
    prob_coalescence = 1 - exp(-tau/(2.0*Ne))
    prob_no_coalescence = 1 - prob_coalescence
    
    # Given that there is no coalescence, each branch will have length tau along the Turkish population history
    # the following formula would apply to each branch (and are the same as the probabilities for the Spanish population)
    expNeg2TauMu = exp(-2*tau*mu)
    prob_no_mut_given_no_coal = (1 + expNeg2TauMu)/2.0
    prob_mut_given_no_coal = 1 - prob_no_mut_given_no_coal
    

    # Here is some very crude averaging over branch lengths that Bryant et al don't do...
    # we'll assume that if there is coalescence it averages out to occurring 1/3 of the way from the present to the 
    #   time of divergence. No real justificaion for this...
    hacky_total_branch_length_given_coal = (4/3) * tau  # hack mentioned above
    expNegHackyBranchLen = exp(-2*hacky_total_branch_length_given_coal*mu)
    prob_no_mut_given_coal = (1 + expNegHackyBranchLen)/2.0
    prob_mut_given_coal = 1 - prob_no_mut_given_no_coal
    
    # 2/3 of tau is contributed by the common ancestral branch,
    # 1/3 of tau is contributed by each terminal (for a total of 4/3)
    # so, given that there was coalescence, and only one mutation in the Turkish branch, and 
    #   our new hack. 50% of mutations will be on the ancestral branch of the gene tree,
    #   and 25% will be on each of the terminal's
    prob_mut_common_anc_branch = 0.5*prob_mut_given_coal # hack mentioned above, again
    prob_mut_terminal_branch = .25*prob_mut_given_coal # hack mentioned above, again
    
    
    if (r_Tb == 1 ):
        #eqn 10 - eqn 14
        if (n_Tt == 2):
            #eqn 10 - eqn 12
            prob_tree_shape = prob_no_coalescence
            if (r_Tt == 1):
                # eqn 11 would be the correct form to use
                #
                # here we'll use the hacky way...
                # same # of lineages in state 0 at the beginning and end. This probability will be dominated 
                #   by the scenario of no mutations...
                prob_two_mutations = prob_mut_given_no_coal * prob_mut_given_no_coal
                prob_zero_mutations = prob_no_mut_given_no_coal * prob_no_mut_given_no_coal
                prob_mutation_scenario = prob_two_mutations + prob_zero_mutations
            else:
                # eqn 10 and 12 are identical
                # hacky way -> one lineage mutated, and the other did not...
                prob_mutation_scenario = prob_mut_given_no_coal * prob_no_mut_given_no_coal
        else:
            #eqn 13 and eqn 14 are the same, so only one branch of code here
            # the hacky way is to assume there was only one mutation, and it must have been on a terminal...
            prob_tree_shape = prob_coalescence
            prob_mutation_scenario = prob_no_mut_given_coal*prob_mut_terminal_branch
    else:
        #eqn 10 - eqn 14
        if (n_Tt == 2):
            #eqn 5 - eqn 5
            prob_tree_shape = prob_no_coalescence
            if (r_Tt == r_Tb):
                # eqn 5 would be the correct way...
                # here is the hack...
                prob_mutation_scenario =  prob_no_mut_given_no_coal * prob_no_mut_given_no_coal
            elif (r_Tt == 1):
                # eqn 6 is the correct way...
                # hacky way -> one lineage mutated, and the other did not...
                prob_mutation_scenario = prob_mut_given_no_coal * prob_no_mut_given_no_coal
            else:
                #eqn 7 is the correct way.
                # if r_Tb is 0 or 2 and r_Tt is 0 or 2, but r_Tt != r_Tb, then both lineages had mutations
                prob_mutation_scenario =  prob_mut_given_no_coal * prob_mut_given_no_coal
        else:
            #eqn 8 or eqn 9
            prob_tree_shape = prob_coalescence
            if (r_Tt == 1):
                if (r_Tb == 2):
                    # eqn 8
                    # no mutations required when a state-0 lineage splits into two state-0 lineages
                    prob_mutation_scenario = prob_no_mut_given_coal
                else:
                    # eqn 9
                    # a mutations before splitting is the easiest way to turn a state-0 lineage splits into two state-1 lineages
                    prob_mutation_scenario = prob_mut_common_anc_branch
            else:
                # eqn 9
                if (r_Tb == 0):
                    # eqn 8
                    # no mutations required when a state-1 lineage splits into two state-1 lineages
                    prob_mutation_scenario = prob_no_mut_given_coal
                else:
                    # eqn 9
                    # a mutations before splitting is the easiest way to turn a state-1 lineage splits into two state-0 lineages
                    prob_mutation_scenario = prob_mut_common_anc_branch
    return prob_tree_shape*prob_mutation_scenario