Small Markov chain Monte Carlo implementation

Question

I've written a small markov chain monte carlo function that takes samples from a posterior distribution, based on a prior and a binomial (Bin(N, Z)) distribution.

I'd be happy to have it reviewed, especially perhaps, regarding how to properly pass functions as arguments to functions (as the function prior_dist() in my code). In this case, I'm passing the function uniform_prior_distribution() showed below, but it's quite likely I'd like to pass other functions, that accept slightly different arguments, in the future. This would require me to rewrite mcmc(), unless there's some smart way around it...

def mcmc(prior_dist, size=100000, burn=1000, thin=10, Z=3, N=10):
    import random
    from scipy.stats import binom
    #Make Markov chain (Monte Carlo)
    mc = [0] #Initialize markov chain
    while len(mc) < thin*size + burn:
        cand = random.gauss(mc[-1], 1) #Propose candidate
        ratio = (binom.pmf(Z, N, cand)*prior_dist(cand, size)) / (binom.pmf(Z, N, mc[-1])*prior_dist(mc[-1], size))
        if ratio > random.random(): #Acceptence criteria
        mc.append(cand)
    else:
        mc.append(mc[-1])
    #Take sample
    sample = []
    for i in range(len(mc)):
        if i >= burn and (i-burn)%thin == 0:
            sample.append(mc[i])
    sample = sorted(sample)
    #Estimate posterior probability
    post = []
    for p in sample:
        post.append(binom.pmf(Z, N, p) * prior_dist(p, size))
    return sample, post, mc

def uniform_prior_distribution(p, size):
    prior = 1.0/size
    return prior

Answer 1

Function Passing

To pass functions with possibly varying argument lists consider using argument unpacking.

By using a dictionary at the call site, we can use the arguments that are relevant to the chosen function whilst discarding those which aren't.

def uniform_prior_distribution(p, size, **args): #this args is neccesary to discard parameters not intended for this function
    prior = 1.0/size
    return prior

def foo_prior_distribution(p, q, size, **args):
    return do_something_with(p, q, size)

def mcmc(prior_dist, size=100000, burn=1000, thin=10, Z=3, N=10):
    import random
    from scipy.stats import binom
    #Make Markov chain (Monte Carlo)
    mc = [0] #Initialize markov chain
    while len(mc) < thin*size + burn:
        cand = random.gauss(mc[-1], 1) #Propose candidate
        args1 = {"p": cand, "q": 0.2, "size": size}
        args2 = {"p": mc[-1], "q": 0.2, "size": size}
        ratio = (binom.pmf(Z, N, cand)*prior_dist(**args1)) / (binom.pmf(Z, N, mc[-1])*prior_dist(**args2))

Depending on how you use the mcmc function you may want to pass the args1 and args2 dicts directly as parameters to mcmc.

However, this approach with only work if the common parameters (in this case p and size) can remain constant between different functions. Unfortunately I don't know enough about prior distributions to know if this is the case.

Other points of note

Use

sample.sort()

Instead of

sample = sorted(sample)

It's clearer and slightly more efficient (it saves a copy).

Instead of:

for i in range(len(mc)):
    if i >= burn and (i-burn)%thin == 0:
        sample.append(mc[i])

Use the more pythonic:

for i,s in enumerate(mc):
    if i >= burn and (i-burn)%thin == 0:
        sample.append(s)

Answer 2

def mcmc(prior_dist, size=100000, burn=1000, thin=10, Z=3, N=10):
    import random
    from scipy.stats import binom

Don't import inside functions. It goes against python convention and is inefficient

    #Make Markov chain (Monte Carlo)
    mc = [0] #Initialize markov chain
    while len(mc) < thin*size + burn:
        cand = random.gauss(mc[-1], 1) #Propose candidate
        ratio = (binom.pmf(Z, N, cand)*prior_dist(cand, size)) / (binom.pmf(Z, N, mc[-1])*prior_dist(mc[-1], size))
        if ratio > random.random(): #Acceptence criteria
            mc.append(cand)
    else:
        mc.append(mc[-1])

Why is this in else clause? It only make sense to use else clauses if you are using break.

    #Take sample
    sample = []
    for i in range(len(mc)):
        if i >= burn and (i-burn)%thin == 0:
            sample.append(mc[i])

Actually something like sample = mc[burn::thin] will have the same effect as this loop

    sample = sorted(sample)
    #Estimate posterior probability
    post = []
    for p in sample:
        post.append(binom.pmf(Z, N, p) * prior_dist(p, size))

I'd suggest post = [binom.pmf(Z, N, p) * prior_dist(p, size) for p in sample]

    return sample, post, mc

def uniform_prior_distribution(p, size):
    prior = 1.0/size
    return prior