4
$\begingroup$

I am trying to implement a simple normal-inverse-Wishart conjugate prior distribution for a multivariate normal with unknown mean and covariance in numpy/scipy such that it can take a data vector and construct a posterior. I'm using the update equations specified by Wikipedia for a NIW: http://en.wikipedia.org/wiki/Conjugate_prior

My distribution class is as follows:

import numpy as np
from scipy.stats import chi2

class NormalInverseWishartDistribution(object):
    def __init__(self, mu, lmbda, nu, psi):
        self.mu = mu
        self.lmbda = float(lmbda)
        self.nu = nu
        self.psi = psi
        self.inv_psi = np.linalg.inv(psi)

    def sample(self):
        sigma = np.linalg.inv(self.wishartrand())
        return (np.random.multivariate_normal(self.mu, sigma / self.lmbda), sigma)

    def wishartrand(self):
        dim = self.inv_psi.shape[0]
        chol = np.linalg.cholesky(self.inv_psi)
        foo = np.zeros((dim,dim))

        for i in range(dim):
            for j in range(i+1):
                if i == j:
                    foo[i,j] = np.sqrt(chi2.rvs(self.nu-(i+1)+1))
                else:
                    foo[i,j]  = np.random.normal(0,1)
        return np.dot(chol, np.dot(foo, np.dot(foo.T, chol.T)))

    def posterior(self, data):
        n = len(data)
        mean_data = np.mean(data, axis=0)
        sum_squares = np.sum([np.array(np.matrix(x - mean_data).T * np.matrix(x - mean_data)) for x in data], axis=0)
        mu_n = (self.lmbda * self.mu + n * mean_data) / (self.lmbda + n)
        lmbda_n = self.lmbda + n
        nu_n = self.nu + n
        psi_n = self.psi + sum_squares + self.lmbda * n / float(self.lmbda + n) * np.array(np.matrix(mean_data - self.mu).T * np.matrix(mean_data - self.mu))
        return NormalInverseWishartDistribution(mu_n, lmbda_n, nu_n, psi_n)

I am running a simple sanity check to see if the posterior converges to the true distribution:

x = NormalInverseWishartDistribution(np.array([0,0])-3,1,3,np.eye(2))
samples = [x.sample() for _ in range(100)]
data = [np.random.multivariate_normal(mu,cov) for mu,cov in samples]
y = NormalInverseWishartDistribution(np.array([0,0]),1,3,np.eye(2))
z = y.posterior(data)

print 'mu_n: {0}'.format(z.mu)

print 'psi_n: {0}'.format(z.psi)

The mean is appropriately converging, but the scale matrix appears to be converging to incorrectly large values along the diagonal, rather than the true value of 1.

As far as I can tell, I'm copying the update rule exactly. Am I implementing something inappropriately here?

Edit: It looks like in fact the posterior is converging, but the sample routine is returning biased samples. Am I doing something wrong in my sampling method?

Edit2: I've confirmed that the same phenomenon happens in the MCMCpack riwish function in R:

> library(MCMCpack)
> samples <- replicate(100000, riwish(3, matrix(c(1,0,0,1),2,2)))
> mean(samples[1,1,])
[1] 4.889211

This leads me to believe I must be misunderstanding something. From the Wikipedia page (http://en.wikipedia.org/wiki/Inverse-Wishart_distribution), we have:

$\newcommand{\E}{\mathrm{E}}$ $\E[{\Sigma}] = \frac{\Psi}{\nu - p - 1}$

However, in my test case, $\nu = p + 1$, so $\E[{\Sigma}] = \Psi$. Thus, if I sample $\Sigma$ a bunch of times, shouldn't I recover $\Psi$ on average?

Edit 3: Realized my problem was simple algebraic oversight: I needed to set $\nu = p+2$. Problem solved

$\endgroup$
4
$\begingroup$

The problem was that I was setting the degrees of freedom too low-- it should be at least P+2, where $\Psi$ is a PxP matrix.

$\endgroup$
1
$\begingroup$

Sorry I am slow, but all that changed was setting nu=4?

x = NormalInverseWishartDistribution(np.array([0,0])-3,1,4,np.eye(2)) # nu > 2 + psi.shape[0]
samples = [x.sample() for _ in range(1000)]
data = [np.random.multivariate_normal(mu,cov) for mu,cov in samples]
y = NormalInverseWishartDistribution(np.array([0,0]),1,4,np.eye(2))
z = y.posterior(data)

print 'mu_n: {0}'.format(z.mu)
print 'psi_n: {0}'.format(z.psi)

Gives me:

mu_n: [-3.00472366 -3.02735843]
psi_n: [[ 1785.25130628   -33.05276129]
        [  -33.05276129  2425.67075978]]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.