AR1 log likelihood

[eph]

I think the log likelihood calculation for the AR1 model is wrong.

class AR1(distribution.Continuous):

    ...

    def logp(self, x):
        k = self.k
        tau_e = self.tau_e

        x_im1 = x[:-1]
        x_i = x[1:]
        boundary = Normal.dist(0., tau=tau_e).logp

        innov_like = Normal.dist(k * x_im1, tau=tau_e).logp(x_i)
        return boundary(x[0]) + tt.sum(innov_like) + boundary(x[-1])

The last term in the log likelihood should not be there.

import pymc3
from scipy.stats import norm

with pymc3.Model() as t:
    my_ar1 = pymc3.AR1('y',0,1,shape=2)

print('pymc3 log lik     :', t['y'].logp({'y':np.zeros(2,)}) )
print('scipy log lik(n=2):', norm.logpdf([0,0]).sum() )
print('scipy log lik(n=3):', norm.logpdf([0,0,0]).sum() )

yields

pymc3 log lik     : -2.756815599614018
scipy log lik(n=2): -1.83787706641
scipy log lik(n=3): -2.75681559961

I will try to write some tests following what's in test_distributions.py. BTW, this is fantastic software -- thanks for all your efforts.

[junpenglao]

I think you are right, the ARI logliklihood is given as

$\log{f_y(y_1;\theta)} + \sum^{T}_{t=2}\log{f_\epsilon(y_t|y_{t-1}; \theta)}$

@twiecki you have more experience of timeserie analysis, thoughts?