Can i sample without any observation but only potential?

Questions
1

trying to sample from a fabricated discrete distribution but with another fabricated likelihood.

but somehow pymc don’t sample for me :frowning:

e.g.

A ~ N(0, NxN).
B = (C . A > 0) . A
def potential(B):
return scalar_function(B)
pm.MCMC([A, B, potential])
pm.sample(100)

2

You should be able to sample in many cases without explicitly using an observation. Would you be willing to post a self-contained code section that describes your problem in a bit more detail?

3

sure, and sorry i was using pymc2. it’s a lot of trouble to do boolean operation with theano

import numpy as np
import pymc as pm
gs = np.random.uniform(size=(4, 3))
seed = pm.Normal('seed', 0, 1, size=[gs.shape[1], 1])
@pm.deterministic(name='label', trace=True)
def label(seed=seed):
  return gs.dot(seed) > 0

@pm.potential
def value(label=label):
  return label.T.dot(gs).sum()

model = pm.MCMC([seed, label, value])
trace = model.sample(10)
assert trace != None
4

I’ve written code for what I think you want to do in PyMC3. I apologize that I can’t answer with PyMC2 because I don’t know much about how to use it.

import pymc3 as pm
import numpy as np

gs = np.random.uniform(size=(4, 3))

with pm.Model() as model:
    seed  = pm.Normal('seed', 0, 1, shape=[gs.shape[1], 1])
    label = pm.Deterministic('label',pm.math.gt(pm.math.dot(gs,seed),0))
    value = pm.Potential('value',pm.math.sum(pm.math.dot(label.T,gs)))
    step  = pm.Metropolis()
    trace = pm.sample(step=step)
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