Regression Prediction Estimation

Question

Say I have a random set of points X, Y:

x = np.array(range(0,50))
y = np.random.uniform(low=0.0, high=40.0, size=200)
y = map((lambda a: a[0] + a[1]), zip(x,y))
plt.scatter(x,y)

Suppose that I model yas Gaussian for each value xusing linear regression , how can I estimate the rear forecast , i.e. for each (possible) value ? p(y|x)x

Is there a direct way to do this with pymcor scikit-learn?

Answer 1

If I understand correctly what you want, you can do this using the git version of PyMC (PyMC3) and the glm submodule. for example

import numpy as np
import pymc as pm
import matplotlib.pyplot as plt 
from pymc import glm 

## Make some data
x = np.array(range(0,50))
y = np.random.uniform(low=0.0, high=40.0, size=50)
y = 2*x+y
## plt.scatter(x,y)

data = dict(x=x, y=y)
with pm.Model() as model:
    # specify glm and pass in data. The resulting linear model, its likelihood and 
    # and all its parameters are automatically added to our model.
    pm.glm.glm('y ~ x', data)
    step = pm.NUTS() # Instantiate MCMC sampling algorithm
    trace = pm.sample(2000, step)


##fig = pm.traceplot(trace, lines={'alpha': 1, 'beta': 2, 'sigma': .5});## traces
fig = plt.figure()
ax = fig.add_subplot(111)
plt.scatter(x, y, label='data')
glm.plot_posterior_predictive(trace, samples=50, eval=x,
                              label='posterior predictive regression lines')

To get something like this

: 1 2, .

Edit y x , glm.

lm = lambda x, sample: sample['Intercept'] + sample['x'] * x ## linear model
samples=50 ## Choose to be the same as in plot call
trace_det = np.empty([samples, len(x)]) ## initialise
for i, rand_loc in enumerate(np.random.randint(0, len(trace), samples)):
    rand_sample = trace[rand_loc]
    trace_det[i] = lm(x, rand_sample)
y = trace_det.T
y[0]

, - , .