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Added predicted values for student t, works well

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This commit is contained in:
Alan Saul 2013-04-02 12:33:01 +01:00
parent 15d5c2f22d
commit ffc168c1d2
2 changed files with 53 additions and 36 deletions
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41
python/likelihoods/likelihood_function.py
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@ -23,6 +23,10 @@ class student_t(likelihood_function):
#FIXME: This should be in the superclass
self.log_concave = False
@property
def variance(self):
return (self.v / float(self.v - 2)) * (self.sigma**2)
def link_function(self, y, f):
"""link_function $\ln p(y|f)$
$$\ln p(y_{i}|f_{i}) = \ln \Gamma(\frac{v+1}{2}) - \ln \Gamma(\frac{v}{2})\sqrt{v \pi}\sigma - \frac{v+1}{2}\ln (1 + \frac{1}{v}\left(\frac{y_{i} - f_{i}}{\sigma}\right)^2$$
@ -79,14 +83,32 @@ class student_t(likelihood_function):
def predictive_values(self, mu, var):
"""
Compute mean, and conficence interval (percentiles 5 and 95) of the prediction
Compute mean, and conficence interval (percentiles 5 and 95) of the prediction
Need to find what the variance is at the latent points for a student t*normal
(((g((v+1)/2))/(g(v/2)*s*sqrt(v*pi)))*(1+(1/v)*((y-f)/s)^2)^(-(v+1)/2))*((1/(s*sqrt(2*pi)))*exp(-(1/(2*(s^2)))*((y-f)^2)))
Need to find what the variance is at the latent points for a student t*normal p(y*|f*)p(f*)
(((g((v+1)/2))/(g(v/2)*s*sqrt(v*pi)))*(1+(1/v)*((y-f)/s)^2)^(-(v+1)/2))
*((1/(s*sqrt(2*pi)))*exp(-(1/(2*(s^2)))*((y-f)^2)))
(((g((v+1)/2))/(g(v/2)*s*sqrt(v*pi)))*(1+(1/v)*((y-f)/s)^2)^(-(v+1)/2))
*((1/(s*sqrt(2*pi)))*exp(-(1/(2*(s^2)))*((y-f)^2)))
"""
#We want the variance around test points y which comes from int p(y*|f*)p(f*) df*
#Var(y*) = Var(E[y*|f*]) + E[Var(y*|f*)]
#Since we are given f* (mu) which is our mean (expected) value of y*|f* then the variance is the variance around this
#Which was also given to us as (var)
#We also need to know the expected variance of y* around samples f*, this is the variance of the student t distribution
#However the variance of the student t distribution is not dependent on f, only on sigma and the degrees of freedom
true_var = var + self.variance
#Now we have an analytical solution for the variances of the distribution p(y*|f*)p(f*) around our test points but we now
#need the 95 and 5 percentiles.
#FIXME: Hack, just pretend p(y*|f*)p(f*) is a gaussian and use the gaussian's percentiles
p_025 = mu - 2.*true_var
p_975 = mu + 2.*true_var
return mu, np.nan*mu, p_025, p_975
def sample_predicted_values(self, mu, var):
""" Experimental sample approches and numerical integration """
#p_025 = stats.t.ppf(.025, mu)
#p_975 = stats.t.ppf(.975, mu)
@ -134,14 +156,13 @@ class student_t(likelihood_function):
def t_gauss_int(mu, var):
print "Mu: ", mu
print "var: ", var
result = integrate.quad(t_gaussian, -np.inf, 0.975, args=(mu, var))
result = integrate.quad(t_gaussian, 0.025, 0.975, args=(mu, var))
print "Result: ", result
return result[0]
vec_t_gauss_int = np.vectorize(t_gauss_int)
p_025 = vec_t_gauss_int(mu, var)
p_975 = vec_t_gauss_int(mu, var)
p = vec_t_gauss_int(mu, var)
p_025 = mu - p
p_975 = mu + p
import ipdb; ipdb.set_trace() ### XXX BREAKPOINT
return mu, np.nan*mu, p_025, p_975