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Fitting logistic regression with PyMC: ZeroProbability error


Fitting logistic regression with PyMC: ZeroProbability error

By : user2955170
Date : November 22 2020, 10:38 AM
like below fixes the issue When you create a your normal stochastastic with pymc.Normal('w0', 0, 0.000001), PyMC2 initializes the value with a random draw from the prior distribution. Since your prior is so diffuse, this can be a value which is so unlikely that the posterior is effectively zero. To fix, just request a reasonable initial value for your Normal:
code :
w0 = pymc.Normal('w0', 0, 0.000001, value=0)
w1 = pymc.Normal('w1', 0, 0.000001, value=0)


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Why is the standard error different in these two fitting methods (R Logistic Regression and Beta Regression) for a commo

Why is the standard error different in these two fitting methods (R Logistic Regression and Beta Regression) for a commo


By : Random Guy
Date : March 29 2020, 07:55 AM
I wish this help you The standard errors are different because the variance assumptions in the two models are different.
Logistic regression assumes the response has a binomial distribution, while beta regression assumes it has a beta distribution.
Logistic Regression in PyMC

Logistic Regression in PyMC


By : user2210621
Date : March 29 2020, 07:55 AM
it fixes the issue You are definitely on the right track. I'm not solving your homework problem am I? It is now a preferred idiom to import pymc as pm instead of mc, so to finish this up with an observed decorator, just use:
code :
import pymc as pm

@pm.observed
def y(logit_p=logit_p, value=df.GotSick):
    return pm.bernoulli_like(df.GotSick, pm.invlogit(logit_p))
Fitting a Binomial distribution with pymc raises ZeroProbability error for certain FillValues

Fitting a Binomial distribution with pymc raises ZeroProbability error for certain FillValues


By : tsssys
Date : March 29 2020, 07:55 AM
I wish this help you I asked the question the pymc developers on GitHub and there I got an answer from fonnesbeck (https://github.com/pymc-devs/pymc/issues/47#issuecomment-129301002):
Logistic regression without an intercept gives fitting warning message

Logistic regression without an intercept gives fitting warning message


By : calledT
Date : March 29 2020, 07:55 AM
Does that help I will try to provide an answer to the question.
What does the warning mean? The warning is given when numerical precision might be in question for certain observations. More precisely, it is given in the case where the fitted model, returns probability of 1 - epsilon or equivalently 0 + epsilon. As standard this bound is 1-10^-8 and 10^-8 respectively (as given by glm.control) for the standard glm.fit function.
Scikit-learn's logistic regression is performing poorer than self-written logistic regression in Python

Scikit-learn's logistic regression is performing poorer than self-written logistic regression in Python


By : randallcp
Date : March 29 2020, 07:55 AM
I wish this helpful for you Your implementation has no regularisation term. The LinearRegression estimator includes by default regularisation with inverse strength C = 1.0. As you set C to higher values, i.e. weaken the regularisation, the decision boundary moves closer to 5.5:
code :
for C in [1.0, 1000.0, 1e+8]:
    lr = LogisticRegression(C=C)
    lr.fit(inp, labels)
    print(f'C = {C}, decision boundary @ {(-lr.intercept_/lr.coef_[0])[0]}')
C = 1.0, decision boundary @ 3.6888430562595116
C = 1000.0, decision boundary @ 5.474229032805065
C = 100000000.0, decision boundary @ 5.499634348989383
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