python - Normalizing Bayesian IRT Model in pymc -

This was the best example of estimating this type of IRT Baician model using MCMC in Python. Below is a reproducible version of code that I got to run I understand that to identify the model, the ability parameter theta is generally constrained to be distributed with 0, standard deviation 1, which I thought was done with this line in the code given below. Samples are from a normal distribution of thita = general ("theta", mu = 0, tau = 1, value = theta_inel, observed = creation)

  # theta (proficiency syllabus) / Code>  
 However, when I run the code, backlit means for theta a b Or with standard deviation is like 1.85e18, i.e. does not mean zero and standard deviation. Why did I get this error and how do I ensure that theta means normal 0, after each iteration, the SD1?  
  #from pylab import * #Pylab will not be established with the PIP, so I just mplot as pym.Matplot import conspiracy import numpy load * import numpy pymc from numpy import * np numpy import as numquestions = 300 # test number of items fake numpeople = 10 # number of participants Numthetas = 1 # generated proficiency variable = 0 theta_initial = zero ((numthetas, numpeople)) Accuracy = np.random.randint ( 2, size = numquestions * numpeople) == 1 # error #correctness = np.random (2, size = numquestions * numpeople) == -1 # all wrong code immediately #correctness = np.random.randint (2, Size = numquestions * numpeople)! = -1 # .shape correct code error message throws Accuracy = (numquestions, numpeople) # Theta (proficiency params) are sampled from a normal distribution theta = normal ( "theta", mu = 0, tau = 1, value = theta_initial , Observed = output) # Question-parameters (IRT parameters) are sampled with normal distribution (though others were tried) a = normal ("a", mu = 1, tau = 1, value = [[0.0] ] = Numthetas] * numquestions) # a = exponential ("a", beta = 0.01, value = [[0.0] * numthetas] * numquestions) = B = normal ("b", mu = 0, tow = 1, value = [0.0] * points) # vectors Eta / A / B take-back of each individual Ditrministik a vector G @ each question correct potential receipt of Diif Sigmayoid (theta = theta, A = A, B = B): BS = repeat (re-size (B, (Lane (b), 1)), number, 1) returns np.xp (1.0 / (1.0 + np.xp (bs-dot (a, theta))) take out the possibility of coming out of sigmoid, and weighted Flip the signal = burnoli ('true', p = sigmoid, value = viable Security is observed) = are not generated) # All Maisimsi including the above variables ([A, B, theta, Sigmoyoid, right]) Create a Piimk simulation object # An interactive MCMC sampling session M. Examples (getter = 20000, burn = 15000) Mydict = m.stats () print (mydict ['theta'] ['mean']) # Get eligibility criteria for each student print (MIDIct ['theta'] [' Mean ']. Mean ()) # should be zero, but gives some links 1.85e18, that is, an absurd big value.  
 

I think you have an extra n.exp According to your sigmoid function, S (t) = 1 / (1 + exp (-t)) . I've changed your line 34 with this alternate version:

  return 1.0 / (1.0 + np.exp (BS-dot (A, theta)))  

With this I get the meaning of the theta of 0.08.