Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Assume that we have a prior distribution for the probability of success as follows.

p = BetaDistribution[6, 14]

which we can graph as this

Plot[Evaluate[PDF[p, k]], {k, 0, 1}]

I would like to use Mathematica to calculate the posterior distribution given that we take a sample and we get 420 successes out of a sample size of 1830.

what would be to calculate it and graph it?

Regards.

share|improve this question
add comment

2 Answers 2

up vote 8 down vote accepted

The posterior distribution for p is a Beta [426, 1424]. You can graph it as you did yourself.

Here is how you can compute in Mathematica.

In[1]:= bin = Binomial[n, x]*p^x*(1 - p)^(n - x)

Out[1]= (1 - p)^(n - x) p^x Binomial[n, x]

In[2]:= prior = p^(a - 1)*(1 - p)^(b - 1)/Beta[a, b]

Out[2]= ((1 - p)^(-1 + b) p^(-1 + a))/Beta[a, b]

In[3]:= den = 
 Integrate[bin*prior, {p, 0, 1}, 
  Assumptions -> {a > 0, b > 0, x >= 0, n >= x}]

Out[3]= (Binomial[n, x] Gamma[b + n - x] Gamma[a + x])/(
Beta[a, b] Gamma[a + b + n])

In[4]:= bin*prior/den // FullSimplify

Out[4]= ((1 - p)^(-1 + b + n - x) p^(-1 + a + x) Gamma[a + b + n])/(
Gamma[b + n - x] Gamma[a + x])
share|improve this answer
    
Yes, that is indeed the answer, but I wanted to know how I can pose the problem in Mathematica for it to come back with that answer –  PatoCriollo Jun 1 '12 at 20:27
add comment

I would like to extend asim's solution, which is a function, to a probability distribution.

ClearAll[evidence, likelihood, prior, posterior];
likelihood = p^x*(1 - p)^(n - x) (* is always unnormalized ! *)
prior = p^(a - 1)*(1 - p)^(b - 1)/Beta[a, b] (* always normalized *)
evidence = Integrate[likelihood*prior, {p, 0, 1},
Assumptions -> {a > 0, b > 0, x >= 0, n >= x}] (* normalisation of posterior *)
posterior = prior*likelihood/evidence // FullSimplify

(* Out[]= *)
(1 - p)^(n - x) p^x
((1 - p)^(-1 + b) p^(-1 + a))/Beta[a, b]
(Gamma[b + n - x] Gamma[a + x])/(Beta[a, b] Gamma[a + b + n])
((1 - p)^(-1 + b + n - x) p^(-1 + a + x) Gamma[a + b + n])/(Gamma[b + n - x] Gamma[a + x])

Nothing new so far. Now for the probability distribution:

ClearAll[posteriorDistribution];
posteriorDistribution[a_, b_, x_, n_] = ProbabilityDistribution[posterior, {p, 0, 1}, 
Assumptions -> {a > 0, b > 0, x >= 0, n >= x}];
PDF[posteriorDistribution[6, 14, 420, 1830], 0.21] (* test at p=0.21 *)

4.6369

Check this answer by asim's solution in BetaDistribution[] form:

PDF[posteriorDistribution[6, 14, 420, 1830], 0.21] == PDF[BetaDistribution[426, 1424], 0.21]

True

This probability distribution can be plotted by:

Plot[PDF[posteriorDistribution[6, 14, 420, 1830], p], {p, 0, 1}, PlotRange -> All]

Several statistical properties work fine:

Mean[posteriorDistribution[6, 14, 420, 1830]] // N
StandardDeviation[posteriorDistribution[6, 14, 420, 1830]] // N

0.23027
0.00978554

But RandomVariate[] does not:

RandomVariate[posteriorDistribution[6, 14, 420, 1830], 5]

RandomVariate[
 ProbabilityDistribution[
 25100175092366680639705706981674366907407305357564535864428789165880\
7774460815201505720742948687064397263210935672027225834571883138571552\
9435073323698076733502611129508985779654636172308803925192878804616791\
4163296382700546574839309987790038929990089862577986787654011552037119\
0043901814728663569787527433974843693348538658624431557074602596292870\
1098499374967102449446582880605618133093761170547891966862458477624898\
28252384878666048 (1 - \[FormalX])^1423 \[FormalX]^425, {\[FormalX], 
0, 1}, Assumptions -> {True, True, True, True}], 5]

Anyone here who knows what is going wrong?

Of course, the BetaDistribution[] works fine:

RandomVariate[BetaDistribution[426, 1424], 5]

{0.224794, 0.246947, 0.225743, 0.241723, 0.219218}
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.