A prior in a uniform 😉

**Introduction** This is a rather interesting combination of the Standard Uniform Distribution and the Binomial Distribution.

As we know the Binomial Distribution has three parameters.

- Number of trials
- Probability for 0 (or 1)
- Number of the occurrences of 0 or 1

For many applications, we assume *p* to be constant and calculate the probabilities of *k* occurrences out of *n* trials. It would be straightforward to calculate and anticipate how the probabilities might look like. How about the case when the probability creating the binomial is itself a distribution? Interesting isn’t it?

*Choices on the probability distribution for the p*.

- Since we need a number between 0 and 1, we need a distribution that has finite limits.
- this makes exponential (and several other) probability distribution function completely useless

- We can choose distributions like
*Standard Uniform Distribution**Triangular distribution with limits between 0 and 1*- there is no restriction on the peak position

- Any other functions whose integral is 1 and the outcomes space is between 0 and 1

- We will look at the simulations of the Binomial outcomes with randomly varying probability

**Calculations**

For the calculations let us recapture a few things

- We will perform
*n*bernoulli trials sequentially - We need to calculate the probability that number of successes is k
- The probability is itself randomly distributed by a uniform distribution

P{ X=k | p=uniform_{[0,1] }} = ∫ Binomial ( n, k, p ) PDF[ uniform_{[0,1]} ] dp

∫ p^{k}(1-p)^{2n-k} dp = ^{Γ(n+1)}⁄_{Γ(n+2)} = ^{n!}⁄_{n+1 !} = ^{1}⁄_{n+1}

Voila! 😮 we have removed the dependence of probability on the *k. *See how the probability is only dependent on the number of trials and for all k: 0 ≤ k ≤ n. Seeing this in another way, any of the *k* successes amongst n trials have the same probability of happening. This is an astounding result.

**Simulations**

Let us do some simulations to see the results we obtained in the previous section. The simulation would only take a few lines but is a very powerful illustration of the concept. We will pass two types of the probability functions to construct the Binomial Distribution.

Uniform probability distribution from [0, 1]

Triangular distribution [0, 1] and peak at 0.5 – Can you guess why the distribution is dome shaped?

**Code**

AppendTo[$Path, "D:\\Mathematica Files 4K\\mathematicaPackages"]; SetOptions[$FrontEndSession, "EvaluationCompletionAction" -> {"ScrollToOutput"}] Needs["utilities`"]; Uniform Priors Gamma[n + 1] / (Gamma[k + 1] Gamma[n - k + 1]) \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\( \*SuperscriptBox[\(p\), \(k\)] \*SuperscriptBox[\((1 - p)\), \(n - k\)]\[DifferentialD]p\)\) ListPlot[{Gamma[# + 1] / Gamma[# + 2] & /@ Range[100], 1 / (# + 1) & /@ Range[100]}, PlotRange -> All, ImageSize -> 700] Module[{p = RandomReal[] & /@ Range[10000], data}, data = KeySort@Counts[RandomVariate[BinomialDistribution[20, #]] & /@ p]; BarChart[data, ChartLabels -> Placed[{Keys@data, Values@data}, {Below, Above}], ImageSize -> 788] ] Module[{p = RandomVariate[TriangularDistribution[{0, 1}], 100000], data}, data = KeySort@ Counts[RandomVariate[BinomialDistribution[20, #]] & /@ p]; BarChart[data, ChartLabels -> Placed[{Keys@data, Values@data}, {Below, Above}], ImageSize -> 788] ]

End of the post 😉