I would like to fit Dirichlet Distribution to all_patterns using MLE in the Distributions package
where alpha to Dirichlet is fill(1,k) and all_patterns is a combination array

all_patterns = (p=(a,c)->c<2 ? a : [[x;y] for x=a for y=p(a,c-1)])(0:1,K)
all_patterns = convert(Array,VectorOfArray(all_patterns)')

Do you need to use floating-point numbers? The code below works:

using Distributions
K = 3
all_patterns = (p=(a,c)->c<2 ? a : [[x;y] for x=a for y=p(a,c-1)])(1:2,K)
all_patterns = hcat(all_patterns...)
julia> fit(Dirichlet, Float64.(all_patterns))
Dirichlet{Float64}(alpha=[1.677728917182107e-5, 1.677728917182107e-5, 1.677728917182107e-5])

My all_patterns is a matrix of binary values (0 & 1) whereas in your code it’s not the case
1.1 While running you code with binary matrix, I am getting the error ArgumentError: Dirichlet: alpha must be a positive vector.

Dirichlet takes a vector (alphas) as an argument. In your code I think all_patterns are being passed as the alphas. I am not sure though.

Did you see my answer on Slack? It would be helpful if you could turn your example into an MWE that people can run, rather than trying to guess what all_patterns is (as K and VectorOfArrays are undefined).

I also don’t understand why in your OP you seem to want to pass a vector of alphas, given that that’s the only parameter of the Dirichlet what are you looking to estimate?

Thanks, that clarifies things a little bit. I still don’t fully understand what you’re trying to do as I haven’t thought through the dimensions of your problem, but it now seems to me that when you say “prior” you mean an initial value for alpha (rather than a prior in the Bayesian sense). In that case you probably need the init keyword:

using Distributions, RecursiveArrayTools
K = 3
all_patterns = (p=(a,c)->c<2 ? a : [[x;y] for x=a for y=p(a,c-1)])(0.0:1.0,K)
all_patterns = convert(Array,VectorOfArray(all_patterns)')
fit_mle(Dirichlet, all_patterns, init = [1. for _ ∈ 1:8])

Note that here I used an init vector of length 8, as your all_patterns matrix ends up being 8x3 - that’s what I meant by I haven’t thought through the dimensions of your problem, it seems from your initial post that you might want 3x8 here?