Add multi-dimensional random variables

[SerenaZwww]

In my problem, I have a series of random variables $d_{i,j,k}$ , where $i\in{1,\dots,10}$, $j\in{0,\dots,2}$ and $k\in{0,\dots,2}$. $d_{i,j,k}$ has a discrete uniform distribution with support ${k\dots,2}$. How can I define such 90 random variables?

[Oscar Dowson]

See https://sddp.dev/stable/guides/add_multidimensional_noise/

You must pass a single vector of the sample space, and an (optional) single vector of the associated probabilities.

There is no direct support for multi-variate random variables.

Given the size (there is something like 6^30 possibilities), you obviously can't add the every possible realization of d, so you'll have to sample.

[SerenaZwww]

Thank you for the reply! I'm reading the Sampling section on the page, and there's something I feel confused about.

distributions = Distributions.Product([
           Distributions.Binomial(100, 0.5),
           Distributions.Geometric(1 / 20),
           Distributions.Poisson(20),
       ]);

Does it mean this is a 3-d random variable, with the marginal distributions being Binomial, Geometric, and Poisson? So, if I have a 90-d random variable, do I have to add 90 lines between square brackets?

[SerenaZwww]

Thank you for the reply! I'm reading the Sampling section on the page, and there's something I feel confused about.

distributions = Distributions.Product([
           Distributions.Binomial(100, 0.5),
           Distributions.Geometric(1 / 20),
           Distributions.Poisson(20),
       ]);

Does it mean this is a 3-d random variable, with the marginal distributions being Binomial, Geometric, and Poisson? So, if I have a 90-d random variable, do I have to add 90 lines between square brackets?

Moreover, is it possible to generate them in a structured form as implied by the indices?

[Oscar Dowson]

If you want to use the 0-indexed indices:

using JuMP
d = Containers.@container([i in 1:10, j in 0:2, k in 0:2], rand(k:2))

Create a list of samples:

N = 20
Ω = [
    Containers.@container([i in 1:10, j in 0:2, k in 0:2], rand(k:2))
    for _ in 1:N
]
P = fill(1 / N, N)

[SerenaZwww]

Thank you for the reply! I think I get it!

[Oscar Dowson]

@variable(subproblem, d)
N = 20
Ω = [
    Containers.@container([i in 1:10, j in 0:2, k in 0:2], rand(k:2))
    for _ in 1:N
]
P = fill(1 / N, N)
SDDP.parameterize(subproblem, Ω, P) do ω
    return JuMP.fix(d, ω)
end

Here d is a scalar. The realizations ω are Containers.DenseAxisArray.

You probably want:

@variable(subproblem, d[i in 1:10, j in 0:2, k in 0:2])
N = 20
Ω = [
    Containers.@container([i in 1:10, j in 0:2, k in 0:2], rand(k:2))
    for _ in 1:N
]
P = fill(1 / N, N)
SDDP.parameterize(subproblem, Ω, P) do ω
    JuMP.fix.(d, ω)  # Note the fix.( 
    return
end

I have another series of random variables

This is not really another set of random variables, but a transformation. Perhaps something like:

@variable(subproblem, d[i in 1:10, j in 0:2, k in 0:2])
@variable(subproblem, r[i in 1:10, j in 0:2, k in 0:2, kp in 0:2])
N = 20
Ω = [
    Containers.@container([i in 1:10, j in 0:2, k in 0:2], rand(k:2))
    for _ in 1:N
]
P = fill(1 / N, N)
SDDP.parameterize(subproblem, Ω, P) do ω
    JuMP.fix.(d, ω)  # Note the fix.( 
    for i in 1:10, j in 0:2, k in 0:2, kp in 0:2
        r[i, j, k, kp] =  d[i, j, k] == k' ? 1 : 0
    end
    return
end