using DiffEqBayes, OrdinaryDiffEq, RecursiveArrayTools, Distributions, Plots, StatsPlots, BenchmarkTools, TransformVariables, CmdStan, DynamicHMC

Let's define our simple pendulum problem. Here our pendulum has a drag term `ω`

and a length `L`

.

We get first order equations by defining the first term as the velocity and the second term as the position, getting:

function pendulum(du,u,p,t) ω,L = p x,y = u du[1] = y du[2] = - ω*y -(9.8/L)*sin(x) end u0 = [1.0,0.1] tspan = (0.0,10.0) prob1 = ODEProblem(pendulum,u0,tspan,[1.0,2.5])

ODEProblem with uType Array{Float64,1} and tType Float64. In-place: true timespan: (0.0, 10.0) u0: [1.0, 0.1]

To understand the model and generate data, let's solve and visualize the solution with the known parameters:

sol = solve(prob1,Tsit5()) plot(sol)

It's the pendulum, so you know what it looks like. It's periodic, but since we have not made a small angle assumption it's not exactly `sin`

or `cos`

. Because the true dampening parameter `ω`

is 1, the solution does not decay over time, nor does it increase. The length `L`

determines the period.

We now generate some dummy data to use for estimation

t = collect(range(1,stop=10,length=10)) randomized = VectorOfArray([(sol(t[i]) + .01randn(2)) for i in 1:length(t)]) data = convert(Array,randomized)

2×10 Array{Float64,2}: 0.0569058 -0.375845 0.136638 0.0720362 … 0.00859386 -0.000752689 -1.20656 0.334965 0.293891 -0.255306 0.0281103 -0.00753411

Let's see what our data looks like on top of the real solution

scatter!(data')

This data captures the non-dampening effect and the true period, making it perfect to attempting a Bayesian inference.

Now let's fit the pendulum to the data. Since we know our model is correct, this should give us back the parameters that we used to generate the data! Define priors on our parameters. In this case, let's assume we don't have much information, but have a prior belief that ω is between 0.1 and 3.0, while the length of the pendulum L is probably around 3.0:

priors = [Uniform(0.1,3.0), Normal(3.0,1.0)]

2-element Array{Distributions.Distribution{Distributions.Univariate,Distrib utions.Continuous},1}: Distributions.Uniform{Float64}(a=0.1, b=3.0) Distributions.Normal{Float64}(μ=3.0, σ=1.0)

Finally let's run the estimation routine from DiffEqBayes.jl with the Turing.jl backend to check if we indeed recover the parameters!

bayesian_result = turing_inference(prob1,Tsit5(),t,data,priors;num_samples=10_000, syms = [:omega,:L])

Chains MCMC chain (9000×15×1 Array{Float64,3}): Iterations = 1:9000 Thinning interval = 1 Chains = 1 Samples per chain = 9000 parameters = L, omega, σ[1] internals = acceptance_rate, hamiltonian_energy, hamiltonian_energy _error, is_accept, log_density, lp, max_hamiltonian_energy_error, n_steps, nom_step_size, numerical_error, step_size, tree_depth Summary Statistics parameters mean std naive_se mcse ess rhat Symbol Float64 Float64 Float64 Float64 Float64 Float64 L 2.4780 0.2113 0.0022 0.0032 3799.5678 1.0001 omega 1.0897 0.2204 0.0023 0.0042 2820.5866 0.9999 σ[1] 0.1596 0.0382 0.0004 0.0006 4229.9547 1.0000 Quantiles parameters 2.5% 25.0% 50.0% 75.0% 97.5% Symbol Float64 Float64 Float64 Float64 Float64 L 2.0473 2.3485 2.4795 2.6056 2.9082 omega 0.7743 0.9501 1.0518 1.1849 1.6446 σ[1] 0.1006 0.1332 0.1537 0.1796 0.2518

Notice that while our guesses had the wrong means, the learned parameters converged to the correct means, meaning that it learned good posterior distributions for the parameters. To look at these posterior distributions on the parameters, we can examine the chains:

plot(bayesian_result)