# An Intro to Expectations via DiffEqUncertainty.jl

## System Model

First, lets consider the following linear model.

$$$u' = p u$$$

f(u,p,t) = p.*u
f (generic function with 1 method)

We then wish to solve this model on the timespan t=0.0 to t=10.0, with an intial condition u0=10.0 and parameter p=-0.3. We can then setup the differential equations, solve, and plot as follows


using DifferentialEquations, Plots
u0 = [10.0]
p = [-0.3]
tspan = (0.0,10.0)
prob = ODEProblem(f,u0,tspan,p)
sol = solve(prob)
plot(sol)
ylims!(0.0,10.0) However, what if we wish to consider a random initial condition? Assume u0 is distributed uniformly from -10.0 to 10.0, i.e.,


using Distributions
u0_dist = [Uniform(-10.0,10.0)]
1-element Array{Distributions.Uniform{Float64},1}:
Distributions.Uniform{Float64}(a=-10.0, b=10.0)

We can then run a Monte Carlo simulation of 100,000 trajectories by solving an EnsembleProblem.


prob_func(prob,i,repeat) = remake(prob, u0 = rand.(u0_dist))
ensemble_prob = EnsembleProblem(prob,prob_func=prob_func)

ensemblesol = solve(ensemble_prob,Tsit5(),EnsembleThreads(),trajectories=100000)
EnsembleSolution Solution of length 100000 with uType:
DiffEqBase.ODESolution{Float64,2,Array{Array{Float64,1},1},Nothing,Nothing,
Array{Float64,1},Array{Array{Array{Float64,1},1},1},DiffEqBase.ODEProblem{A
rray{Float64,1},Tuple{Float64,Float64},false,Array{Float64,1},DiffEqBase.OD
EFunction{false,typeof(Main.##WeaveSandBox#751.f),LinearAlgebra.UniformScal
ing{Bool},Nothing,Nothing,Nothing,Nothing,Nothing,Nothing,Nothing,Nothing,N
othing,Nothing,Nothing,Nothing},Base.Iterators.Pairs{Union{},Union{},Tuple{
},NamedTuple{(),Tuple{}}},DiffEqBase.StandardODEProblem},OrdinaryDiffEq.Tsi
t5,OrdinaryDiffEq.InterpolationData{DiffEqBase.ODEFunction{false,typeof(Mai
n.##WeaveSandBox#751.f),LinearAlgebra.UniformScaling{Bool},Nothing,Nothing,
Nothing,Nothing,Nothing,Nothing,Nothing,Nothing,Nothing,Nothing,Nothing,Not
hing},Array{Array{Float64,1},1},Array{Float64,1},Array{Array{Array{Float64,
1},1},1},OrdinaryDiffEq.Tsit5ConstantCache{Float64,Float64}},DiffEqBase.DES
tats}

Plotting the first 250 trajectories produces


plot(ensemblesol, vars = (0,1), lw=1,alpha=0.1, label=nothing, idxs = 1:250) Given the ensemble solution, we can then compute the expectation of a function $g\left(\cdot\right)$ of the system state u at any time in the timespan, e.g. the state itself at t=4.0 as


g(sol) = sol(4.0)
mean([g(sol) for sol in ensemblesol])
1-element Array{Float64,1}:
-0.007573120887912955

Alternatively, DiffEqUncertainty.jl offers a convenient interface for this type of calculation, expectation().


using DiffEqUncertainty
expectation(g, prob, u0_dist, p, MonteCarlo(), Tsit5(); trajectories=100000)
1-element Array{Float64,1}:
-0.0037433989352062547

expectation() takes the function of interest $g$, an ODEProblem, the initial conditions and parameters, and an AbstractExpectationAlgorithm. Here we use MonteCarlo() to use the Monte Carlo algorithm. Note that the initial conditions and parameters can be arrays that freely mix numeric and continuous distribution types from Distributions.jl. Recall, that u0_dist = [Uniform(-10.0,10.0)], while p = [-0.3]. From this specification, the expectation is solved as

$$$\mathbb{E}\left[g\left(X\right)\vert X\sim Pf\right]$$$

where $Pf$ is the "push-forward" density of the initial joint pdf $f$ on initial conditions and parameters.

Alternatively, we could solve the same problem using the Koopman() algorithm.


expectation(g, prob, u0_dist, p, Koopman(), Tsit5())
u: 1-element Array{Float64,1}:
0.0

Being that this system is linear, we can analytically compute the solution as a deterministic ODE with its initial condition set to the expectation of the initial condition, i.e.,

$$$e^{pt}\mathbb{E}\left[u_0\right]$$$

exp(p*4.0)*mean(u0_dist)
0.0

We see that for this case the Koopman() algorithm produces a more accurate solution than MonteCarlo(). Not only is it more accurate, it is also much faster


@time expectation(g, prob, u0_dist, p, MonteCarlo(), Tsit5(); trajectories=100000)
2.280767 seconds (79.32 M allocations: 7.160 GiB, 71.23% gc time)
1-element Array{Float64,1}:
-0.0050382269528445305

@time expectation(g, prob, u0_dist, p, Koopman(), Tsit5())
0.000798 seconds (12.26 k allocations: 1.111 MiB)
u: 1-element Array{Float64,1}:
0.0

Changing the distribution, we arrive at


u0_dist = [Uniform(0.0,10.0)]
@time expectation(g, prob, u0_dist, p, MonteCarlo(), Tsit5(); trajectories=100_000)
1.027060 seconds (79.30 M allocations: 7.160 GiB, 44.15% gc time)
1-element Array{Float64,1}:
1.5059653813564504

and


@time expectation(g, prob, u0_dist, p, Koopman(), Tsit5())
0.004259 seconds (14.02 k allocations: 1.221 MiB)
1.5059722133001539

where the analytical solution is


exp(p*4.0)*mean(u0_dist)
1.5059710595610105

Note that the Koopman() algorithm doesn't currently support infinite or semi-infinite integration domains, where the integration domain is determined by the extrema of the given distributions. So, trying to using a Normal distribution will produce NaN


u0_dist = [Normal(3.0,2.0)]
expectation(g, prob, u0_dist, p, Koopman(), Tsit5())
u: 1-element Array{Float64,1}:
NaN

Here, the analytical solution is


exp(p*4.0)*mean(u0_dist)
0.9035826357366062

Using a truncated distribution will alleviate this problem. However, there is another gotcha. If a large majority of the probability mass of the distribution exists in a small region in the support, then the adaptive methods used to solve the expectation can "miss" the non-zero portions of the distribution and errantly return 0.0.


u0_dist = [truncated(Normal(3.0,2.0),-1000,1000)]
expectation(g, prob, u0_dist, p, Koopman(), Tsit5())
u: 1-element Array{Float64,1}:
0.0

whereas truncating at $\pm 4\sigma$ produces the correct result


u0_dist = [truncated(Normal(3.0,2.0),-5,11)]
expectation(g, prob, u0_dist, p, Koopman(), Tsit5())
u: 1-element Array{Float64,1}:
0.9035833577709517

If a large truncation is required, it is best practice to center the distribution on the truncated interval. This is because many of the underlying quadrature algorithms use the center of the interval as an evaluation point.


u0_dist = [truncated(Normal(3.0,2.0),3-1000,3+1000)]
expectation(g, prob, u0_dist, p, Koopman(), Tsit5())
u: 1-element Array{Float64,1}:
0.903584360812248

## Vector-Valued Functions

expectation() can also handle vector-valued functions. Simply pass the vector-valued function and set the nout kwarg to the length of the vector the function returns.

Here, we demonstrate this by computing the expectation of u at t=4.0s and t=6.0s


g(sol) = [sol(4.0), sol(6.0)]
expectation(g, prob, u0_dist, p, Koopman(), Tsit5(); nout = 2)
u: 2-element Array{Float64,1}:
0.903584360812248
0.49589556820916314

with analytical solution


exp.(p.*[4.0,6.0])*mean(u0_dist)
2-element Array{Float64,1}:
0.9035826357366062
0.4958966646647597

this can be used to compute the expectation at a range of times simultaneously


saveat = tspan:.5:tspan
g(sol) = Matrix(sol)
mean_koop = expectation(g, prob, u0_dist, p, Koopman(), Tsit5(); nout = length(saveat), saveat=saveat)
u: 1×21 Array{Float64,2}:
3.0  2.58213  2.22246  1.91289  1.64644  …  0.201619  0.173536  0.149364

We can then plot these values along with the analytical solution


plot(t->exp(p*t)*mean(u0_dist),tspan..., xlabel="t", label="analytical")
scatter!(collect(saveat),mean_koop.u[:],marker=:o, label=nothing) ### Benefits of Using Vector-Valued Functions

In the above examples we used vector-valued expectation calculations to compute the various expectations required. Alternatively, one could simply compute multiple scalar-valued expectations. However, in most cases it is more efficient to use the vector-valued form. This is especially true when the ODE to be solved is computationally expensive.

To demonstrate this, lets compute the expectation of $x$, $x^2$, and $x^3$ using both approaches while counting the number of times g() is evaluated. This is the same as the number of simulation runs required to arrive at the solution. First, consider the scalar-valued approach. Here, we follow the same method as before, but we add a counter to our function evaluation that stores the number of function calls for each expectation calculation to an array.


function g(sol, power, counter)
counter[power] = counter[power] + 1
sol(4.0)^power
end

counters = [0,0,0]
x_koop = expectation(s->g(s,1,counters), prob, u0_dist, p, Koopman(), Tsit5())
x2_koop = expectation(s->g(s,2,counters), prob, u0_dist, p, Koopman(), Tsit5())
x3_koop = expectation(s->g(s,3,counters), prob, u0_dist, p, Koopman(), Tsit5())
counters
3-element Array{Int64,1}:
375
405
375

Leading to a total of 1155 function evaluations.

Now, lets compare this to the vector-valued approach


function g(sol, counter)
counter = counter + 1
v = sol(4.0)
[v, v^2, v^3]
end

counter = 
expectation(s->g(s,counter), prob, u0_dist, p, Koopman(), Tsit5(); nout = 3)
counter
1-element Array{Int64,1}:
405

This is 35.06% the number of simulations required when using scalar-valued expectations. Note how the number of evaluations used in the vector-valued form is equivelent to the maximum number of evaluations for the 3 scalar-valued expectation calls.

## Higher-Order Moments

Leveraging this vector-valued capability, we can also efficiently compute higher-order central moments.

### Variance

The variance, or 2nd central moment, of a random variable $X$ is defined as

$$$\mathrm{Var}\left(X\right)=\mathbb{E}\left[\left(X-\mu\right)^2\right]$$$

where

$$$\mu = \mathbb{E}\left[X\right]$$$

The expression for the variance can be expanded to

$$$\mathrm{Var}\left(X\right)=\mathbb{E}\left[X^2\right]-\mathbb{E}\left[X\right]^2$$$

Using this, we define a function that returns the expectations of $X$ and $X^2$ as a vector-valued function and then compute the variance from these


function g(sol)
x = sol(4.0)
[x, x^2]
end

koop = expectation(g, prob, u0_dist, p, Koopman(), Tsit5(); nout = 2)
mean_koop = koop
var_koop = koop - mean_koop^2
0.36287237175498144

For a linear system, we can propagate the variance analytically as

$$$e^{2pt}\mathrm{Var}\left(u_0\right)$$$

exp(2*p*4.0)*var(u0_dist)
0.36287181315765005

This can be computed at multiple time instances as well


saveat = tspan:.5:tspan
g(sol) = [Matrix(sol)'; (Matrix(sol).^2)']

koop = expectation(g, prob, u0_dist, p, Koopman(), Tsit5(); nout = length(saveat)*2, saveat=saveat)
μ = koop.u[1:length(saveat)]
σ = sqrt.(koop.u[length(saveat)+1:end] - μ.^2)

plot(t->exp(p*t)*mean(u0_dist),tspan..., ribbon = t->-sqrt(exp(2*p*t)*var(u0_dist)), label="Analytical Mean, 1 std bounds")
scatter!(collect(saveat),μ,marker=:x, yerror = σ, c=:black, label = "Koopman Mean, 1 std bounds") ### Skewness

A similar approach can be used to compute skewness


function g(sol)
v = sol(4.0)
[v, v^2, v^3]
end

koop = expectation(g, prob, u0_dist, p, Koopman(), Tsit5(); nout = 3)
mean_koop = koop
var_koop = koop - mean_koop^2
(koop - 3.0*mean_koop*var_koop - mean_koop^3) / var_koop^(3/2)
3.7952822389774445e-9

As the system is linear, we expect the skewness to be unchanged from the inital distribution. Becasue the distribution is a truncated Normal distribution centered on the mean, the true skewness is 0.0.

### nth Central Moment

DiffEqUncertainty provides a convenience function centralmoment around this approach for higher-order central moments. It takes an integer for the number of central moments you wish to compute. While the rest of the arguments are the same as for expectation(). The following will return central moments 1-5.


g(sol) = sol(4.0)
centralmoment(5, g, prob, u0_dist, p, Koopman(), Tsit5(),
ireltol = 1e-9, iabstol = 1e-9)
5-element Array{Float64,1}:
0.0
0.3628723692985327
4.026725619610261e-10
0.39502906894683987
1.2523240222606091e-9

## Batch-Mode

It is also possible to solve the various simulations in parallel by using the batch kwarg and a batch-mode supported quadrature algorithm via the quadalg kwarg. To view the list of batch compatible quadrature algorithms, refer to Quadrature.jl. Note: Batch-mode operation is built on top of DifferentialEquation.jl's EnsembleProblem. See the EnsembleProblem documentation for additional options.

The default quadtrature algorithm used by expectation() does not support batch-mode evaluation. So, we first load dependencies for additional quadrature algorithms

using Quadrature, Cuba

We then solve our expectation as before using a batch=10 multi-thread parallelization via EnsembleThreads() of Cuba's SUAVE algorithm. However, in this case we introduce additional uncertainty in the model parameter.


u0_dist = [truncated(Normal(3.0,2.0),-5,11)]
p_dist = [truncated(Normal(-.7, .1), -1,0)]

g(sol) = sol(6.0)

expectation(g, prob, u0_dist, p_dist, Koopman(), Tsit5(), EnsembleThreads();
quadalg = CubaSUAVE(), batch=10)
0.05397860786456136

Now, lets compare the performance of the batch and non-batch modes


using BenchmarkTools

@btime expectation(g, prob, u0_dist, p_dist, Koopman(), Tsit5();
quadalg = CubaSUAVE())
44.556 ms (1006187 allocations: 91.55 MiB)
0.05397860786456136

@btime expectation(g, prob, u0_dist, p_dist, Koopman(), Tsit5(), EnsembleThreads();
quadalg = CubaSUAVE(), batch=10)
21.936 ms (1019426 allocations: 93.00 MiB)
0.05397860786456136

It is also possible to parallelize across the GPU. However, one must be careful of the limitations of ensemble solutions with the GPU. Please refer to DiffEqGPU.jl for details.

Here we load DiffEqGPU and modify our problem to use Float32 and to put the ODE in the required GPU form


using DiffEqGPU

function f(du, u,p,t)
@inbounds begin
du = p*u;
end
nothing
end

u0 = Float32[10.0]
p = Float32[-0.3]
tspan = (0.0f0,10.0f0)
prob = ODEProblem(f,u0,tspan,p)

g(sol) = sol(6.0)

u0_dist = [truncated(Normal(3.0f0,2.0f0),-5f0,11f0)]
p_dist = [truncated(Normal(-.7f0, .1f0), -1f0,0f0)]

@btime expectation(g, prob, u0_dist, p_dist, Koopman(), Tsit5(), EnsembleGPUArray();
quadalg = CubaSUAVE(), batch=1000)
6.861 ms (76204 allocations: 3.71 MiB)
0.056093966910433016

The performance gains realized by leveraging batch GPU processing is problem dependent. In this case, the number of batch evaluations required to overcome the overhead of using the GPU exceeds the number of simulations required to converge to the quadrature solution.

## Appendix

This tutorial is part of the SciMLTutorials.jl repository, found at: https://github.com/SciML/SciMLTutorials.jl. For more information on doing scientific machine learning (SciML) with open source software, check out https://sciml.ai/.

To locally run this tutorial, do the following commands:

using SciMLTutorials
SciMLTutorials.weave_file("DiffEqUncertainty","01-expectation_introduction.jmd")

Computer Information:

Julia Version 1.4.2
Commit 44fa15b150* (2020-05-23 18:35 UTC)
Platform Info:
OS: Linux (x86_64-pc-linux-gnu)
CPU: Intel(R) Core(TM) i7-9700K CPU @ 3.60GHz
WORD_SIZE: 64
LIBM: libopenlibm
LLVM: libLLVM-8.0.1 (ORCJIT, skylake)
Environment:
JULIA_DEPOT_PATH = /builds/JuliaGPU/DiffEqTutorials.jl/.julia
JULIA_CUDA_MEMORY_LIMIT = 2147483648


Package Information:

Status /builds/JuliaGPU/DiffEqTutorials.jl/tutorials/DiffEqUncertainty/Project.toml
[6e4b80f9-dd63-53aa-95a3-0cdb28fa8baf] BenchmarkTools 0.5.0
[8a292aeb-7a57-582c-b821-06e4c11590b1] Cuba 2.1.0
[071ae1c0-96b5-11e9-1965-c90190d839ea] DiffEqGPU 1.6.0
[41bf760c-e81c-5289-8e54-58b1f1f8abe2] DiffEqSensitivity 6.31.1
[ef61062a-5684-51dc-bb67-a0fcdec5c97d] DiffEqUncertainty 1.5.0
[0c46a032-eb83-5123-abaf-570d42b7fbaa] DifferentialEquations 6.15.0
[31c24e10-a181-5473-b8eb-7969acd0382f] Distributions 0.23.8
[f6369f11-7733-5829-9624-2563aa707210] ForwardDiff 0.10.12
[76087f3c-5699-56af-9a33-bf431cd00edd] NLopt 0.6.0
[1dea7af3-3e70-54e6-95c3-0bf5283fa5ed] OrdinaryDiffEq 5.42.3
[91a5bcdd-55d7-5caf-9e0b-520d859cae80] Plots 1.6.0
[67601950-bd08-11e9-3c89-fd23fb4432d2] Quadrature 1.3.0