Hi,

I’m currently coding up a particle filter (a.k.a. Sequential Monte Carlo, among other names) for approximate inference in Chemical Reaction Networks. My dynamics for the propagation of particles will be based on the SSA algorithm, as implemented within DifferentialEquations.jl.

The plan is to eventually use this filter within a Particle MCMC scheme (with some extra details), and so I need to be able to run the SSA algorithm many times, with many different initial conditions, and with different rate constants. For the moment, let’s just focus on varying the initial conditions.

My main concern is how I should construct the `DiscreteProblem, JumpProblem`

objects which will be `solve`

d in the main loop of the filter. Below is a toy version of the problem in which I am interested:

```
using Catalyst, DifferentialEquations, Distributions
lotvol_model = @reaction_network begin
c1, x1 --> 2x1
c2, x1 + x2 --> 2x2
c3, x2 --> 0
end c1 c2 c3
# Generate synthetic data
x0 = floor.(100 * rand(10))
y0 = [rand(Poisson(1)) for i in 1:10]
# Advance MJP from x by time Δt, with rates c
function MJPStep_LotVol(x, c, Δt)
MJP = DiscreteProblem(lotvol_model, x, (0.0, Δt), c)
jump_prob = JumpProblem(lotvol_model, MJP, Direct())
sol = solve(jump_prob, SSAStepper())
return sol(Δt)
end
# Importance Sampling with
# prior (independently across time) x[1], ... , x[T],
# evolution by MJP with rates c, and
# data y[1], ..., y[T]
function ImportanceSampling(x, y, c)
T = length(y)
Z = zeros(T)
W = zeros(T)
for t = 1:T
Z[t] = MJPStep_LotVol(x[t], c, 1) # propagate particle using SSA
W[t] = logpdf(Poisson(Z[t]), y[t]) # weight particle by observation likelihood
end
end
```

The situations in which I am interested correspond (roughly) to `T`

in the above being very large.

My main worry is that the way in which I define `MJPStep_LotVol(x, c, Δt)`

is probably quite inefficient - I’m guessing that forming the `DiscreteProblem, JumpProblem`

at each step in the inner loop could be quite costly, or might be in some way unnecessary?

On the other hand, the problems which the function is trying to solve at each step are very similar (in this example, they only differ in initial conditions) and so it seems reasonable that there’s a simpler solution which isn’t so costly.