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# The tau-leap method for simulating stochastic kinetic models

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### The tau-leap method for simulating stochastic kinetic models

1. 1. Approximate accelerated stochastic simulation of chemically reacting systems Colin Gillespie October 16, 2013
2. 2. Overview Stochastic kinetic models Issues with the Direct method τ -leap method (Gillespie, 2001) Leap conditions Mid-point estimation Examples A Source code (R and LTEX of these slides): https://github.com/csgillespie/talks/
3. 3. Stochastic kinetic models A biochemical network is represented as a set of pseudo-biochemical reactions: u species & v reactions Ri : ci p11 X1 + p12 X2 + · · · + p1u Xu − q11 X1 + q12 X2 + · · · + q1u Xu → Stochastic rate constant ci . Hazard/instantaneous rate: hi (Xt , ci ) where Xt = (X1,t , . . . , Xu ,t ) is the current state of the system. Under mass-action stochastic kinetics, the hazard function is proportional to a product of binomial coefﬁcients, with u hi (Xt , ci ) = ci j =1 Xj ,t pij .
4. 4. Stochastic kinetic model Describe the SKM by a Markov jump process (MJP) The effect of reaction Rk is to change the value of each species Xi by qji − pji The stoichiometry matrix S has elements sij = qji − pji It can be shown that the time to the next reaction is v t ∼ Exp(h0 (Xt , c )) where h0 (Xt , c ) = hi (Xi , ci ) i =1 and the reaction is of type i with probability hi (Xt , ci )/h0 (Xt , c ) The process is easily simulated using the Direct method (Gillespie algorithm)
5. 5. Example: Lotka-Volterra system Stochastic realisations 500 c1 Population 400 R1 : Prey reproduction X1 − → 2X1 − Predator 300 200 100 Prey 0 0 10 0 20 10 30 20 30 500 R2 : Prey death, predator reproduction X1 + X2 − → 2X2 − Population 400 c2 300 200 100 R3 : Predator death 0 c3 X2 − → ∅ − Time X(0) = (100, 100) c = (0.5, 0.0025, 0.3)
6. 6. The direct method 1 Initialisation: initial conditions, reactions constants, and random number generators 2 Propensities update: Update each of the v hazard functions, hi (x ) 3 Propensities total: Calculate the total hazard h0 = 4 Reaction time: τ = −ln[U (0, 1)]/h0 and t = t + τ 5 Reaction selection: A reaction is chosen proportional to it’s hazard 6 Reaction execution: Update species 7 Iteration: If the simulation time is exceeded stop, otherwise go back to step 2 Typically there are a large number of iterates v i =1 hi (x )
7. 7. The direct method 1 Initialisation: initial conditions, reactions constants, and random number generators 2 Propensities update: Update each of the v hazard functions, hi (x ) 3 Propensities total: Calculate the total hazard h0 = 4 Reaction time: τ = −ln[U (0, 1)]/h0 and t = t + τ 5 Reaction selection: A reaction is chosen proportional to it’s hazard 6 Reaction execution: Update species 7 Iteration: If the simulation time is exceeded stop, otherwise go back to step 2 Typically there are a large number of iterates v i =1 hi (x )
8. 8. The direct method 1 Initialisation: initial conditions, reactions constants, and random number generators 2 Propensities update: Update each of the v hazard functions, hi (x ) 3 Propensities total: Calculate the total hazard h0 = 4 Reaction time: τ = −ln[U (0, 1)]/h0 and t = t + τ 5 Reaction selection: A reaction is chosen proportional to it’s hazard 6 Reaction execution: Update species 7 Iteration: If the simulation time is exceeded stop, otherwise go back to step 2 Typically there are a large number of iterates v i =1 hi (x )
9. 9. The direct method 1 Initialisation: initial conditions, reactions constants, and random number generators 2 Propensities update: Update each of the v hazard functions, hi (x ) 3 Propensities total: Calculate the total hazard h0 = 4 Reaction time: τ = −ln[U (0, 1)]/h0 and t = t + τ 5 Reaction selection: A reaction is chosen proportional to it’s hazard 6 Reaction execution: Update species 7 Iteration: If the simulation time is exceeded stop, otherwise go back to step 2 Typically there are a large number of iterates v i =1 hi (x )
10. 10. Approximations Relax some assumptions (e.g. discreteness and stochasticity) in order to make simulation faster and more scalable Diffusion approximation / chemical Langevin equation (CLE) Linear noise approximation (LNA) / moment closure (2MA) ODE Hybrid discrete-continuous models
11. 11. Poisson leap If all reactions are zeroth-order, then the model is a homogeneous Poisson process Hence the number of reactions in (t0 , t1 ) follows a Poisson distribution For a more general model, if we consider a small time interval, (t , t + ∆t ), then: the hazard rates should be approximately constant the number of reactions (of a given type) can be sampled from a Poisson distribution A balance between speed and accuracy
12. 12. Poisson leap method 1 Set t = 0. Initialise the rate constants and the initial molecule numbers x 2 Calculate hi (x , ci ), for i = 1, . . . , v , and simulate the v -dimensional reaction vector r , with i th entry a Po(hi (x , ci )∆t ) random quantity 3 Update the state according 4 Update t := t + ∆t 5 Output t and x. If t < Tmax return to step 2
13. 13. Poisson leap method 1 Set t = 0. Initialise the rate constants and the initial molecule numbers x 2 Calculate hi (x , ci ), for i = 1, . . . , v , and simulate the v -dimensional reaction vector r , with i th entry a Po(hi (x , ci )∆t ) random quantity 3 Update the state according 4 Update t := t + ∆t 5 Output t and x. If t < Tmax return to step 2 How do you chose ∆t?
14. 14. Example: Lotka-Volterra Gillespie simulations 100 80 Population Suppose we are interested in parameter inference 60 40 20 A possible prior could be independent Uniform priors over U (−8, 8) for each log(ci ) 0 20 30 0 10 20 30 0 10 20 30 1600 1200 800 400 0 Probability of extinction by time t = 30, is around 0.86 10000 Population Each simulation would require a very different ∆t 10 2000 Population Three samples from this prior yield very different realisations 0 8000 6000 4000 2000 0
15. 15. Basic τ -leap method Suppose a temporal leap τ will result in a state change λ Choose a value of τ that satisﬁes the leap condition. For each reaction, Rj , we want |hj (x + λ) − hj (x)| to be small Sample kj ∼ P (hj (x)τ ) Compute λ Set t := t + τ and x := x + λ
16. 16. Choosing τ If the reactions don’t depend on x, the leap condition will be satisﬁed exactly for any τ (and so exact) If population numbers are large, then it would take a large number of reactions to “noticeably” change the hazard functions If satisfying the leap condition requires a very small value τ << 1/h0 (x), then we may as well use the exact SSA
17. 17. A procedure for selecting τ (Gillespie 2001) The expected net change in (t , t + τ ) will be: v ¯ λ= [hj (x)τ ]sj = τ ξ(x) j =1 So we require that the expected changes in the propensity functions in time τ , are bounded by some fraction of all propensity functions, i.e. |hj (x + λ) − hj (x)| < h0 (x) for j = 1, . . . , v . Estimate the difference using a Taylor expansion: u hj (x + λ) − hj (x) τ ξi (x) i =1 ∂ hj (x) ∂ xi Deﬁning bji (x) ≡ ∂ hj (x) ∂ xi (j = 1, . . . , v ; i = 1, . . . , u )
18. 18. A procedure for selecting τ The requirement becomes u ξi (x)bji (x) ≤ h0 (x) (j = 1, . . . , v ) τ i =1 The largest value of τ consistent with this condition (and hence optimal) is τ = min j ∈[1,v ] Typical values of are around 0.05 h0 ( x) u i =1 ξi (x)bji (x)
19. 19. Example: Lotka-Volterra 200 150 Hazard h2 h1 100 h3 50 0 0 10 20 30
20. 20. Example: Lotka-Volterra 200 150 Hazard h2 h1 100 h3 50 0 0 10 20 30 0 10 20 30 1 0.8 τ 0.6 0.4 0.2
21. 21. Example: Lotka-Volterra 200 150 Hazard h2 h1 100 h3 50 0 0 10 20 30 0 10 20 30 1 0.8 τ 0.6 0.4 0.2
22. 22. Example: Lotka-Volterra 1000 200 150 800 h3 Frequency Hazard h2 h1 100 50 0 0 10 20 600 400 30 200 1 0.8 0 0.0 0.6 0.1 0.2 0.3 0.4 0.5 0.3 0.4 0.5 τ τ 0.4 0.2 Mean τ 0 10 20 30 0.0 0.1 0.2
23. 23. The estimated-midpoint technique The leap condition requires that the hazards functions do not “appreciably” change in the course of a leap But we want to take large leaps, so we will inevitably get computational errors This is similar to solving the ODE dX (t ) dt = f (X ) using an Euler scheme A standard technique is to use a second-order Runge-Kutta or modiﬁed Euler method X (t + ∆t ) = X (t ) + f [X (t ) + 0.5f (X (t ))∆t ]∆t i.e. we use an Euler method to estimate the midpoint during [t , t + ∆t ], then calculate the increment in X by evaluating the slope function f at that estimated midpoint
24. 24. Example: Death model The death model contains a single reaction µ X −→ ∅ − and has hazard function h1 (x , µ) = µx and state change vector s = −1. The solution to the CME is: Pr(X = x ; t ) = x0 x e−µτ (x0 −x ) (1 − e−µτ )x
25. 25. Death model Method τ = 0.1 Exact τ = 0.9 τ = 0.4 0.15 Pr(k) 0.10 0.05 0.00 0 50 100 150 0 50 100 150 0 Population Pr(X = x ; τ ) = x0 x e−µτ (x0 −x ) (1 − e−µτ )x 50 100 150
26. 26. Death model: p-leap Method Exact τ = 0.1 Poisson−leap τ = 0.9 τ = 0.4 0.15 Pr(k) 0.10 0.05 0.00 0 50 100 150 0 50 100 150 0 50 100 Population If we perform a single leap of length τ , the number of executed reactions are Pp (k ; µx0 , τ ) = eµx0 τ (µx0 τ )k k! for k = 0, . . . 150
27. 27. Death model: τ -leap + mid-point Method Exact Poisson−leap τ = 0.1 Mid−point τ = 0.9 τ = 0.4 0.15 Pr(k) 0.10 0.05 0.00 0 50 100 150 0 50 100 150 0 50 Population We estimate the mid-point to be: x ≡ x0 − 0.5τ µx0 so the number of reactions executed is Pp (k ; µx0 , τ ) = eµx τ (µx τ )k k! for k = 0, . . . 100 150
28. 28. Example: Lotka-Volterra Prey Predator 400 q q ∆t q q 0.09 q q 0.18 q q 0.27 q 0.36 q 0.45 q 0.54 q 0.63 Population 300 q 200 q 100 0 q q q q q q p−leap τ−leap τ−leap + mid Simulator p−leap τ−leap τ−leap + mid
29. 29. Example: Lotka-Volterra Prey Predator 400 q q ∆t q q 0.09 q 0.18 q 0.27 q 0.36 q 0.45 q 0.54 q 0.63 300 q q Population q q q 200 q q q q q q 100 q q q q q q p−leap 0 q q q q q q τ−leap τ−leap + mid p−leap τ−leap τ−leap + mid Simulator For these parameter values and this model, we have the approximate relationship (obtained via simulation) 0.556 × ∆t
30. 30. Example: Lotka-Volterra Prey Predator 400 q q ∆t q q 0.09 q 0.18 q 0.27 q 0.36 q 0.45 q 0.54 q 0.63 300 q q Population q q q 200 q q q q q q q q q 100 q q q q q q p−leap 0 q q q q q q τ−leap q τ−leap + mid p−leap τ−leap τ−leap + mid Simulator For these parameter values and this model, we have the approximate relationship (obtained via simulation) 0.556 × ∆t
31. 31. Summary Similar issues arise when solving SDEs with the Euler-Maruyama scheme In fact, if we substitute Poisson with Gaussian random numbers in the p-leap scheme, we get the Euler-Maruyama algorithm Choosing a ﬁxed ∆t for a wide range of parameter combinations doesn’t make sense Is it possible to these ideas when constructing bridges for SDEs?
32. 32. References Gillespie DT. Exact stochastic simulation of coupled chemical reactions. The Journal of Physical Chemistry, 1977, 81: 2340 – 2361. Direct method Gillespie, DT. Approximate accelerated stochastic simulation of chemically reacting systems. The Journal of Chemical Physics, 2001, 115: 1716. τ -leap method Sandmann, W. Streamlined formulation of adaptive explicit-implicit tau-leaping with automatic tau selection. Simulation Conference (WSC), Proceedings of the 2009 Winter. IEEE, 2009. Nice overview of the various τ -leap ﬂavours. Golightly, A & Gillespie, CS. Simulation of Stochastic Kinetic Models. In Silico Systems Biology. Humana Press, 2013, 169 – 187. Overview of various SSA. https://github.com/csgillespie/In-silico-Systems-Biology