4 Deterministic and stochastic modelling Flashcards
deterministic models
ODE or a set of ODEs for
X(t) = vector
(N(t))
P(t))
ODE:
dX(t)/ dt = F(X(t))
X(t) ∈ IR^n
F:IR^n→IR^n
Given X(t) at some time t, we can obtain X(t + ∆t) from
X(t + ∆t) =
X(t) + F(X(t))∆t +O(∆t²).
(O(∆t²) means roughly constant times ∆t² for small ∆t)
Spacial models
individuals more likely to interact dep on location x(t)
stochastic models
If randomness is important then we should, instead of ODEs, use a stochastic description. A
fundamental property of a stochastic model is that it considers an ensemble of possible histories
of the system. If we know N(t), we can’t say with certainty what N(t + ∆t) will be, no matter
how small ∆t is. Instead, we talk in terms of “probabilities”. In particular, we consider the
quantities
p_n(t) = P (N(t) = n),
for each integer n ≥ 0. You should read this as:
“The probability that there are exactly n individuals in the population at time t, is
denoted pn(t).”
Example: pure linear birth process
N(t)
simple model assumptions
(stochastic)
N(t) Number of cells at time t
cells divide and dont die- exp growth
We assume each cell divides with rate β indep of history and other cells
Example: pure linear birth process
Probabilities for model
P (one of the cells divides before t + ∆t)
find p_n(t+∆t)
remember we have n cells…
P (one of the cells divides before t + ∆t) = N(t)β∆t + O(∆t²)
*at most we assume 1 division in ∆t
Thus
p_n((t + ∆t) =
P (N(t) = n and** no cell divides before** t + ∆t)
+ P (N(t) = n − 1 and one cell divides before t + ∆t)
= p_n(t)(1 − nβ∆t) + (n − 1)βp_{n−1}(t)∆t
+ O(∆t²)
Example: pure linear birth process
using Probabilities for model for ODE
How do we find an ODE from the probabilities?
p_n(t + ∆t) − p_n(t)
[(− nβp_n(t) + (n−1)βp_{n−1}(t))*∆t] + O(∆t²)
Take ∆t → 0 we find
dp_n(t)/ dt
= −nβp_n(t) + (n−1)βp_{n−1}(t).
Set of ODES for which ode of p_n(t) dep on prev p_(n-1)
Example: pure linear birth process
N(0)=n_0 gives
p_n(0)
dp_n(0)/dt
(P(N_0 /= n_0)=0)
p_n(0)= (
1 if n = n_0
0 otherwise ,
initial condition is “deterministic
p_n(t)=0 for all t ≥ 0 if n < n_0
dp_n(t)/dt = −βn_0p_{n_0}(t)
which has solution
p_{n_0}(t) = exp(−βn_{0}t)
Example: pure linear birth process
Use dp_n(t)/dt with n ={n_0}+1
dp_n(t)/dt
= −nβp_{n}(t) + (n − 1)β* p_{n−1}(t)
**
give dp_{n₀+1}(t)/dt
and solve**
dp_{n₀+1}(t)/dt
= −β(n₀ + 1)p_{n₀+1}(t) + βn₀p_{n₀}(t)
= −β(n₀ + 1)p_{n₀+1}(t) + βn₀e^{−βn₀t}
The solution is
p_{n₀+1}(t) = n₀(e^{−βn₀t} − e^{−β(n₀+1)t)
Probability that we have one division more than n_0 at time t
(solution is found using variation of constants)
For the simple PURE LINEAR BIRTH PROCESS what do the solutions of p_{n₀} and p_{n₀+1} look like
with n₀=4 and β=1
p_{n₀} : at t=0 start is large (1-certain) then decreases(exponentially) with time, for these values close to 0 at 1 and hits 0 after a long time)
p_{n₀+1} this is 0 immediately at t=0 but increases straight away, the rate of increase dereases and it never gets too high and rate of increase starts to decrease then it decreases (for these values slightly above p_{n₀} for the last part after it starts to decrease but towards 0 after a long time.
Gets to 0 later than p_{n₀
p_{n₀+1}(t) = n₀(e^{−βn₀t} − e^{−β(n₀+1)t)
p_{n_0}(t) = exp(−βn_{0}t)
For the simple PURE LINEAR BIRTH PROCESS quick recap
We found the p_n (t)
from finding an ODE from the probabilities of ∆t
solved this for individual ODEs
We could solve for p_{n_0+2}(t) next
OR could obtain all p
_n(t) at once using the pgf
X(t) = IE(N(t))
then dX/dt=?
X(t) = E(N(t))
=Σ _[n=1,..∞] n*pₙ(t)
Then dX(t)/dt =
Σ _[n=1,..∞] n*dpₙ(t)/dt
= dp/dt+ 2dp₂/dt+ 3dp₃/dt+ 4dp₄/dt+ ….
Example: pure linear birth process
Use dp_n(t)/dt to find the mean E(N(t))
dp_n(t)/dt
= −nβp_{n}(t) + (n − 1)β* p_{n−1}(t)
dX(t)/dt =
Σ _[n=1,..∞] n*dpₙ(t)/dt
= dp₁/dt+ 2dp₂/dt+ 3dp₃/dt+ 4dp₄/dt+ ….
= β(-p₁ + 2(p₁-2p₂) +3(p₂-3p₃)+4(3p₃-4p₄)
= β(p₁+2p₂+3p₃+4p₄)+…..)
=βX(t)
d( E(N(t))) /dt = βX(t)
solution
E(N(t)) = n₀e^(βt)
Example: pure linear birth process
Use dp_n(t)/dt
dp_n(t)/dt
= −nβp_{n}(t) + (n − 1)β* p_{n−1}
(t)
To find : d/dt of
E(N²(t)) = Σ _[n] n²pₙ(t),
solution?
dE(N²(t)) /dt
= 2β*E(N²(t)) + βX(t).
solution is
E(N²(t)) = n₀ (e^{2βt} − e{βt})
+ n₀²e^{2βt}
(using variation of constants again)
so that
var(N²(t)) =
E(N(t)) − (IE(N(t)))²
= n₀(e^{2βt }− e{βt})
Pure linear birth process mean behaviour
*we have found that the mean # cells at time t X(t) = n_0e^{βt}
Shows the mean grows exponentially with time
For a general model
(pure linear birth process)
Say p_{n₀} (0)=1 then initially we hae no cells
p_{n₀} (t) = exp(-)
VARIATION OF CONSTANTS METHOD
y’(x) + p(x)y(x) = q(x)
so dp_{n₀+1}/dt =
P_{n₀+1}(t) =
sub into s.t satisfies ODE…… notes
