# Finite state Markov model notes

simple switching model. Two systems, 1 and 2, each two states A and B.

``````   fi
Ai <==> Bi
bi

Bi --> Bi  + mRNA

A1 <==> A2

[*   f1   s12   0
b1  *     0     0
s2  0    *      f2
0   0     b2   * ]
``````

Previous notes:

Maybe I can clarify the question a bit as well.

First, it might be useful to consider the simpler, more concrete problem: Let Sx and Sy be Poisson processes. We switch determinsitically between these two processes so we spend a fixed amount of time in Sx, where we produce a random number of molecules X, then switch to Sy and produce Y molecules. The total number of molecules produced is:
M = X1 + Y1 + X2 + Y2…. = NX + NY ?

(q1) what are the first two moments of M?
mean = lambda_x(fraction of time spent in Sx) + lambda_y(fraction of time spent in Sy).
Not so sure about the variance. E[M^2] = N^2*( E[X^2] +2 E[XY] + E[Y^2] ) . What is E[XY]? The X1 is determined just by process Sx so I am tempted to write E[XY] = E[X][Y] as if they were independent. But given that the process is in Sx it is definetely not in Sy so maybe E[XY] should be zero.

(q2) How about if Sx and Sy are Poisson, and we hop between them will exponentially distributed lifetimes in each state?

For the more general case, I’d like to assume the two production processes come to equilibrium rapidly (relative to the length of each cycle). I also want to assume that the death rate is slow relative to switching rate (and hence also to the kinetics of each process). Since if the death rate was really fast the system would just look like process 1 for a while and then switch to being process 2 for a while and there would be no mRNA molecules carried over between regimes. Maybe it would be best to ignore decay entirely for the present.

The intuitive (qualitative) answer, which I can also get from simulation is based on the following observation:
if process Sx has a high mean and a small variance, and Sy has a low mean and a low variance, the mixed process should produce some hig

I tried working out the simplier case where the switching between the states is deterministic, so it spends time fixed time tx in state Sx and produces a random number Xi products, then fixed time ty in state Sy, producing a random number Yi products. I write this out below to illustrate more clearly the assumptions I think are appropriate:

The random number signifying the total output M = X1+Y1+X2+Y2+… where all the Xi are iid and the Yi are iid.
Or for N cycles it is: E[M] =
E[M] = NE[X]+NE[Y]
and
E[M^2] = N^2*( E[X^2] +2 E[XY] + E[Y^2] )

Let’s call the total time T = Ntx+Nty. We can solve this for N. N=T/(tx+ty)
E[X] is equal to the average synthesis rate of Sx times tx, which I write as E[X] = rxtx
similarly for E[Y].
combining these:
E[M] = T(rx
tx/(tx+ty) + ry*ty/(tx+ty)). So the average synthesis rate is just the sum of the independent synthesis rates weighted by the fraction of the time spent in each state. The total production in time T is this average synthesis rate times T.

Accounting for exponential distributed decay at rate delta, we observe that the average steady state population of state Sx is: mux = rx/delta.
E[M]/T is the average synthesis rate during the switching. E[M]/T/delta is the sort of steady state mRNA level in the switching regime which is
= muxtx/(tx+ty) + muyty/(tx+ty).

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