Stochastic Processes Nancy Griffeth January 10 2014 Funding

Stochastic Processes Nancy Griffeth January 10, 2014 Funding for this workshop was provided by the program “Computational Modeling and Analysis of Complex Systems, ” an NSF Expedition in Computing (Award Number 0926200).

Motivation n n Continuous solutions depend on large numbers (e. g. , ODE’s) Stochastic solutions are better for small numbers n Strange and surprising things happen!

Game n n Two players each have a pile of pennies Taking turns: n n Player flips a penny Heads, he gets a penny from the other player Tails, he gives the penny to the other player Until one player runs out of pennies

Questions n n What each player’s chance of winning? What is the expected value of the game to a player? How long will the game go on before a player runs out of pennies? What is the probability after m moves that player A has a pennies and player B has b pennies? n Or the probability distribution over (a, b)?

Chance of winning n n If A and B have the same number of coins, they’re equally likely to win. With m coins to player A and n coins to player B n n n A wins with probability m/(m+n) B wins with probability n/(m+n) The player with more coins is more likely to win.

Expected Value n Definition: n n Exp(winnings)=Prob(winning)*value of winning +Prob(losing)*value of losing Value of winning is m+n Value of losing is 0 So, if A starts with m coins and B with n: n n E(A winnings) = m/(m+n)*(m+n) + n/(m+n)*0 =m E(B winnings) = n/(m+n)*(m+n) + m/(m+n)*0 =n

Markov Chain 3, 3 2, 4 1, 5 0, 6 The probability of each transition is ½, except 0, 6 ->0, 6 and 6, 0 ->6, 0, where the probability is 1

How long n n n Relatively hard from the UCLA tutorial Sum over n of n*[p((5, 1); n)*0. 5+p((1, 5); n)*0. 5] = Sum over n of n*p((5, 1; n) Starting from (3, 3), about 8 turns

Probability Matrix 0, 6 1, 5 2, 4 3, 3 4, 2 5, 1 6, 0 0, 6 1 0 0 0 1, 5 0 0 0 0 2, 4 0 0. 5 0 0 0 3, 3 0 0 0. 5 0 0 4, 2 0 0. 5 0 5, 1 0 0 0. 5 6, 0 0 0 0 1

Two Steps 0, 6 1, 5 2, 4 3, 3 4, 2 5, 1 6, 0 0, 6 1 0 0 0 1, 5 0. 25 0 0 0 2, 4 0. 25 0 0 3, 3 0 0. 25 0 4, 2 0 0 0. 25 5, 1 0 0. 25 0. 5 6, 0 0 0 0 1

Three Steps 0, 6 1, 5 2, 4 3, 3 4, 2 5, 1 6, 0 0, 6 1 0 0 0 1, 5 0. 625 0 0. 125 0 0 2, 4 0. 25 0 0. 375 0 0. 125 0 3, 3 0. 125 0 0. 375 0 0. 125 4, 2 0 0. 125 0 0. 375 0 0. 25 5, 1 0 0 0. 125 0 0. 625 6, 0 0 0 0 1

64 Steps 0, 6 1, 5 2, 4 3, 3 4, 2 5, 1 6, 0 0, 6 1 0 0 0 1, 5 0. 833 0 0 0. 167 2, 4 0. 667 0 0 0. 333 3, 3 0. 5 0 0 0. 5 4, 2 0. 333 0 0 0. 667 5, 1 0. 167 0 0 0. 833 6, 0 0 0 0 1

Steady States n n n Sooner or later, somebody wins (with probability=1) Either party can win The system has two stable steady states (bistable)

The “Game” with Molecules L 2 L 1 L 0 R 10 R 9 R 8 L. R 0 L. R 1 L. R 2 L. R 0 L. R. R. L 1

Chemical Master Equation n n A set of first-order differential equations describing the change in the probability of states with time t. With time-independent reaction rates, the process is Markovian.

The Chemical Master Equation We are interested in p(x; t), the probability that the chemical system will be in state x at time t. The time evolution of p(x; t) is described by the Chemical Master Equation: Where sμ is the stoichiometric vector for reaction μ, giving the changes in each type of molecule as a result of μ, and aμ is the propensity for reaction μ

Bistability n n n A steady state is a state that a system tends to stay in once it reaches it A steady state can be stable or unstable A system that has two stable steady states is called bistable.