Group problem solutions 1.(a) (b)

2. In order to be reversible we need or equivalently Now divide by h and let h go to Assuming (as in Holgate, Math. Gazette, 1967) that the intensity of amalgamation is proportional to the number of herds (perhaps it should be proportional to the number of pairs of herds) and the intensity of splitting a herd of size x i is proportional to x i -1, the overall intensity of splitting will be the sum of these. Hence the G-matrix has

so the equations for the stationary distribution are (using Holgate’s model) whence the stationary distribution is 1+Bin(K-1,  ).

Brown’s motion Richard Brown ( ) Scottish botanist First systematic description of Australian fauna Named the cell nucleus Observed irregular pollen motion in liquid (first described by Ingenhousz in 1784)

Observed motion

Interpretation The basic unit of living matter (molecule) Same motion with dust particles (or carbon on oil) Ongoing motion in closed box “forever” Explained by: fluctuating temperature due to heating by incident light; electrical forces; temperature difference in the liquid. Delsaux (1877): impact of molecules of liquid on immersed particles–lots of small bumps

Einstein’s contribution 1905: Annus mirabilis 1.Photoelectric effect 2.Special relativity 3.E=mc 2 4.“On the Motion — Required by the Molecular Kinetic Theory of Heat— of Small Particles Suspended in a Stationary Liquid”

Some physics Two forces act on suspended particles: osmotic pressure (tendency towards equal concentration) diffusion (particles hit by heat-induced movement of liquid molecules) If q(x) is the density of particles per unit volume, the osmotic force is (R gas constant, T abs temp, N Avogadro’s number) This force causes flux

More physics The flux due to diffusion with diffusion coefficient D is These two fluxes must be equal in equilibrium, so If we know D and r we can compute N (which was not known in 1905).

Einstein’s stochastic model Over a short time period , a particle is displaced by a random quantity Y. It should have mean 0 in equilibrium. Assume for simplicity with probability 1/2 each. Let x be a point. Only particles in (x-y,x) can pass x from left to right in a time interval . Only half of those particles will in fact move from left to right, i.e. about particles. Similarly, about particles move from right to left.

Net flux is then and the quantity diffusing in unit time is so D=y 2 /2 . The displacement over unit time will be the sum of 1/  iid (why?) random variables, so has variance  2 =Var(Y)/ . Since Var(Y)=y 2 we have D=  2 /2. Perrin used this to estimate Avogadro’s number. Nobel prize in physics, 1926.

The distribution of Brownian motion A particle starting at the origin jumps Y units in time . The pgf of Y is In time t, the particle takes t/  independent steps. The total displacementhas pgf with mean 0 and variance ty 2 / . In order to have the variance converge to  2 t we need y=  1/2. Then

An excursion Let X~N(  2 ). Then

Let s=e . This yields the moment generating function. We get Exponentiating back we get the mgf for a N(0,  2 t)-random variable. One can use this type of limit to calculate the forward equation for the process, which turns out to be the heat equation.

A general definition A Brownian motion process is a stochastic process having: (1)Independent increments (2)Stationary increments (3)Continuity: for any 

Some Brownian motion process paths

Properties of Brownian motion process X(t)~N(  t,  2 t) Continuous paths Finite squared variation Not bounded variation Not differentiable paths Derivative of location is velocity, so for small time intervals this is not defined (not a very accurate model!) Is it Markov?

More realistic assumption Stationarity is the probability counterpart to conservation of momentum (mass x velocity) Instead of independent increments of location, we could consider independent increments of momentum Velocity change in particle would be twice the speed of the molecule times the ratio of molecule to particle mass

The Ornstein-Uhlenbeck approach Let U(t) be the velocity of a particle. Langevin’s equation says where K*(t) is a stochastic process with mean zero and very quickly decreasing covariance (collisions from independent molecules) Divide through by m to get  =  */m and K(t)=K*(t)/m. Multiply by exp(  t), so

A formal solution The integral makes sense if K(t) is continuous, but it probably is not. Write formally K(t)=dB(t)/dt to get The problem is now to make sense of the stochastic integral on the rhs (492). The result is the Ornstein-Uhlenbeck process.

Properties of the O-U process It is the only stationary Gaussian Markov process in continuous time and space.

The displacement is another Gaussian process with mean X(0) and variance For large t this behaves like a constant times t, as Einstein found, but for small t this behaves like t 2. It has a derivative (velocity). Why?