Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 6B (Mathematical Biology)

Table of contents
- i 6B
  - Solution i
- ii 6B
  - Solution ii
- iii 6B
  - Solution iii

i

Solution

For drift

A_{i} (x)

and constant diffusion matrix

b_{ij}

, the Fokker-Planck equation is

\frac{\partial P}{\partial t} = - \frac{\partial}{\partial x _{i}} (A_{i} P) + \frac{1}{2} \frac{\partial ^{2}}{\partial x _{i} \partial x _{j}} (b_{ij} P) .

Here

A_{1} = - x_{1} + x_{2}, A_{2} = - 2 x_{1} - x_{2}, b = I,

so

\frac{\partial P}{\partial t} = \frac{\partial}{\partial x _{1}} [(x_{1} - x_{2}) P] + \frac{\partial}{\partial x _{2}} [(2 x_{1} + x_{2}) P] + \frac{1}{2} (\frac{\partial ^{2} P}{\partial x _{1}^{2}} + \frac{\partial ^{2} P}{\partial x _{2}^{2}}) .

ii

Solution

The drift is the linear vector field

A (x_{1}, x_{2}) = (- x_{1} + x_{2}, - 2 x_{1} - x_{2}) .

The eigenvalues of its matrix are

- 1 \pm i 2

, so the origin is a stable spiral. The field rotates clockwise while contracting toward the origin:

Figure 1.
The drift vector field of a two-dimensional Ornstein-Uhlenbeck process
.

iii

Solution

For this two-dimensional Ornstein-Uhlenbeck process, the covariance matrix

C (t) = E [x (t) x (t)^{T}]

satisfies

\frac{d C}{d t} = a C + C a^{T} + b .

Thus the stationary covariance solves the continuous Lyapunov equation

a C + C a^{T} + b = 0 .

Write

C = (u v v w) .

Substitution gives

- 2 u + 2 v + 1 = 0, w - 2 u - 2 v = 0, - 4 v - 2 w + 1 = 0.

Solving,

u = \frac{5}{12}, v = - \frac{1}{12}, w = \frac{2}{3} .

Therefore

C = (5/12 - 1/12 - 1/12 2/3) .

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version