stochastic differential equations in matrix notation

Question

How is Matrix K determined?

$ dY_1 = \frac{-1}{2} Y_1dt - Y_2dB_t $

$ dY_2 = \frac{-1}{2} Y_2dt - Y_1dB_t $

In matrix notation, the above equations can be written as:

$ dY(t) = \frac{-1}{2}Y(t)dt + KY(t)dB_t $

where K is equal to

$\begin{pmatrix}0&&-1\\1&&0\end{pmatrix}$

score 1 · Answer 1 · answered Jul 10 '16 at 15:39

1

We have $$\begin{pmatrix}dY_1\\dY_2\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}Y_1 + 0Y_2\\0Y_1 -\frac{1}{2}Y_2\end{pmatrix}dt + \begin{pmatrix}0Y_1 - 1Y_2\\ -1Y_1 + 0Y_2\end{pmatrix}dB_t $$ $$= \begin{pmatrix}-\frac{1}{2} & 0\\ 0 & -\frac{1}{2}\end{pmatrix}Ydt + \begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}YdB_t,$$

so $K = \begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}$. I assume you switched a sign somewhere.

answered Jul 10 '16 at 15:39

Anon

5,489

McFry: How did you determine the coefficient in the first equation? – flyhigh1 Jul 12 '16 at 03:03
McFry: True I made a mistake in my original second equation. It should be -ve then +ve. – flyhigh1 Jul 12 '16 at 03:08
Got it. You arrange the coefficients of dt and dBt in the order Y_1 Y_2 – flyhigh1 Jul 12 '16 at 03:10
McFry: So in my third equation why is it $ -1/2Y(t) $ and not a matrix say S – flyhigh1 Jul 12 '16 at 03:13
$-1/2Y(t)$ is the same as $-1/2IY(t)$, where $I$ is an identity matrix. – Anon Jul 12 '16 at 07:21

score 0 · Answer 2 · answered Jul 10 '16 at 06:43

0

$$\begin{pmatrix}dY_1\\dY_2\end{pmatrix}=\begin{pmatrix}-\frac{Y_1}{2}&&-Y_2\\ \frac{-1}{2} Y_2&&-Y_1\end{pmatrix}\begin{pmatrix}dt\\dB_t\end{pmatrix}$$

answered Jul 10 '16 at 06:43

Nebo Alex

1,974

Boris: how do you find the values of Matrix K. I have already given the answer but I want to understand how. – flyhigh1 Jul 10 '16 at 12:49
how you want to write it its the only way matrix is written from the equation.@flyhigh1 – Nebo Alex Jul 10 '16 at 13:30

stochastic differential equations in matrix notation

2 Answers2