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Kloeden Platen Schurz 1994 states the linear 2D SDE:

$$dX_t = A X_t dt + B X_t dW_t$$

where $a=5$ and $b=0.01$ and

$$A = \Big( \begin{matrix} -a & a \\ a & -a \end{matrix} \Big)$$ $$B = \Big( \begin{matrix} b & 0 \\ 0 & b \end{matrix} \Big)$$

They give explicit solution

$$X(t) = X_0 \exp{\Big( \big(A - \frac{B^2}{2} \big) t + B W_t \Big)}$$

As an example starting point, they give $X_0=\big(\begin{matrix} 1 \\ 0\end{matrix}\big)$. However, at $t=0$, $W_t=0$ and

$$\Big( \big(A - \frac{B^2}{2} \big) 0 + B 0 \Big) = \Big( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \Big)$$

and $$\exp{\Big( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \Big)} = \Big( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \Big)$$

So

$$X(0) = X_0 \Big( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \Big) = \big(\begin{matrix} 1 \\ 0\end{matrix}\big) \Big( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \Big) = \big(\begin{matrix} 1 \\ 1\end{matrix}\big) \neq X_0 $$

Please help. I should be able to choose any starting point for $X_0$ and then evolve it according to the SDE. The given explicit solution seems wrong in this case. Is it wrong or am I missing something obvious?

Lars Ericson
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  • The exponent of the zero matrix is the identity matrix. – C. Hamster Feb 13 '20 at 13:19
  • That would imply that the log of the identiy matrix is the 0 matrix. Python's NumPy package, for what it's worth, is giving back (1 1; 1 1) as the exponent of the 0 matrix. Any thoughts? – Lars Ericson Feb 13 '20 at 14:32
  • Also Python's numpy is giving $\Big( \begin{matrix} 0 & -\infty \ -\infty & 0 \end{matrix} \Big)$ as the log of the identity matrix. Is that a bad choice? Maybe I should put in a separate question for that. – Lars Ericson Feb 13 '20 at 14:50

1 Answers1

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It is pointed out in comments to this question that NumPy's exp and log are doing element-wise operations, which are incorrect in this case. I should have been using SciPy's expm and logm.

Lars Ericson
  • 160