Transition rate matrix from transition probability matrix

Question

If I have a $2 \times 2$ continuous time Markov chain transition probability matrix (generated from a financial time series data), is it possible to get the transition rate matrix from this and if Kolmogorov equations can assist, how would I apply them.

Answer 1

If $P_t=\begin{pmatrix}1-a(t)&a(t)\\ b(t)&1-b(t)\end{pmatrix}$ and $Q=\begin{pmatrix}-\alpha&\alpha\\ \beta&-\beta\end{pmatrix}$ are such that $P_t=\mathrm e^{Qt}$, then $$P'_t=P_tQ,$$ hence $$a'(t)=\alpha(1-a(t))-\beta a(t),\qquad b'(t)=\beta(1-b(t))-\alpha b(t).$$ Thus, $$a'(t)+(\alpha+\beta)a(t)=\alpha,\qquad b'(t)+(\alpha+\beta)b(t)=\beta.$$ Since $a(0)=b(0)=0$, one gets $$a(t)=\frac{\alpha}{\alpha+\beta}(1-\mathrm e^{-(\alpha+\beta)t}),\qquad b(t)=\frac{\beta}{\alpha+\beta}(1-\mathrm e^{-(\alpha+\beta)t}). $$ Finally, $$ P_t=\frac{1}{\alpha+\beta}\begin{pmatrix}\beta+\alpha\mathrm e^{-(\alpha+\beta)t}&\alpha(1-\mathrm e^{-(\alpha+\beta)t})\\ \beta(1-\mathrm e^{-(\alpha+\beta)t})&\alpha+\beta\mathrm e^{-(\alpha+\beta)t}\end{pmatrix}. $$ Note that by homogeneity there are only two parameters here, namely $A=\alpha t$ and $B=\beta t$, since $$ P_t=\frac{1}{A+B}\begin{pmatrix}B+A\mathrm e^{-A-B}&A(1-\mathrm e^{-A-B})\\ B(1-\mathrm e^{-A-B})&A+B\mathrm e^{-A-B}\end{pmatrix}. $$ If the observed matrix is $$ P_{\mathrm{obs}}=\begin{pmatrix}1-U& U\\ V&1-V\end{pmatrix}, $$ it is necessary that $U+V\lt1$ (otherwise, $P_{\mathrm{obs}}$ does not correspond to a continuous time transition matrix), then one can estimate $A$ and $B$ by $$ \hat A=-\log(1-U-V)\cdot\frac{U}{U+V},\qquad \hat B=-\log(1-U-V)\cdot\frac{V}{U+V}. $$