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We have the following transition rate matrix for a two-state Markov Chain: $$ Q = \begin{pmatrix} -\lambda_1 & \lambda_2 \\ \lambda_1 & -\lambda_2 \end{pmatrix} $$ Note I use the convention where the columns add to zero. I want to find the transition probability matrix P(t) using the differential equation: $$ \frac{dP(t)}{dt} = Q P(t) \ . $$ We can find $P(t)$ by letting $P(0)=I$ the identity matrix such that we get: $$ P(t) = \exp(Qt) \ . $$ By writing $Q = U D U^{-1}$ where $D$ is a diagonal matrix consisting of eigenvalues of $Q$ and $U $ is a matrix consisting of eigenvectors of $Q$. The eigenvalues of $Q$ are $0$ and $-(\lambda_1 + \lambda_2)$ and the eigenvectors are $(\lambda_2 , \lambda_1)^T$ and $(1 , -1)^T$. Then we get the following: \begin{align*} D &= \begin{pmatrix} 0 & 0 \\ 0 & -(\lambda_1 + \lambda_2) \end{pmatrix} \\[10pt] U &= \begin{pmatrix} \lambda_2 & 1 \\ \lambda_1 & -1 \end{pmatrix} \\[10pt] U^{-1} &= \frac{1}{\lambda_1+\lambda_2}\begin{pmatrix} 1 & 1 \\ \lambda_1 & -\lambda_2 \end{pmatrix} \ . \end{align*} Lets check that $Q = U D U^{-1}$ is correct: \begin{align*} U D U^{-1} &= \begin{pmatrix} \lambda_2 & 1 \\ \lambda_1 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & -(\lambda_1 + \lambda_2) \end{pmatrix} \frac{1}{\lambda_1+\lambda_2}\begin{pmatrix} 1 & 1 \\ \lambda_1 & -\lambda_2 \end{pmatrix} \\[10pt] &= \begin{pmatrix} \lambda_2 & 1 \\ \lambda_1 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ -\lambda_1 & \lambda_2 \end{pmatrix} \\[10pt] &= \begin{pmatrix} -\lambda_1 & \lambda_2 \\ \lambda_1 & -\lambda_2 \end{pmatrix} \\[10pt] &= Q \ . \end{align*} Now we calculate $P(t) = \exp(Qt)$ using $\exp(UDU^{-1}t) = U \exp(Dt)U^{-1}$ then we get: \begin{align*} P(t) &= U \exp(Dt)U^{-1} \\[10pt] &= \begin{pmatrix} \lambda_2 & 1 \\ \lambda_1 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & \exp(-(\lambda_1 + \lambda_2)t) \end{pmatrix} \frac{1}{\lambda_1+\lambda_2}\begin{pmatrix} 1 & 1 \\ \lambda_1 & -\lambda_2 \end{pmatrix} \\[10pt] &= \frac{\exp(-(\lambda_1 + \lambda_2)t)}{\lambda_1+\lambda_2} \begin{pmatrix} \lambda_2 & 1 \\ \lambda_1 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ \lambda_1 & -\lambda_2 \end{pmatrix} \\[10pt] &= \frac{\exp(-(\lambda_1 + \lambda_2)t)}{\lambda_1+\lambda_2} \begin{pmatrix} \lambda_2 & 1 \\ \lambda_1 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ \lambda_1 & -\lambda_2 \end{pmatrix}\\[10pt] &= \frac{\exp(-(\lambda_1 + \lambda_2)t)}{\lambda_1+\lambda_2} \begin{pmatrix} \lambda_1 & -\lambda_2 \\ -\lambda_1 & \lambda_2 \end{pmatrix} \ . \end{align*} The problem with this results is that the columns do not add to one and therefore this is not a valid transition probability matrix. What have I done wrong?

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Most of your work is right. The only mistake is that you should have $$ \exp(Dt) = \begin{pmatrix} \color{red}1 & 0 \\ 0 & \exp(-(\lambda_1 + \lambda_2)t) \end{pmatrix}. $$ In the end, this means that you'll have to add $$ \frac 1{\lambda_1 + \lambda_2} \begin{pmatrix} \lambda_2 & 1 \\ \lambda_1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ \lambda_1 & -\lambda_2 \end{pmatrix} = \frac 1{\lambda_1 + \lambda_2} \pmatrix{\lambda_2 & \lambda_2\\ \lambda_1 & \lambda_1} $$ to your final answer.

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  • $\begingroup$ Damn the exp(0) got me! Thanks :) $\endgroup$
    – Max
    Commented Mar 6, 2023 at 19:51

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