Solving a first order matrix differential equation

[docnet]

Homework Statement

TL;DR: solving for $P(t)=e^{Qt}$ doesn't go as planned, ending up with an answer that is similar to the given answer, but with missing terms in each entry.

Relevant Equations

.

Let X be a continuous-time Markov chain that hops between two states $\{1, 2\}$ with rates $\lambda, \mu>0$, so its generator is
$$Q = \begin{pmatrix}
-\mu & \mu\\
\lambda & -\lambda
\end{pmatrix}.$$
Solve $\pi Q = 0$ for the stationary distribution, and verify that $P_{ij}(t)\longrightarrow \pi_j$ as $t\longrightarrow \infty.$
\\\\
By looking at $Q,$ it is easy to see that $\text{Row }1 \cdot \lambda +\text{Row }2\cdot \mu =0$. This implies $\pi = (\lambda,\mu) C$ for a constant $C$ such that $\pi_1+\pi_2=1.$ Solving for $C,$ we have
$$(\lambda + \mu)C = 1,$$
$$C = \frac{1}{\lambda + \mu}.$$
And so
$$\pi = \Big(\frac{\lambda}{\lambda + \mu}, \frac{\mu}{\lambda + \mu}\Big).$$
\\\\
We know from Wiki that the transition matrix $P(t)$ is
$$P(t)=Ce^{Qt}$$
where $C$ is a constant matrix that is determined by the initial condition $P(0)=I.$ In order to find $e^{Qt},$ we use the eigendecomposition of $Qt.$ The eigenvectors of $Q$ are
$$\begin{pmatrix}
\mu & -\lambda
\end{pmatrix}
\text{ and }
\begin{pmatrix}
1 & 1
\end{pmatrix}$$
with respective eigenvalues $-(\mu+\lambda)$ and $0$. And so we have the eigendecomposition
$$
Qt =
\begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}
\begin{pmatrix}
-(\mu+\lambda)t & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
\lambda & \mu\\
1 & 1
\end{pmatrix}^{-1}.$$
And so,
\begin{align*}
e^{Qt}&=
\begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}
\begin{pmatrix}
e^{-(\mu+\lambda)t} & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}^{-1}\\
&= \begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}
\begin{pmatrix}
\frac{e^{-(\mu+\lambda)t}}{\mu+\lambda} & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
1 & -1\\
\lambda & \mu
\end{pmatrix}\\
&= \begin{pmatrix}
\frac{\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}&\frac{-\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}\\
\frac{-\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}& \frac{\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}
\end{pmatrix}.
\end{align*}
However, this solution does not satisfy the initial condition $P(0)=I$. My answer differs from the true solution given by wiki, which is given by Wiki as
$$P(t)=
\begin{pmatrix}
\frac{\lambda}{\mu+\lambda} + \frac{\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}&\frac{\mu}{\mu+\lambda} +\frac{-\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}\\
\frac{\lambda}{\mu+\lambda} +\frac{-\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}& \frac{\mu}{\mu+\lambda} + \frac{\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}
\end{pmatrix}.$$

My solution is lacking the terms
$$\begin{pmatrix}
\frac{\lambda}{\lambda + \mu} & \frac{\mu}{\lambda + \mu}\\
\frac{\lambda}{\lambda + \mu} & \frac{\mu}{\lambda + \mu}
\end{pmatrix}.$$
Which is $$\lim_{t \to \infty}P(t).$$

I've tried for a while and don't see where these terms could have come from, and where I made a mistake in computing $e^{Qt}.$

 

[Orodruin]

$e^0 = 1 \neq 0$

 

[docnet]

How embarrassing..

 

[docnet]

I am just winging it.

 

[docnet]

Thank you for pointing out my error.

 

[Orodruin]

How embarrassing..

We all do those things from time to time.

 

[docnet]

$$P(t)=e^{Qt}$$
We use the eigendecomposition of $Qt.$ The eigenvectors of $Q$ are
$$\begin{pmatrix}
\mu & -\lambda
\end{pmatrix}
\text{ and }
\begin{pmatrix}
1 & 1
\end{pmatrix}$$
with respective eigenvalues $-(\mu+\lambda)$ and $0$. And so we have the eigendecomposition
$$
Qt =
\begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}
\begin{pmatrix}
-(\mu+\lambda)t & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}^{-1}.$$
And so,
\begin{align*}
e^{Qt}&=
\begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}
\begin{pmatrix}
e^{-(\mu+\lambda)t} & 0 \\
0 & e^0
\end{pmatrix}
\begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}^{-1}\\
&=\frac{1}{\mu+\lambda} \begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}
\begin{pmatrix}
e^{-(\mu+\lambda)t} & 0 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & -1\\
\lambda & \mu
\end{pmatrix}\\
&=\frac{1}{\mu+\lambda} \begin{pmatrix}
\mu & 1\\
-\lambda & 1
\end{pmatrix}
\begin{pmatrix}
e^{-(\mu+\lambda)t} & -e^{-(\mu+\lambda)t}\\
\lambda & \mu
\end{pmatrix}\\
P(t)&=
\begin{pmatrix}
\frac{\lambda}{\mu+\lambda} + \frac{\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}&\frac{\mu}{\mu+\lambda} +\frac{-\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}\\
\frac{\lambda}{\mu+\lambda} +\frac{-\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}& \frac{\mu}{\mu+\lambda} + \frac{\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}
\end{pmatrix}.
\end{align*}
We take the limit of $P(t)$ as $t\longrightarrow \infty$.
\begin{align*}
\lim_{t \to \infty}P(t)&= \lim_{t \to \infty}\begin{pmatrix}
\frac{\lambda}{\mu+\lambda} + \frac{\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}&\frac{\mu}{\mu+\lambda} +\frac{-\mu e^{-(\mu+\lambda)t}}{\mu+\lambda}\\
\frac{\lambda}{\mu+\lambda} +\frac{-\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}& \frac{\mu}{\mu+\lambda} + \frac{\lambda e^{-(\mu+\lambda)t}}{\mu+\lambda}
\end{pmatrix}\\
&=
\begin{pmatrix}
\frac{\lambda}{\lambda + \mu} & \frac{\mu}{\lambda + \mu}\\
\frac{\lambda}{\lambda + \mu} & \frac{\mu}{\lambda + \mu}
\end{pmatrix}.
\end{align*}
And so $P_{ij}(t)\longrightarrow \pi_j$ as $t\longrightarrow \infty$.


Edit: $LATEX$ Embedding.