$\lambda_t$ is binary with $\lambda_H$ and $\lambda_L$, with instantaneous transition probailities of $\mu_H$ and $\mu_L$.
What is $\mathbb{E}_t[\lambda_T]$, assuming $\lambda_t=\lambda_H$ or $\lambda_t=\lambda_L$?
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- $\begingroup$ Hi, what are your thoughts on this problem? $\endgroup$Wei– Wei2024-10-01 12:46:30 +00:00Commented Oct 1, 2024 at 12:46
- $\begingroup$ I though the solution is $\mathbb{E}_t[\lambda_T]=\lambda_He^{-\mu_H(T-t)}+\lambda_L(1-e^{-\mu_H(T-t)})$, assuming $\lambda_t=\lambda_H$. Is that correct? $\endgroup$fincecon– fincecon2024-10-01 14:19:35 +00:00Commented Oct 1, 2024 at 14:19
- $\begingroup$ I believe the long run fractions of time spent in the two states are $\frac{\mu_L}{\mu_L+\mu_H},\frac{\mu_H}{\mu_L+\mu_H}$. For ex when the transition rates are equal you spend 50% of the time in each state. $\endgroup$nbbo2– nbbo22024-10-01 18:20:42 +00:00Commented Oct 1, 2024 at 18:20
- $\begingroup$ that makes sense. $\endgroup$fincecon– fincecon2024-10-01 19:20:40 +00:00Commented Oct 1, 2024 at 19:20
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1 Answer
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5 This is a continuous-time Markov chain with rate matrix $$Q=\begin{pmatrix} -\mu_H & \mu_H\\ \mu_L & -\mu_L \end{pmatrix}.$$ The transition matrix associated with going from time $t$ to time $T$ is then (using WolframAlpha) $$P(t,T) = \exp((T-t)Q)=\frac{1}{\mu_H+\mu_L}\begin{pmatrix} \mu_He^{-(T-t)(\mu_H+\mu_L)} + \mu_L & \mu_H(1-e^{-(T-t)(\mu_H+\mu_L)})\\ \mu_L(1-e^{-(T-t)(\mu_H+\mu_L)}) & \mu_H + \mu_Le^{-(T-t)(\mu_H+\mu_L)} \end{pmatrix}.$$ The top row of this matrix gives the distribution of $\lambda_T$ given $\lambda_t=\lambda_H$. The bottom row of this matrix gives the distribution of $\lambda_T$ given $\lambda_t=\lambda_L$. You can see that it's the identity matrix when $t=T$, and as $T\to \infty$ it converges to the stationary distribution given by nbbo2 in the comments of your question.
- $\begingroup$ Wonderful! Thank you so much! I have a question though, how should I interpret the instantaneous transition probability? To be specific, I am thinking something like the following. For example, at time $t$, $\lambda_t = \lambda_H$, then at time $t+\Delta t$, the probability of entering $H$ state is what? I am not sure I understand it in continuous time. Relatedly, what does $Q$ represents intuitively? $\endgroup$fincecon– fincecon2024-10-01 21:40:39 +00:00Commented Oct 1, 2024 at 21:40
- 1$\begingroup$ Good question! $\frac d{dT} P(t,T) = P(t,T)Q$ (Kolmogorov's forward equation). Evaluate it at $T=t$, and it shows that the transition matrix of going from $t$ to $t+ \Delta t$ is $I+Q\Delta t$ to first order. $\endgroup$Wei– Wei2024-10-01 22:32:33 +00:00Commented Oct 1, 2024 at 22:32
- $\begingroup$ I have another question, maybe it is due to my misunderstanding of the notations. If the top row of $P(t,T)$ corresponds to the distribution of $\lambda_T$ conditional on $\lambda_t=\lambda_H$, then is it correct that $E_t[\lambda_T | \lambda_t=\lambda_H] = P(t,T)[1,1]\lambda_H + P(t,T)[1,2]\lambda_L$. But I am confused here $P(t,T)[1,1]+P(t,T)[1,2]=\mu_H+\mu_L \neq 1$... $\endgroup$fincecon– fincecon2024-10-05 17:45:50 +00:00Commented Oct 5, 2024 at 17:45
- $\begingroup$ There's a factor $\frac{1}{\mu_H + \mu_L}$ outside the matrix. And yes your equation for $E_t[\ldots]$ is correct. $\endgroup$Wei– Wei2024-10-05 18:10:18 +00:00Commented Oct 5, 2024 at 18:10
- $\begingroup$ oh it is my negligence... $\endgroup$fincecon– fincecon2024-10-05 18:27:44 +00:00Commented Oct 5, 2024 at 18:27