Markov process in continuous time: meaning of derivative wrt time

Question

Suppose we have a Markov process $X=(X_t)_{t\geq 0}$, taking values on $\mathcal{N}$, with this Q-matrix (Example 17.27 of Probability Theory, 3rd version, by A. Klenke): $$ q(x,y) = \begin{cases} x &\text{if } y=x+1\\ -x &\text{if } y=x\\ 0 &\text{otherwise} \end{cases} $$

Let $f_n(t):=P_1[X_t>n]$. How can I obtain in a formal way and give a meaning to this?

$$ \frac{d}{dt}f_n(t)=nP_1[X_t=n] $$

I know that $f_n(0)=0$ for all $n\in \mathcal{N}$ and that $X$ jumps from $n$ to $n+1$ at rate $n$. This is what the author is saying to obtain the expression but it is not enough for me to see what he's doing.

Thanks for the help. Let me know if more context is needed.

Answer 1

Here it looks like the CTMC never decreases. You fix $\delta>0$ with $\delta \approx 0$ and use the Law of Total Probability: \begin{align} P[X(t+\delta)>n] &=\sum_{i=1}^{\infty} P[X(t+\delta)>n|X(t)=i]P[X(t)=i]\\&=\sum_{i=1}^{n} P[X(t+\delta)>n|X(t)=i]P[X(t)=i]+P[X(t)>n]\end{align} where we have used the nondecreasing property of $X(t)$ to obtain $$P[X(t+\delta)>n|X(t)=i]=1 \quad \forall i>n$$

Then:

• $P[X(t+\delta)>n|X(t)=n]=n\delta+O(\delta^2)$
• $P[X(t+\delta)>n|X(t)=i]=O(\delta^2) \quad \forall i \in \{1, ..., n-1\}$