I have been studying continuous time markov chains through Dobrow's book. Everything went fine until the author introduced the concept of infinitesimal generator, which he refers to as $\textbf{Q}$. More specifically, i cannot figure out a statement regarding the diagonal entries of $\textbf{Q}$, which is derived as shown below:

$$ Q_{i i}=P_{i i}^{\prime}(0)=\lim _{h \rightarrow 0^{+}} \frac{P_{i i}(h)-P_{i i}(0)}{h}=\lim _{h \rightarrow 0^{+}} \frac{P_{i i}(h)-1}{h} $$ which equals (this is where i got stuck)

$$\lim _{h \rightarrow 0^{+}}-\frac{\sum_{j \neq i} P_{i j}(h)}{h}$$

  1. Why does the $\text{-}$ (minus) sign appears in the last equality?
  2. Why does $$\lim _{h \rightarrow 0^{+}} \frac{P_{i i}(h)-1}{h}=\lim _{h \rightarrow 0^{+}}-\frac{\sum_{j \neq i} P_{i j}(h)}{h}$$I don't get where the summation comes from.

Furthermore, $\textbf{P}(t)$ is the transition function with respect to a continuous-time Markov chain (CTMC) Finally, $$ P_{i j}(t)=P\left(X_{t}=j \mid X_{0}=i\right) $$

Note: $$ P_{i j}(0)=\left\{\begin{array}{l} 1, \text { if } i=j \\ 0, \text { if } i \neq j \end{array}\right. $$

Can someone help? I am really struggling. Thanks in advance, Lucas

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    $\begingroup$ Hi: For a given $i$, $\sum_{j=1}^{n} P_{ij} = 1 $ so that's where the negative sign comes from. $\endgroup$ – mlofton Nov 29 '20 at 5:08
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    $\begingroup$ Note, just in case above was too short and not clear, note that, when you sum over the $j$, one of those $j$ is equal to $i$. $\endgroup$ – mlofton Nov 29 '20 at 5:10
  • $\begingroup$ I am afraid i am missing something simple. Why does the equality $\sum_{j=1}^{n} P_{i j}=1$ makes the minus sign appear? $\endgroup$ – Lucas Nov 29 '20 at 20:41
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    $\begingroup$ Hi: No problem. Take my equation and take the case of i = j out of the sum and leave that one element on the left hand side. Then take the rest of the sum and put it on the right hand side. The take the 1.0 on the right hand side and put it on the right hand side. Does that make sense ? If not, I can write it out all out but it's just a matter of moving things over to the other sides of the equals sign. $\endgroup$ – mlofton Nov 30 '20 at 15:17

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