Reading about birth-death processes i encountered a formula which i can't derive It's here, page 4, number (7)-(8).
$P_{i,i+1}(h)=P(X(t+h)-X(t)=1|X(t)=1)=\frac{(\lambda_ih)^1e^{-\lambda_ih}}{1!}\frac{(\mu_ih)^0e^{-\mu_ih}}{0!}+o(h)$
I understand that here we have 2 (independent?) nonhomogenous Poisson processes for births and for deaths, and that they have distributions respectively
$P(N_b(t+h)-N_b(t)=k|N_b(t)=i)=\frac{(\lambda_ih)^ke^{-\lambda_ih}}{k!}$ $P(N_d(t+h)-N_d(t)=k|N_d(t)=i)=\frac{(\mu_ih)^ke^{-\mu_ih}}{k!}$
So the initial formula, as i suppose, should somehow lead to
$P_{i,i+1}(h)=P(X(t+h)-X(t)=1|X(t)=1)=P(N_b(t+h)-N_b(t)=1|N_b(t)=i)P(N_d(t+h)-N_d(t)=k|N_d(t)=i)+\sum_{}^{}$
Where some $\sum_{}^{} \in o(h)$
But how do i get to this point?
\begin{align} P_{i,i+1}(h) &= P(X(t+h)-X(t)=1\mid X(t) = i) \\ &= P\left(\bigcup_{j=1}^{\infty} \left[N_b(t+h)-N_b(t)=j \;\cap\; N_d(t+h)-N_d(t)=j-1\right]\;\bigg|\; X(t)=i\right) \\ & \qquad\qquad\qquad\text{i.e. $1$ birth, $0$ deaths, or $2$ births, $1$ death, or $3$ births, $2$ deaths, etc. } \\ &= \sum_{j=1}^{\infty} P\left(\left[N_b(t+h)-N_b(t)=j \;\cap\; N_d(t+h)-N_d(t)=j-1\right]\;\bigg|\; X(t)=i\right) \\ & \qquad\qquad\qquad\text{since the union is disjoint} \\ &= \sum_{j=1}^{\infty} P\left(N_b(t+h)-N_b(t)=j\mid X(t)=i\right) P\left(N_d(t+h)-N_d(t)=j-1\mid X(t)=i\right) \\ & \qquad\qquad\qquad\text{by independence of $N_b$ and $N_d$} \\ & \\ &= P\left(N_b(t+h)-N_b(t)=1\mid X(t)=i\right) P\left(N_d(t+h)-N_d(t)=0\mid X(t)=i\right) + o(h) \\ & \\ &=\frac{(\lambda_ih)^1e^{-\lambda_ih}}{1!}\frac{(\mu_ih)^0e^{-\mu_ih}}{0!}+o(h). \end{align}