= Solution
In a <loss network> with <fixed routing>, calls of type $r$ arrive as independent <Poisson processes> with rates $\nu_r$. Every accepted call occupies $A_{jr}$ units of resource $j$ throughout its holding time. The <link-route incidence matrix> $A$ has nonnegative integer entries, $C$ is the resource-capacity vector, $n_r$ counts calls currently present, and $e_r$ is the $r$th coordinate unit vector. The feasible occupancy space is
$$
S(C)=\{n\in\mathbb Z_{\geq0}^{R}:An\leq C\}.
$$
The inequalities are coordinatewise. A call that would leave this space is rejected immediately, without queuing or changing $n$. With independent unit-rate <exponential distributions> for holding times, the <memoryless property> makes $n$ a <continuous-time Markov chain>, whose nonzero off-diagonal <transition rates> are
$$
\boxed{q(n,n+e_r)=\nu_r\quad(n+e_r\in S(C)),
\qquad q(n,n-e_r)=n_r\quad(n_r>0).}
$$
There are $n_r$ independent clocks of rate one for the second transition. Rejected arrivals are not jumps and do not appear in the off-diagonal <transition rates>.
Let $a(n)=\prod_r\nu_r^{n_r}/n_r!$. For every feasible adjacent pair,
$$
a(n)\nu_r=a(n+e_r)(n_r+1).
$$
Thus <detailed balance for a continuous-time Markov chain> gives the <product-form stationary distribution of a loss network>:
$$
\boxed{\pi(n)=G(C)\prod_r\frac{\nu_r^{n_r}}{n_r!},\qquad
G(C)=\left[\sum_{n\in S(C)}\prod_r\frac{\nu_r^{n_r}}{n_r!}\right]^{-1}.}
$$
In this convention $G(C)$ is the reciprocal normalizing sum. If every route consumes some resource, finite capacities make $S(C)$ finite. Even zero-resource routes cause no normalization difficulty: the sum is at most $\exp(\sum_r\nu_r)$. With positive arrival rates the reachable feasible class is an <irreducible Markov chain>, since calls can depart to the empty state and any feasible occupancy can be built from it. The <stationary distribution> is therefore unique on that class.
The vector $m=C-An$ records the free units of each resource. Its <probability distribution> $\pi'(m)$ sums $\pi(n)$ over all occupancies with the same free resources. The map from $n$ to $m$ generally loses information: different call mixtures may consume exactly the same resources.
For $\pi'(m)>0$, the <conditional expectation> can be computed by removing one call. The factorial identity $n_r\pi(n)=\nu_r\pi(n-e_r)$ gives
$$
\sum_{n:\,An=C-m} n_r\pi(n)
=\begin{cases}
\nu_r\pi'(m+Ae_r),&Ae_r\leq C-m,\\
0,&\text{otherwise}.
\end{cases}
$$
Indeed, $n\mapsto n-e_r$ is a bijection from the occupancies in this sum with $n_r>0$ to occupancies using $C-m-Ae_r$ units. Conversely, adding one call to any such occupancy is feasible because its final resource use is $C-m$. Consequently
$$
\boxed{\mathbb E[n_r\mid m]=
\frac{\nu_r\pi'(m+Ae_r)}{\pi'(m)}
\quad\text{if }Ae_r\leq C-m,}
$$
and the <conditional expectation> is zero otherwise. Conditioning the identity $\sum_r A_{jr}n_r=C_j-m_j$ and multiplying by $\pi'(m)$ now yields
$$
\boxed{(C_j-m_j)\pi'(m)=
\sum_{r:\,Ae_r\leq C-m}A_{jr}\nu_r\pi'(m+Ae_r).}
$$
Alternatively, multiply the preceding unnormalized factorial identity by $A_{jr}$ and sum over $r$ directly; this proves the same formula even when $\pi'(m)=0$, without dividing by a zero <probability>. This gives both the <conditional expectation> argument and the direct summation argument.
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