= Solution
The protein production rate is proportional to the current <messenger RNA> <concentration>, not an independently constant $\beta_P$: $\beta_P(t)=rX_R(t)$. Setting the two coupled rate equations to zero gives
$$
X_{R,\mathrm{st}}=\frac{\beta_R}{\alpha_R},\qquad
\boxed{X_{P,\mathrm{st}}=\frac{r\beta_R}{\alpha_R\alpha_P}
=\frac{b\beta_R}{\alpha_P}}.
$$
With first-order removal, a <messenger RNA> molecule has mean lifetime $1/\alpha_R$. At <biological translation> rate $r$ per molecule, it produces an expected $r/\alpha_R=b$ <proteins> before its removal. Thus the <translational burst size> is a dimensionless mean number of offspring <proteins> per <messenger RNA>, not a concentration and not a fixed number produced in every event. When production and removal are independent Poisson events, mixing a Poisson number of <biological translation> events over an exponentially distributed lifetime makes the burst count geometrically distributed on the nonnegative integers. Its <mean> is $b$ and its <variance> is $b(1+b)$.
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