 Solution

If $\mathbb{E}{S}^2<\infty$, then $g^{\prime \prime }(1)={\lambda }^2\mathbb{E}{S}^2$. Expand at $s=1$:Substitution into the previous quotient givesThus the stationary mean isOnly a finite first service moment was stated. If the second moment is infinite and $\lambda >0$, the stationary mean is infinite; the formula has that extended-value interpretation. This follows as well by writing the second-order remainder as an integral of $g^{\prime \prime }(s)$ and taking $s\uparrow 1$, where monotone convergence gives $g^{\prime \prime }(s)\uparrow {\lambda }^2\mathbb{E}{S}^2$. A finite stationary distribution need not have a finite mean.