I am working on problem 3.16 in Sheldon Ross's Stochastic Processes book. The problem is, "Consider a renewal process whose interarrival distribution is the gamma distribution with parameters $(n,\lambda)$. Show that $$ \lim_{t\rightarrow\infty}E[Y(t)] = \frac{n+1}{2\lambda}$$
Now explain how this could have been obtained without any computations."
I could solve it analytically using the fact that "If the interarrival distribution is nonlattice and $E[X^2] < \infty$, then $$\lim_{t\rightarrow\infty}E[Y(t)] = \frac{E[X^2]}{2\mu}$$
But I am having hard time thinking how to get the same result without any computation. Any hint/help would be really helpful to understand the whole thing intuitively
I'm not sure that the following reasoning is the simplest, but ..
Firstly assume that we consider the stationary situation (limiting case, or $t$ is sufficiently large).
Inter-arrival gamma distribution with parameters $(n,\lambda)$ is the distribution of sum of $n$ independent exponential r.v.'s with the same expectations $\frac1\lambda$. Consider this r.v.'s as "sub"-interarrival times: $X_1+\ldots+X_n=X$, $X\sim \Gamma(n,\lambda)$. Conditioning on the event that given $t$ lie within some interarrival time $X$, there are equal chances $1/n$ that given $t$ will lie within any of these "sub"-interarrival times $X_i$.
Memoryless property of exponential distribution guarantees that the residual time of $X_i$ to the next "sub-renewal" is distributed like $X_i$. In this case the total residual time $Y(t)$ is equal to sum $X_i+\ldots+X_n$ with mean $(n-i+1)\frac1\lambda$.
So, $$ E[Y(t)]=\frac{1}{n}(1+2+\ldots+n)\frac1\lambda = \frac{n(n+1)}{2n}\frac1\lambda = \frac{n+1}{2\lambda}. $$