# Expected Residual life in Renewal process with gamma interarrival distribution

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

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, $$E[Y(t)]=\frac{1}{n}(1+2+\ldots+n)\frac1\lambda = \frac{n(n+1)}{2n}\frac1\lambda = \frac{n+1}{2\lambda}.$$