Consider a renewal process with the lifetimes $X_1,X_2,\ldots$ having the probability density function $f(x)=0.125\cdot (4-x)\;$ for $\;0<x<4$
Determine the asymptotic expression for the probability distribution of excess life $\gamma_t$: $\displaystyle\lim_{t\to\infty} \mathbb{P}(\gamma_t\leq 1.45)$
Determine the limiting mean excess life: $\displaystyle\lim_{t\to\infty} \mathbb{E}[\gamma_t]$
So far I have correctly calculated that $\displaystyle\lim_{t\to\infty} \mathbb{P}(\gamma_t\leq 1.45) = \frac{3}{4}\cdot\frac{379349}{384000}$ and my attempt to calculating $\displaystyle\lim_{t\to\infty} \mathbb{E}[\gamma_t] = \frac{M(t)}{t} + \frac{1}{t}-\frac{1}{\mu} = \frac{1}{\mu} + \frac{1}{t} - \frac{1}{\mu} = \frac{1}{t} = 0$ by using the fact that $\frac{M(t)}{t} \rightarrow \frac{1}{\mu}$ as $t \rightarrow \infty$. But this answer was incorrect, I would appreciate any help in figuring this out.
Edit: $\gamma_t$ is the residual life or excess (at time t) where $\gamma_t = W_{N(t) +1} - t$ and $\gamma_t \sim Exp(\lambda)$ .$E[\gamma_t] = \frac{M(t)}{t} + \frac{1}{t} - \frac{1}{\mu}$ where $M(t)$ is the renewal function where $\frac{M(t)}{t} \rightarrow \frac{1}{\mu}$ as $t \rightarrow \infty$ and $W_{N(t)+1}$ is the first renewal time after time $t$. $\mu$ is the mean of the given distribution