Let $\gamma_t$ be the excess life and $\delta_t$ the age in a renewal process having interoccurence distribution function $F(x)$. Determine the conditional probability $\mathbb{P}(\gamma_t>y|\delta_t=x)$ and the conditional mean $\mathbb{E}(\gamma_t|\delta_t=x)$.
Solving for the expectation should follow directly from solving for the tail probability, and so I'm not going to worry myself with that part of the problem. As of right now, I've simplified the problem down with some basic conditional probability:
$$\mathbb{P}(\gamma_t>y|\delta_t=x) = \frac{1-F(x+y)}{\mathbb{P}(\delta_t=x)}$$ The denominator is just the density of the age of a renewal process at time $t$ evalulated at $x$. I tried using tail probabilities to solve for this in the beginning, but that didn't work out. Another approach I tried was using a fact from Durett's Essentials of Stoachstic Processes:
The limiting distribution for the age and residual life have the density function given by: $$g(z) = \frac{\mathbb{P}(t_i>z)}{\mathbb{E}(t_i)}$$
The only problem is that we haven't actually covered this fact in my stochastic processes class yet, and so I'm pretty sure that I'm supposed to solve this problem without using it. How would you guys approach this problem?
Let $T$ be the time of the next renewal, then as the distribution of $\gamma_t$ conditioned on ${\delta_t=x}$ for $0\leqslant y$ is independent of $y$ we have $$ \mathbb P(\gamma_t>y\mid\delta_t=x) = \mathbb P(T-x>y\mid T>x) = \frac{1-F(t+x)}{1-F(x)}. $$ Differentiating with respect to $t$, we find that the above is equal to $$ \frac{f(t+x)}{1-F(x)}\cdot\mathsf 1_{\{t>0\}}. $$ It follows then that the conditional mean is $$ \frac1{1-F(x)}\int_0^\infty t f(x+t)\ \mathsf dt. $$