 Solution

A renewal process counts arrivals $N(t)=\max \{j:A_j\le t\}$, where $A_0=0,A_j=S_1+\cdots +S_j$ and the holding times are independent, identically distributed positive random variables. Their sum tends to infinity almost surely, so the count is finite on bounded time intervals. Use $N(s)=X_s$ in the question's notation. The index $J=N(s)+1=\min \{j:A_j>s\}$ is a stopping index for the holding-time sequence. Conditional on $J=j$ and the first $j$ holding times, every later $S_{j+k}$ retains its original probability distribution. Summing over $j$ provesThe straddling time $T_1=S_J$ is different. Let $\overline{F}(t)=P(S_1>t)$ and let $U$ be the renewal measure $U(B)={\sum }_{j\ge 0}P(A_j\in B)$. Independence of the next holding time from its start givesFor two numbers in $[0,1]$, their minimum is at least their product. Since $\overline{F}(\max (t,a))=\min (\overline{F}(t),\overline{F}(a))$, subtracting $\overline{F}(t)$ times the second integral provesThese formulae handle holding-time atoms with the stated renewal convention $A_j\le s<A_{j+1}$; sums may be interchanged by nonnegativity.

For rate-$\lambda$ exponentials, the renewal process is Poisson. The elapsed age $B_s=s-A_{N(s)}$ has probability distribution $\min (E,s)$ with $E\sim \mathrm{Exp}(\lambda )$: for $0\le b<s$, $P(B_s>b)$ is the probability of no arrivals in $(s-b,s]$, namely $e^{-\lambda b}$, with an atom $e^{-\lambda s}$ at $s$. The forward residual $R_s$ is an independent rate-$\lambda$ exponential by independent future increments. Thus $T_1=B_s+R_s$. Conditioning on the age gives the exact survival lawFor $s,t>0$ the extra term is strictly positive, proving strict inequality. For fixed $t$ and $s\to \infty$, the limit is $e^{-\lambda t}(1+\lambda t)$, which equals the survival probability of the sum of two independent rate-$\lambda$ exponentials. This is the finite-time form of the stochastic length bias of a renewal interval.