I am trying to find the following probability in a renewal poisson process: $P[Y(t)>x/A(t+ \frac{x}{2})=s] $, where $A(t)=t-S_{N(t)}$ and $Y(t)=S_{N(t)+1}-t$.
I assumed that $X_{N(t)+1}=A(t+ \frac{x}{2}) +Y(t)- \frac{x}{2}=s- \frac{x}{2}+Y(t)>s- \frac{x}{2} +x=s+ \frac{x}{2}$.
Thus, $P[Y(t)>x/A(t+ \frac{x}{2})=s] =\frac{P[Y(t)>x,A(t+ \frac{x}{2})=s] }{P(A(t+ \frac{x}{2})=s)}=\frac{P(X_{N(t)+1}>s+\frac{x}{2})}{P(A(t+ \frac{x}{2})=s)}=\frac{\overline{F}(\frac{x}{2} +s)}{P(A(t+ \frac{x}{2})=s)}$,
and $\overline{F}(s)=1-F(s)$.
I do not know if all that are correct, but if they are I do not know how to deal with $P(A(t+ \frac{x}{2})=s)$ .
Is there any concrete relation with $A(t), Y(t)$ and the distribution function $F(t)$?
Both the backward and forward recurrence times are exponentially distributed with parameter $\lambda$, and are independent due to the memoryless property. Therefore $\mathbb P(Y(t)>x\mid A(t)=s)=\mathbb P(Y(t)>x)$. However, if $N\left(t+\frac x2\right)>N(t)$, then the conditional probability $\mathbb P\left(Y(t)>x\mid A\left(t+\frac x2\right)=s\right)$ is zero. It follows that \begin{align} \mathbb P\left(Y(t)>x\mid A\left(t+\frac x2\right)=s\right) &= \mathbb P\left(Y(t)>x\mid A\left(t\right)=s\right)\mathbb P\left(N\left(t+\frac x2\right) - N(t) = 0\right)\\ &= \mathbb P(Y(t>x)\mathbb P\left(N\left(\frac x2\right)=0\right)\\ &= \left(\int_x^\infty \lambda e^{-\lambda u}\ \mathsf du\right)\cdot e^{-\lambda\frac x2}\\ &= e^{-\lambda x}e^{-\lambda\frac x2}\\ &= e^{-\lambda\left(x+\frac x2\right)}. \end{align}