Let $T\sim~U[-t_0,t_0]$ and let $\epsilon\sim~U[0,1]$ and $T\perp \epsilon$.

We define $R=T+\epsilon$. I was trying to compute $f(T=t|R=r,T\geq t_1)$, where $t_1\in[-t_0,t_0]$ is some pre-specified number.

My attempt is \begin{equation*} \begin{aligned} f(T=t|R=r,T\geq t_1)&=\frac{f(T=t,R=r,T\geq t_1)}{f(R=r,T\geq t_1)}\\ &=\frac{f(T=t,\epsilon = r-t,T\geq t_1)}{f(R=r,T\geq t_1)}\\ &=\frac{f_\epsilon(r-t)f(T=t,T\geq t_1)}{f(R=r,T\geq t_1)},~\because~\epsilon\perp T\\ &=\frac{\mathbf{1}_{\{r-1\leq t\leq r\}}\frac{1}{2t_0}\mathbf{1}_{\{t\geq t_1\}}}{f(R=r,T\geq t_1)}\\ &=\frac{\mathbf{1}_{\{r-1\leq t\leq r\}}\frac{1}{2t_0}\mathbf{1}_{\{t\geq t_1\}}}{f(\epsilon=r-T,T\geq t_1)} \end{aligned} \end{equation*} I think the above derivation is correct, but I stuck at the last step, how to derive the explicit form of $f(R=r,T\geq t_1)$?


Yes, looks good. Regarding the last step: $$ f(R=r, T \ge t_1) = f(T + \epsilon = r, T \ge t_1) = \int_{t_1}^{t_0} f(T + \epsilon = r \mid T =t) f_T(t)dt = \int_{t_1}^{t_0}f_{\epsilon}(r-t)f_T(t) dt. $$

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