Simple Renewal Process Question The question reads:
Suppose the lifetime of a component $T_i$ in hour is uniformly distributed on $[100, 200]$.  Components are replaced as soon as one fails and assume that this process has been going on long enough to reach equilibrium.  Suppose it is known that the current component has been in operation for exactly 90 hours.  What is the probability that it will last at least 50 more hours?
Essentially, we are being asked to compute:
$$P(B_t \geq 50\ |\ A_t = 90),$$
where $B_t$ is the time remaining in the component's life and $A_t$ is the current age of the component.  The computation can be rewritten as 
$$P(B_t \geq 50\ |\ A_t = 90)=\frac{P(B_t \geq 50, A_t = 90)}{P(A_t = 90)}$$
by Bayes' Theorem.  I know that $A_t$ and $B_t$ aren't independent, but I'm not seeing how to factor that into the computation.  I know that both follow uniform distributions with mean $\mu=150$, but honestly I can't get much farther than that.  I know how to calculate both $P(B_t \geq 50)$ and $P(A_t = 90)$, but I can't put the two together.  Any help is greatly appreciated.
Thanks!
 A: In the stationary regime, $A+B$ is distributed like a size-biased version $S$ of the lifetime of a component (more about this later) and $(A,B)$ is distributed like $(US,(1-U)S)$, where $U$ is uniform on $[0,1]$ and independent of $S$. Thus, to compute $p=\mathbb P(B\geqslant b\mid A=a)$, one needs to know the density $f_{A,B}$ of the joint distribution of $(A,B)$, since then $p=q(b)/q(0)$, where
$$
q(z)=\int_{y\geqslant z}f_{A,B}(a,y)\mathrm dy.
$$
The distribution of $(A,B)$ is entirely determined by the above if one adds a description of the distribution of $S$: this is the unique distribution such that $\mathbb E(\varphi(S))=\mathbb E(T\varphi(T))/\mathbb E(T)$ for every bounded measurable function $\varphi$.
Thus, to compute $f_{A,B}$, consider any bounded measurable function $\varphi$ and use the identity
$$
(*)=\iint\varphi(x,y)f_{A,B}(x,y)\mathrm dx\mathrm dy=\mathbb E(\varphi(A,B))=\mathbb E(\varphi(US,(1-U)S)),
$$
which yields
$$
(*)=\mathbb E(T\varphi(UT,(1-U)T))/\mathbb E(T).
$$
Measuring $T$ in hundreds of hours, $T$ is uniform on $[1,2]$ hence $\mathbb E(T)=\frac32$ and
$$
(*)=\tfrac23\int_1^2t\int_0^1\varphi(ut,(1-u)t)\mathrm du\mathrm dt.
$$
The change of variables $(x,y)=(ut,(1-u)t)$ with Jacobian $\mathrm dx\mathrm dy=t\mathrm dt\mathrm du$ yields
$$
(*)=\tfrac23\iint\varphi(x,y)\mathbf 1_{1\leqslant x+y\leqslant2}\mathbf 1_{x\geqslant0,y\geqslant0}\mathrm dx\mathrm dy,
$$
which implies that
$$
f_{A,B}(x,y)=\tfrac23\mathbf 1_{1\leqslant x+y\leqslant2}\mathbf 1_{x\geqslant0,y\geqslant0}.
$$
In other words, $(A,B)$ is uniform on the inside of the quadrilateral with vertices $(1,0)$, $(2,0)$, $(0,2)$ and $(0,1)$.
In particular, conditionally on $[A=a]$ for some $a$ in $[0,1]$, $B$ is uniformly distributed on $[1-a,2-a]$ and, for every $b$ in  $[1-a,2-a]$, $p=\mathbb P(B\geqslant b\mid A=a)=2-a-b$. Your case is $a=\frac9{10}$, $b=\frac12$, hence $p=\frac35$.
Note finally that $A$ and $B$ are identically distributed and not independent (this is always the case) and that, in the present case, their common distribution has density 
$$\tfrac23(\mathbb 1_{0\leqslant x\leqslant1}+(2-x)\mathbb 1_{1\lt x\leqslant2}),
$$
hence $\mathbb E(A)=\mathbb E(B)=\frac79$. In the general case, 
$$
f_{A,B}(x,y)=\frac{f_T(x+y)}{\mathbb E(T)}\mathbf 1_{x\geqslant0,y\geqslant0},
$$
where $f_T$ is the density of $T$. Furthermore, $\mathbb E(A)=\mathbb E(B)$ and their common value is $\frac12\mathbb E(S)$ where $\mathbb E(S)=\mathbb E(T^2)/\mathbb E(T)$, hence $\mathbb E(S)\gt\mathbb E(T)$ as soon as the distribution of $T$ is not degenerate.
