Suppose a machine operates with a certain component which comes in two different colors, purple and blue. When the component fails it is replaced immediately. The Probability a purple component is chosen during replacement is $\frac 34$, otherwise a blue component is chosen. The purple components have a lifetime $T\sim \mathcal{U}(0,40)$(uniformly distributed). The blue component has a lifetime of $T\sim \mathcal{U}(10,20)$.(also uniformly distributed)
Let $A_t$ denote the time elapsed since the last renewal and $C_t$ denote the total life time. Suppose at some time $A_t=5$, what is the probability the component is purple? Find $P(\text{component purple} | A_t = 5)$. Also find $P(C_t<15 | A_t=5)$
What I did is $P(\text{component purple} | A_t = 5) = \frac{P(\text{component purple}, A_t = 5)}{ P(A_t=5)}$.
And $P(A_t = 5) = \text{density of} (A_t = 5) = (\frac 1\mu)(1-F(x))$
How do I solve mixed renewal process with two components? Thank you very much.
In the homogenous case based on i.i.d. exponential lifetimes $T_i$, let $A_t$ denote the time elapsed since the last renewal, $B_t$ the residual lifetime and $C_t=A_t+B_t$ the total life time. Then $A_t=a$ corresponds to the event $T\geqslant a$ hence, conditionally on $A_t=a$, $B_t$ is distributed like $T-a$ conditionally on $T\geqslant a$.
This remark is the starting point of the analysis of the homogenous case. Much more is known, for example, when $t\to\infty$, $(A_t,B_t,C_t)$ converges in distribution to $(US,(1-U)S,S)$ where $(U,S)$ is independent, $U$ is uniform on $(0,1)$ and $S$ follows the so-called size-biased transform of the distribution of $T$.
My guess is that the purpose of your homework is to adapt the proofs of these to the "bicolor" setting, if only to check that you master the "monocolor" setting.