Renewal process large time behaviour I have the following question:
I have two types of battery, $1$ and $2$. Suppose the lifetime of $1$ is uniformly distributed on the interval $(0,3)$, battery $2$ uniformly distributed on the interval $(10,13)$. $p=90\%$ of my batteries are type 1, $1-p=10\%$ are type $2$. Battery $1$ costs £$1$ per battery, Battery $2$ costs £$5$ per battery. I have a device which runs on one battery until the battery runs out, then I replace it. At each renewal, I pick a battery at random.


*

*For a large time $t$, what is the approximate probability that my device is being powered by a Type $2$ battery at time $t$?

*Would you recommend a different mix of batteries? What's the long term average cost per unit time for your preferred mix of batteries?


The first part sort of reminds of the inspection paradox/waiting time paradox, but I don't really know what to do. I feel like it is to do with the limiting distributions of the age $A_t=T_{X_t+1}-t$ and excess lifetime $E_t=t-T_{X_t}$.
 A: Consider the Markov renewal-reward process $\{(X_n, T_n) : n = 1,2,3,\ldots\}$ where 


*

*$\tau_n$ is the lifetime of the $n^{\mathrm{th}}$ battery, $n\geqslant 1$ 

*$T_n=\sum_{k=1}^n \tau_k$, $n\geqslant 1$

*$\{X_n\}$ is a Markov chain on $\{1,2\}$ with transitions according to the arrival process $\{T_n\}$,

*$\{Y_n\}$ is the reward sequence, i.e. $Y_n = -B_n - 5(1-B_n)$ where $B_n:=\mathsf 1_{\{X_n=1\}}$. 


Then $\tau_n=B_nU_n + (1-B_n)V_n$ where $\{U_n\}$ is an iid sequence of $\mathcal U(0,3)$ random variables and $\{V_n\}$ is an iid sequence of $\mathcal U(10,13)$ random variables. As the $\{X_n\}$ are iid with $$\mathbb P(X_n=1)=p=1-\mathbb P(X_n=2) $$
for all $n$, it follows that the interrival times $\tau_n$ are iid with mixture density
$$f(t) = \frac9{10}\left(\frac13\cdot\mathsf 1_{(0,3)}(t)\right) + \frac1{10}\left(\frac13\cdot\mathsf 1_{(10,13)}(t)\right) = \frac3{10}\cdot\mathsf 1_{(0,3)}(t) + \frac1{30}\cdot\mathsf 1_{(0,3)}(t), $$
and the $\{Y_n\}$ are iid with distribution
$$
\mathbb P(Y_n=j) = \begin{cases}
\frac9{10}(-1)\mathbb P(B_n=1) = -\frac9{10},&j=1\\
\frac1{10}(-5)\mathbb P(B_n=2) = -\frac12,& j=2.
\end{cases}
$$
Question $(a)$ concerns the asymptotic distribution of $\{\mathcal B(t):t\geqslant 0\}$ where $\mathcal B(t) = X_{\sup\{n\ :\ T_n<t\}}$, i.e. $\mathcal B(t)=j$ iff the last renewal $T_n$ before time $t$ was of battery type $j$. If we define $Y_n=T_nB_n$ and $Z_n=T_n(1-B_n)$, that is, splitting the process into an alternating renewal process which changes states when the type of battery changes, then we may compute
\begin{align}
\mathbb E[Y_n] &= \mathbb E[\tau_n\mid B_n=1] + \mathbb P(B_n=1)\mathbb E[Y_n]\\
&= \frac32 + \frac9{10}\mathbb E[Y_n]\\
\end{align}
so that $\mathbb E[Y_n] = 15$, and similarly
$$\mathbb E[Z_n] = \frac{23}2 + \frac1{10}\mathbb E[Z_n] $$
so that $\mathbb E[Z_n] = \frac{115}9$. It follows from the renewal-reward theorem that
$$\lim_{t\to\infty} \mathcal B(t) = \frac{\mathbb E[Z_n]}{\mathbb E[Z_n]+\mathbb E[Y_n]} = \frac{\frac{115}9}{\frac{115}9+\frac{23}2}=\frac{23}{50}. $$
Note that this is the same as if we apply a similar logic to a single renewal epoch:
$$\frac{\frac1{10}\cdot \frac{23}2}{\frac1{10}\cdot \frac{23}2 +\frac9{10}\frac32}=\frac{23}{50}. $$
As for $(b)$, we compute the expected interrenewal times $$\mathbb E[\tau_n] = \frac3{10}\int_0^3 t\ \mathsf dt + \frac1{30}\int_{10}^{13} t\ \mathsf d t = \frac52$$ and expected reward per epoch $$\mathbb E[Y_n] = \frac9{10}(-1) + \frac1{10}(-5) = -\frac75. $$ (The reward is negative because it is a cost.) Let $N(t)=\sum_{n=1}^\infty \mathsf 1_{(0,t]}(\tau_n)$ be the number of arrivals up to time $t$, and $W(t)=\sum_{n=1}^{N(t)}Y_n$ be the total award accumulated up to time $t$. Then by the renewal-reward theorem, the asymptotic long run average reward per unit time is
$$\lim_{t\to\infty}\frac1t W(t) = \frac{\mathbb E[Y_1]}{\mathbb E[T_1]} = \frac{-7/5}{5/2} =  -\frac{14}{25}.$$
If we are allowed to change $p$, then the above becoves
$$R(p):=\frac{10-8p}{20p-23}, $$
which has strictly negative derivative on $(0,1)$, so in order to minimize cost, we should choose $p$, i.e. choose battery type 2.
