2 states, 2 interarrival distribution Renewal Process.

Question

Karlin and Taylor (1975):

18. Consider a stochastic process $X(t)$, $t \geq 0$, which alternates in 2 states $A$ and $B$. Denote by $\xi_1, \eta_1, \xi_2, \eta_2, \ldots,$ the successive sojourn times spent in states $A$ ind $B$, respectively, and suppose $X(0)$ is in $A$. Assume $\xi_1, \xi_2, \ldots,$ are i.i.d. r.v.’s with distribution function $F(\xi)$ and $\eta_1, \eta_2, \ldots$ are i.i.d. r.v.'s with distribution function $G(\eta)$. Denote by $Z(t)$ and $W(t)$ the total sojourn time spent in states $A$ and $B$ during the time interval $(0,t)$. Clearly $Z(t)$ and $W(t)$ are random variables and $Z(t) + W(t) = t$. Let $N(t)$ be the renewal process generated by $\xi_1,\xi_2,\ldots$. Define $$\theta(t) = \eta_1 + \eta_2 + \cdots + \eta_{N(t)}.$$ Prove $$P\{W(t) \leq x\} = P\{\theta\left(t-x\right) \leq x\},$$ and express this in terms of the distributions $F$ and $G$.

Answer: $$\Pr\{W(t) \leq x\} = \sum_{n=1}^\infty G_n(t-x) \left[F_n(x) - F_{n+1}(x)\right],$$ where $F_n$ and $G_n$ are the usual convolutions.

I need some hint, not the answer, how to solve the problem. Thank you.

Answer 1

According to these notes on Renewal Theory, the "usual" convolutions are the n-fold convolutions.

$$ F_n(x) = \underbrace{F \ast F\ast \dots \ast F}_{n}(x) = \mathbb{P}[N(x) \geq n]$$

We then an identity for the probability the renewal process has value $n$:

$$ \mathbb{P}[N(x)=n] = F_n(x) - F_{n+1}(x)$$

So we could re-write the identity in a more common sense fashion

$$ \mathbb{P}\{W(t) \leq x\} = \sum_{n=1}^\infty G_n(t-x) \cdot \; \mathbb{P}[N(x) = n], $$

The sets $\{ t: X(t) = A\} \cup \{ t: X(t) = B\} = [0,t]$ interlace.

Why were we able to act as if $X(t)=A$ was in a for up to time $x$ and then $X(t)=B$ for until time $t$?

Answer 2

Not a full answer but a hint that something might be amiss in the "Answer" provided by the authors.

Let $A=[W(t)\leqslant x]$, $B=[\theta(t-x)\leqslant x]$, $U_0=V_0=0$ and, for every $n\geqslant0$, $U_n=\xi_1+\cdots+\xi_n$ and $V_n=\eta_1+\cdots+\eta_n$. Then, for every $n\geqslant0$, $$ B\cap[N(t-x)=n]=[V_n\leqslant x,U_n\leqslant t-x\lt U_{n+1}]. $$ The processes $(U_n)$ and $(V_n)$ are independent, the CDF of each $U_n$ is $F_n$ and the CDF of each $V_n$ is $G_n$, hence $$ P(B)=\sum_{n\geqslant0}G_n(x)\cdot(F_n(t-x)-F_{n+1}(t-x)). $$ If indeed $P(A)=P(B)$, then two aspects of the formula suggested for $P(A)$ need explanation:

• The $n=0$ term of the summation is missing.
• The arguments $x$ and $t-x$ are exchanged.