Long time reader, first time writer here. I am dealing with a very specific type of an exercise in stochastic processes. Calls are arriving according to the renewal process, where interarrival times are a mix of $Exp(λ)$ with the weight $1/3$ and $Exp(2*λ)$ with the weight $2/3$. Telephone operator decides to go on a strike and to respond to a call with probability $p$, independently from everything else. Find distribution and expected value of the first responded call $T_1$.
So here's how far I have come. Calls are arrivals of the process, they are distributed as follows: $T' ∼ \frac{1}{3} * Exp(λ) + \frac{2}{3} * Exp(2λ)$. The operator throws a coin at each arrival, the coin shows heads with probability $p$ and tails with probability $(1-p)$. If the coin shows heads, the operator will accept the call - meaning that we can count the arrival in our process and $T_1 | heads ∼ T'$.
If the coin shows tails, the operator will not accept the call and the process will restart from the beginning. I am struggling to find the conditional distribution of $T_1 | tails ∼$ ? I'm thinking it's something like $$T_1 | tails ∼ T' + p * T_1 | heads + (1-p) * T_1 | tails,$$ however I doubt that that is correct.
(Partial answer)
Let $S_n$ be the sequence of renewal times, which have interrenewal distribution $$ f(t) = \frac13 \lambda e^{-2 \lambda t} \left(e^{\lambda t}+4\right), \ t>0. $$ Let $N$ be the first call that the operator accepts, then $N\sim\mathsf{Geo}(p)$ and by Wald's identity the mean time of the first responded call is
$$ \mathbb E\left[\sum_{n=1}^N S_n\right] = \mathbb E[N]\mathbb E[S_1] = \frac1p\cdot\frac{2}{3 \lambda }= \frac 2{3\lambda p}. $$ The distribution of $T_1$ could be computed by the law of total probability: $$ f_{T_1}(t) = \sum_{n=1}^\infty f_{T_1\mid N=n}(t)\mathbb P(N=n), $$ but I was not able to find a tractable closed form for the density of $T_1$ conditioned on $N=n$ (that is, the $n$-fold convolution of the interrenewal density).