Htydjtk

Let $tau_isimexpleft(lambdaright)$ be independent and identically distributed exponentials with parameter $lambda$. Then, for given $n$, the sum of these values
$$T_n := sum_{i=0}^n tau_i$$
follows an Erlang-Distribution with probability density function
$$pi(T_n=T| n,lambda)={lambda^n T^{n-1} e^{-lambda T} over (n-1)!}quadmbox{for }T, lambda geq 0.$$

I am interested in the distribution of $T_tilde n$ where $tilde n$ is a random variable such that for $tau_a sim exp(lambda_a)$ exponentially distributed, it holds that
$$T_tilde n leq tau_a \T_{tilde{n}+1} > tau_a.$$

In other words, $T_{tilde n}$ is truncated by an exponential distribution. I fail in deriving the distribution of $tilde n$ but maybe there is an easier way:
$$pileft(tilde n = kright) = pileft(T_n < tau_a|n=kright)\
=1-intlimits_{mathbb{R}^+}sumlimits_{n=0}^{k-1}frac{1}{n!}expleft(-(lambda+lambda_a)tau_aright)(taulambda_a)^nlambda_adtau_a.$$

However, just sampling and eye-balling looks to me like this density isn't that ugly:

iter <- 20000



lambda_a <- 1

lambda <- 2



df <- data.frame(tau=rep(NA, iter), a=rep(NA, iter))



for(i in 1:iter){

    set.seed(i)

    a <- rexp(1, rate = lambda_a)

    s <- cumsum(rexp(500, rate = lambda))



    df[i,] <- c(max(s[1], s[s<a]), a)

}



library(tidyverse)



ggplot(df %>% gather(), aes(x = value, fill = key)) +

geom_density(alpha = .3) + theme_bw()

As detailed in this X validated answer, waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed one produces an Poisson $mathcal P(lambda)$ variate $N$. Hence waiting for a sum of iid exponential $mathcal E(lambda)$ variates to exceed $tau_a$ produces an Poisson $mathcal P(tau_alambda)$ variate $N$, conditional on $tau_a$ (since dividing the sum by $tau_a$ amounts to multiply the exponential parameter by $tau_a$. Therefore
begin{align*}
mathbb P(N=n)&=int_0^infty mathbb P(N=n|tau_a) ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&= int_0^infty dfrac{(lambdatau_a)^n}{n!},e^{-tau_alambda} ,lambda_a e^{-lambda_atau_a},text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},int_0^infty tau_a^n,e^{-tau_a(lambda+lambda_a)} ,text{d}tau_a\
&=dfrac{lambda_alambda^n}{n!},dfrac{Gamma(n+1)}{(lambda_a+lambda)^{n+1}}=dfrac{lambda_alambda^n}{(lambda_a+lambda)^{n+1}}
end{align*}
which is a Geometric $mathcal G(lambda_a/{lambda_a+lambda})$ random variable. (Here the Geometric variate is a number of failures, meaning its support starts at zero.)

Considering now $N$ as a Geometric number of trials, $Nge 1$, the distribution of$$zeta=sum_{i=1}^N tau_i$$the moment generating function of $zeta$ is
$$mathbb E[e^{zzeta}]=mathbb E[e^{z{tau_1+cdots+tau_N}}]=mathbb E^N[mathbb E^{tau_1}[e^{ztau_1}]^N]=E^N[{lambda/(lambda-z)}^N]=E^N[e^{N(ln lambda-ln (lambda-z))}]$$and the mgf of a Geometric $mathcal G(p)$ variate is
$$varphi_N(z)=dfrac{pe^z}{1-(1-p)e^z}$$Hence the moment generating function of $zeta$ is $$dfrac{pe^{ln lambda-ln (lambda-z)}}{1-(1-(lambda_a/{lambda_a+lambda}))e^{ln lambda-ln (lambda-z)}}=dfrac{p lambda}{ lambda-z-lambda^2/{lambda_a+lambda}}$$
where $p=lambda_a/{lambda_a+lambda}$, which leads to the mfg
$$dfrac{lambdalambda_a/{lambda_a+lambda}}{ lambda-z-lambdalambda_a/{lambda_a+lambda}^2}=dfrac{1}{1-z(plambda)^{-1}}$$meaning that $zeta$ is an Exponential $mathcal{E}(lambdalambda_a/{lambda_a+lambda})$ variate.