3

We can write $\xi(t)$ as $$\xi(t)=\sum_{k=1}^{N(t)} \xi_k$$ where $\{N(t)\}_{t\in\mathbb{R}}$ is a counting Poisson process with rate $\lambda$. It is not specified in your question but I assume that $\lambda>0$ otherwise we would have $N(t)=0$ for all $t$ and thus $\xi(t)=0$ which contradicts what we are trying to prove. I am also going to assume that ...


2

Suppose the additional service time for the existing customer time is $S \sim \text{Exp}(\mu)$. Then the number of arrivals in time $S$ is $Z\sim \text{Pois}(\lambda S)$ This implies $$\mathbb P(Z=a) = \int_0^\infty e^{-\lambda s} \frac{\lambda^a s^a}{a!} \mu e^{-\mu s}\, ds = \frac{\lambda^a \mu}{(\lambda+\mu)^{a+1}}= \frac{\frac{\mu}{\lambda}}{\left(1+\...


1

Just to flesh out @HagenvonEitzen's sketch, because I always find fleshed-out solutions helpful. To state it explicitly, geometrically, the relevant fact is that a set of lengths will be able to form a polygon as long as none of the lengths are bigger than the sum of all the others. Taking, as Hagen says, for all $k \leq n$, $Y_k$ to be $Y_k:=X_1+\cdots+X_{...


1

By denoting $x = \xi(7)$ you get the denominator is simply $x+1$, so your only task is to find $\Bbb E[\xi(5)|\xi(7) = x]$. By definition you have that $$ \Bbb E[\xi(5)|\xi(7) = x] = \sum_{k=0}^\infty k\cdot \Bbb P(\xi(5) = k|\xi(7) = x) $$ so you need to compute the latter conditional probabilities. I hope you are good with $\Bbb P(\xi(7) = x|\Bbb \xi(5) = ...


1

$$\sum_{i=0}^k \frac {\rho^i}{i!}=\frac{e^{\rho }\, \Gamma (k+1,\rho )}{k!}\implies f(k)=\frac{e^{-\rho } \rho ^{k+1}}{(k+1)\, \Gamma (k+1,\rho )}$$ You will find here a very good paper about uniform bounds for the incomplete gamma function. To my knowledge, this is one of the most recent paper on this topic. If $\rho$ is small, you can use the expansion $$ \...


1

The first one is a fact that is generally taught in class and not expected to be derived (based on my limited experience searching for similar courses). Were you not taught that the arrival times are i.i.d. uniform on $[0,T]$ conditioned on the number of arrivals up to a time $T$? It is 2.4.3 on Durrett EOSP (available online free). The proof of it is not ...


1

1. For $\alpha>0$, the integral, $$ \int_0^\infty(1-e^{-\lambda x}){dx\over x^{1+\alpha}} $$ converges if and only if $0<\alpha<1$, because the integrand is $\sim \lambda x^{-\alpha}$ for $x\to 0+$. This is precisely the range of $\alpha$s for which $X_t<\infty$ a.s. For $\alpha\in(0,1)$, the change of variables $u=\lambda x$ results in $$ \...


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