Email arrivals in rates and time intervals according to a Poisson process Regular (not junk) emails arrive at your inbox according to a Poisson process with rate $r$; and junk emails arrive at your inbox according to an independent Poisson process with rate $j$. Assume both processes have been going on forever. Fix a time $t$ to be 8 o'clock.
According to my caluclations, the expected length of the interval that $t$ belongs to is:
$$\frac{1}{(r + j)}$$
Furthermore, for the probability that $t$ belongs to an RR interval I've got:
$$\big(\frac{r}{(r + j)}\big)^2$$
However, I'm having troubles finding the probability between two time intervals, i.e.
What is the probability that between $t$ and $t+1$, that exactly 2 emails, a regular email followed by a junk email, arrive?
I suspect it might be:
$$\big(\frac{rj}{(r + j)^2}\big)\big(\frac{(r + j)^2*e^{-r-j}}{2}\big)$$
But I'm not certain about that. Maybe someone would shed some light on it.
Thanks!
 A: Let $J_1,J_2,...$ denote the arrival times of the junk emails since 8am and $R_1,R_2,...$ the arrival times of the regular emails since 8am. You're tasked to compute $\mathbb{P}\left(R_1<J_1<1,R_2>1,J_2>1\right)$. If you set $$X=R_1,Y=J_1,Z=R_2-R_1,W=J_2-J_1$$ then $X,Z\sim \text{Exponential}(r)$ while $Y,Z\sim \text{Exponential}(j)$ are independent. Therefore the joint density of $(X,Y,Z,W)$ factors as $$r^2j^2\exp\big\{rx+jy+rz+jw\big\}\cdot 1_{\mathbb{R}_{\geq 0}^4}(x,y,z,w)$$ The sought after probability of $\mathbb{P}\left(X<Y<1,X+Z>1,Y+W>1\right)$ is the integral $$\int_0^1 \int _x^1 \int_{1-x}^{\infty}\int_{1-y}^{\infty}r^2j^2\text{exp}\big\{rx+jy+rz+jw\big\}\mathrm{d}w\mathrm{d}z\mathrm{d}y\mathrm{d}x$$ This evaluates (nicely) to $\frac{1}{2}rje^{-\left(r+j\right)}$ and agrees with your response.
Another approach involves a slight adjustment of the event whose probability we seek.
Take $J,R$ as the events that exactly one junk/regular email arrive on $[t,t+1]$ respectively. We get with independence that  $$\mathbb{P}(J \cap R)=\mathbb{P}(J)\mathbb{P}(R)=rje^{-(r+j)}$$ If $X,Y$ are the corresponding arrival times, it is known that $X|J,Y|R\sim \mathcal{U}[0,1]$ so using independence again we see $(X,Y)|J\cap R \sim \mathcal{U}([0,1]^2)$. Hence $$\mathbb{P}\left(Y<X<1|J\cap R \right)=\int_0^1 \int_0^x \mathrm{d}y\mathrm{d}x = \frac{1}{2}$$ Finally, $$\mathbb{P}\left(Y<X<1,J,R\right)=\mathbb{P}\left(Y<X<1|J\cap R\right)\mathbb{P}\left(J\cap R\right)=\frac{1}{2}rje^{-(r+j)}$$
