Question:
Chris counts the number of new patients with Covid-42 entering some hospital. Over a long
period of time, he observes that there is in average 2 new cases per day entering the hospital.
He decides to model the situation by a Poisson process.
Given that exactly 2 new Covid-42 patients arrived within a time window of 12 hours,
what is in expectation the interval of time between the two arrivals?
My niave attempt:
Since $T_1$ and $T_2$ are disjoint i.i.d variables with distribution $T_n\sim exp(\lambda),\space\forall n\ge 1.$
$\Rightarrow \mathbb E[T_2]=\frac{1}{\lambda}=\frac{1}{2}$ ($\equiv$ 12hrs)
Comments:
I realise my attempt is incorrect- it's the best I could come up with. I also see that I need to incorporate the fact we are given that only 2 patients arrived in the 12hr interval, but I am unsure how to use that- perhaps some kind of use of the Law Of Total Expectation?
Furthermore, I have posted this in hopes that someone could give me a few hints or their version of a solution as I am at a bit of a dead-end and need some new ideas 🙂
Best Answer
I'll illustrate two different methods for solving this problem.
Method #1: As I wrote in my comments, let $X\sim \exp(2)$ time of the first case, $Y\sim \exp(2)$ as the interarrival time between the first and second case, and $Z\sim \exp(2)$ as the interarrival time between the second and third case. Define the following event $A$ by
$$A=\Big\{X+Y\leq12,X+Y+Z>12\Big\}$$ The conditional density of $(X,Y,Z)|A$ has pdf $f_{A}$ and is given by the following formula $$f_{A}(x,y,z)=\frac{2^3\exp\left(-2(x+y+z)\right)}{\mathbb{P}(A)}\cdot 1_{A}$$ Here, $1_{A}$ is the indicator variable of the event $A$ whereas $$\mathbb{P}(A)=\int_0^{12}\int_0^{12-x}\int_{12-x-y}^{\infty}2^3\exp\left(-2(x+y+z)\right)\mathrm{d}z\mathrm{d}y\mathrm{d}x$$ You're asked to evaluate $\mathbb{E}(Y|A).$ We'll do so using the conditional pdf of $(X,Y,Z)|A$: $$\mathbb{E}(Y|A)=\int_0^{12}\int_0^{12-x}\int_{12-x-y}^{\infty}yf_{A}(x,y,z)\mathrm{d}z\mathrm{d}y\mathrm{d}x=4$$
Method #2: Let $T_1,T_2$ denote the first two arrival times we're interested in. It is known that the distribution of the random vector $(T_1,T_2)$ has the same distribution as $(X_{(1)},X_{(2)})$ where $X_1,X_2\sim \mathcal{U}[0,12]$. To find the joint density of $(X_{(1)},X_{(2)})$ we'll use the fact that $f_{X_{(1)}}(x)=\frac{12-x}{72}\cdot 1_{[0,12]}$ and $X_{(2)}|X_{(1)}=x\sim \mathcal{U}[x,12]$. We get for any $(x,y)\in \mathbb{R}$ with $0 \leq x\leq y \leq 12$ that $$f_{X_{(1)},X_{(2)}}(x,y)=f_{X_{(2)}|X_{(1)}=x}(y|x)f_{X_{(1)}}(x)=\frac{1}{72}$$ In other words, the joint density of the random vector $(X_{(1)},X_{(2)})$ is uniform on the triangular region $$\mathcal{T}=\{(x,y):0\leq x\leq y \leq 12\}$$ So we get $$\mathbb{E}\left(T_2-T_1\right)=\iint_{\mathcal{T}}\frac{y-x}{72}\mathrm{d}A=4$$