# Computing covariance with the help of the memoryless property

Question: Suppose that in a certain town earthquakes occur as a Poisson point process with an average of 3 per decade, floods are a Poisson process with an average of 2 per decade, and meteor strikes are a Poisson process with an average of 1 per decade. Consider the present to be time zero, and write $E$, $F$ and $M$ for the time in decades between the present and the first earthquake, flood, and meteor strike (respectively). Compute $\mbox{Cov}(\min\{E, F, M\}, M)$.

By reading the brief solution, I understand that the process starts by "adding nothing" to obtain

$\mbox{Cov}(\min\{E, F, M\}, M) = \mbox{Cov}(\min\{E, F, M\}, M - \min\{E, F, M\} + \min\{E, F, M\})$.

Then it splits this up using bilinearity into

$\mbox{Cov}(\min\{E, F, M\}, \min\{E, F, M\}) + \mbox{Cov}(\min\{E, F, M\}, M - \min\{E, F, M\})$

The left component is just $\mbox{Cov}(\min\{E, F, M\}, \min\{E, F, M\}) = \mbox{Var}(\min\{E, F, M\}) = 6^2$ (or $1/6^2$ depending on what definition you use). I understand that.

But it also says that, due to the memoryless property, the right hand component is zero becuse the two random variables are independent (that's a simple fact about covariances). But I'm not intuitively seeing the connection between the memoryless property and the fact that $\min\{E,F,M\}$ and $M - \min\{E,F,M\}$ are independent.

UPDATE: Just to make things clear, the solutions manual gives the final answer as $1/6^2$.

• Where you wrote $\mbox{Var}(E, F, M)$, I think you need $\mbox{Var}(\min\{E, F, M\})$. – Michael Hardy Aug 27 '12 at 18:16
• There is positive probability that one of those two allegedly independent random variables is $0$. I'm not sure what the import of that fact is and right now I have other things to attend to. I shall return.......... – Michael Hardy Aug 27 '12 at 18:28
• Yes I did, thanks Michael Hardy – TakeS Aug 27 '12 at 18:40
• $6^2$ (or $1/6^2$ depending on what definition you use)... What do you mean? It cannot be both and the choice is NOT a matter of definitions. – Did Aug 27 '12 at 19:54
• On the MIT OCW and Wiki, I see that variance is $\lambda^{-2}$. But on another set of notes that my professor showed me, variance is $\lambda^2$ and he specifically says that there is inconsistency in how $\lambda$ is used, so I am not sure which one I should use. However, I will stick with the former definition for variance (of exponential random variables). – TakeS Aug 27 '12 at 20:19

Let $D = \min(E,F,M)$ be the time until the first "disaster" (where a disaster is either an earthquake, a flood, or a meteor strike). Disasters form a Poisson process with rate $3+2+1=6$ per decade, and each disaster is (independent of the time it occurs) an earthquake, flood or meteor strike with probabilities $3/6, 2/6, 1/6$ respectively. Given $D=x$, $M = x$ with probability $1/6$; if not, $M - D$ is exponential with rate $1$ per decade. Thus $E[M-D | D=x] = (1/6) 0 + (5/6) 1 = 5/6$. So $E[M | D] = D + 5/6$, $E[MD] = E[E[MD | D]] = E[D^2 + (5/6) D] = 7/36$, and $\text{Cov}(M,D) = E[MD] - E[M]E[D] = 1/36$.
• $E(D^2)=2E(D)^2$ and $E(D)=1/6$ hence $E(MD)=2(1/6)^2+(5/6)(1/6)=7/36$. – Did Aug 27 '12 at 19:53
The memoryless property asserts that the joint distribution of $D=\min\{E,F,M\}$ and $M$ is as follows: conditionally on $D=x$, $M=x$ with probability $1-a$ and $M\gt x$ with probability $a$, in which case the conditional distribution of $M$ is $\nu$, where $a$ and $\nu$ are independent on $x$.
Let $A$ denote a Bernoulli random variable with $P(A=1)=a$ and $P(A=0)=1-a$, and $N$ a random variable with distribution $\nu$, both independent of $D$. A reformulation of the preceding property is that $(D,M)$ is distributed like $(D,D+AN)$.
In the case at hand, $D$ is exponential with parameter $6$ and $N$ is exponential with parameter $1$, hence $\mathrm E(M\mid D)=D+aE(N)$, thus $\mathrm E(DM)=\mathrm E(D\mathrm E(M\mid D))=\mathrm E(D^2)+a\mathrm E(D)\mathrm E(N)$, which leads to $\mathrm{Cov}(D,M)=\mathrm{Var}(D)=1/36$.
Edit: In the case at hand, the distribution of $N$ and the value of $a$ are well known. However, it is a nice feature (among many others!) of Markov queues that these do not enter into the final result $\mathrm{Cov}(D,M)=\mathrm{Var}(D)$.