Inferring transition rates from continuous markov chain question

Question

A house has 2 rooms of similar sizes with identical air conditioners equipped with thermostats which turn on and off as needed to maintain the temperature in each room to a desired level of 22 degrees. Suppose that a thermostat remains on or off for exponential amounts of time with means $1/\mu$ and $1/\lambda$, respectively, independently of other thermostats. Consider the Markov process $\{X(t), t \ge 0\}$ whose state space is the number of active air conditioners. Write down the matrix of transition rates.

I'm not sure how to exactly approach this type of question. My working is as follows but if someone could clarify my confusion that would be good.

Working:

So clearly there are 3 states, 0 for no air conditioners are on, 1 for one air conditioner is on (active), and 2 for two air conditioners are on (active). Now to work out $q_{01}$, i.e., the transition rate from state 0 to 1, assume currently no air conditioners are on. Consider the two independent poisson processes both with parameter $\lambda$, where the interarrival time is the duration of "off", then merging these two processes gives a poisson process with parameter $2\lambda$, so the transition rate from state 0 to state 1 is $2\lambda$.

Now what about the transition rate from state 0 to 2? I am told that it's 0, but why? Isn't it possible for both air conditioners to both go from "off" to "on"? What is the argument that $q_{02} = 0$?

Answer 1

Consider two independent Poisson processes, let $(T_n)$ and $(S_n)$ denote the times when they increase. Then the probability that they both increase at a common time is bounded by $$ \sum_{n,m}P(T_n=S_m). $$ Here is a technical result which may help:

Lemma: Assume that $X$ and $Y$ are independent and that $Y$ has no atom, that is, $P(Y=y)=0$ for every $y$. Then, $P(X=Y)=0$.

Applying this to each $(T_n,S_m)$ shows that the probability that two independent Poisson processes both increase at a common time is zero.

To prove the lemma, note that, by definition of the distribution $P_{(X,Y)}$ of $(X,Y)$, then by independence of $(X,Y)$, and finally by Fubini, one has $$ P(X=Y)=\iint\mathbf 1_{x=y}\,\mathrm dP_{(X,Y)}(x,y)=\int\left(\int\mathbf 1_{x=y}\,\mathrm dP_Y(y)\right)\,\mathrm dP_X(x), $$ that is, $$ P(X=Y)=\int P(Y=x)\mathrm dP_X(x), $$ and, since $P(Y=x)=0$ for every $x$, we are done.