Equivalence between exponential clocks that ring in different ways

Question

Could someone explain to me why the following facts are equivalent.

1) I have a clock that rings with exponential distribution of parameter $1$. When the click rings a transition to the state $s_1$ occurs with probability $p$ and a transition to $s_2$ occurs with probability $1-p$.

2) I have two independent clocks that ring with exponential distribution of parameter $p$ and $1-p$ respectively. If the clock with parameter $p$ rings first then a transition to $s_1$ occurs, if the clock with parameter $1-p$ rings first then a transition to the state $s_2$ occurs.

Why the two systems are equivalent?

Answer 1

This equivalence comes down to the following two facts (which you can try proving as an exercise). Suppose that $X \sim \mathrm{Exponential}(a)$ and $Y\sim \mathrm{Exponential}(b)$ (where $a$ and $b$ are rate parameters) and $X$ and $Y$ are independent. Then:

1. $\min(X,Y)\sim \mathrm{Exponential}(a + b)$
2. $\mathbb{P}\left(X \leq Y \right) = \dfrac{a}{a + b}$.

Using these, you can see that in both your formulations, the time to transition is $\mathrm{Exponential}(1)$ and the probability of transitioning to $s_{1}$ is $p$.