Discrete-time stochastic process - waiting time

Question

Suppose you are given a discrete-time stochastic process which startes in state $C$ and can attain two further states $A$ and $B$.

In state $C$ you take two independent exponential times $T_1 \sim E(2)$ and $T_2 \sim E(4)$, where $E(\lambda)$ denotes the exponential distribution with parameter $\lambda$. If $T_1$ is the smaller you go to position $A$ after time $T_1$, and if $T_2$ is the smaller you go to $B$ after time $T_2$.

Now my questions refers to the following statement:

The time spent in state $C$ is exponential of parameter $6$.

How can I show that the distribution of this random variable is $E(6)$?

Answer 1

First, you generate $T_1\sim E(2)$ and $T_2 \sim E(4)$ independently. Then, you leave the state C after $T_1$ or $T_2$ time has passed — whichever is sooner.

By the definition of the exponential distribution (with parameter $\lambda$), we have that

$$\Pr\left[X \geq t\right] = \exp{(-\lambda t)}.$$

Let $T \equiv \min(E(2), E(4))$ be the amount of time before you leave the state C. We have that for any $t$:

$$\begin{align*} \Pr[T \geq t] &= \Pr[E(2) \geq t \text{ and }E(4) \geq t]& \text{defn of }T\\ &=\Pr[E(2) \geq t] \cdot \Pr[E(4)\geq t]&\text{independence}\\ &= \exp{(-2t)} \cdot \exp{(-4t)}&\text{exponential distribution}\\ &= \exp{(-6t)}&\text{arithmetic}\\ \end{align*}$$

which shows, according to our first equation above, that $T$ is distributed as an exponential variable E(6) with parameter $\lambda = 6$.

Answer 2

What you are asking about is really not anything specifically to do with stochastic processes. You just want to prove that if $X$ is Exp(2) (where the parameter here is rate; sometimes the parameter of the exponential distribution is in fact the mean time), $Y$ is Exp(4), and the two are independent, then $Z:=\min \{ X,Y \}$ is Exp(6). One way to do this would be to prove that $P(Z>z)=e^{-6z}$ for $z \geq 0$ and $1$ otherwise. But $\{ Z>z \}=\{ X>z \} \cap \{ Y>z \}$, so this probability is easy to do using the independence and the known formulas for $P(X>z)$ and $P(Y>z)$.