# Math Help - Interesting theorem for exponential dist.

1. ## Interesting theorem for exponential dist.

I have a problem proving the following theorem...

Let $T_1, T_2, ...$ be independent random variables of exponential distributions with parameters $q_i > 0$. Let $T(\omega) = \inf_{j \geq 1} T_j$. Provided $\sum_{i=1}^\infty q_i =q < \infty$ prove that for every $\omega \in \Omega$ (except for zero-probability set) exists exactly one index $j = J(\omega)$ such that $T(\omega) = T_j(\omega)$. What is more $T$ and $J$ are independent, and $P(J = j) = \frac{q_j}{q}$.

2. I'm pretty sure, that "exponential with parameter $q_j$ " means that cumulative distribution function is
$F_j(t) = 1 - e^{q_j t}$ (with support $[0, \infty)$)

Hence probability density function is $f_j(t) = q_j e^{q_j t}$
And the expected value is $ET_j = \frac{1}{q_j}$

Because series is convergent, we know that $q_j \rightarrow 0$
Hence $ET_j \rightarrow \infty$

3. ## Solution!

To solve the problem we shall use the following theorem

Let S, T be random variables of exponential distributions with parameters $\lambda_S$ and $\lambda_T$. The random variable $\min\{S(\omega), T(\omega)\}$ has exponential distribution with parameter $\lambda = \lambda_S + \lambda_T$. Probability $P(S \leq T) = \frac{\lambda_S}{\lambda}$. What is more, events $\{S \leq T\}$, $\bigl\{\min\{S(\omega), T(\omega)\} \geq t\bigr\}$ are indpendent for all $t \in \mathbb{R}$

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The theorem implies that for random variables $T_1, T_2,...,T_n$ random variable $\min\{T_1(\omega), T_2(\omega),...,T_n(\omega)\}$ has exponential distribution with parameter $\sum_{k = 1}^{n} q_k$. For $n \rightarrow \infty$ this distribution converges to exponential distribution with parameter $q = \sum_{k = 1}^{\infty} q_k$, which is also distribution of T (because $\min\{T_1(\omega), T_2(\omega),...,T_n(\omega)\}$ converges to T almost surely).

Let $J(\omega) = j$ where $j \in \mathbb{N}$ is the number for which $T_j(\omega) = T(\omega)$. If such (unique!) j does not exist we put $J(\omega) = \infty$

First of all we will calculate $P(J = j)$. We can easily see that $\{J = j\} = \{T_j < \inf_{i \neq j}\{T_i\} \}$.
$\inf_{i \neq j}\{T_i\}$ has exponential distribution with parameter $q - q_j$. Thus $P(J = j) = \frac{q_j}{q - q_j + q_j} = \frac{q_j}{q}$ by the theorem. What is more, we can see that $T$ and $J$ are independent (we use the theorem putting $S \leftarrow T_j$, $T \leftarrow \inf_{i \neq j}\{T_i\}$).

Now, it is easy to prove that J is finite for almost all $\omega$. Indeed, we see that:
$P(J = \infty) = 1 - P(J < \infty) = 1 - \sum_{j=1}^{\infty}P(J = j) = 1 - \sum_{j=1}^{\infty}\frac{q_j}{q} = 0$