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Thread: Interesting theorem for exponential dist.

  1. #1
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    Interesting theorem for exponential dist.

    I have a problem proving the following theorem...

    Let $\displaystyle T_1, T_2, ...$ be independent random variables of exponential distributions with parameters $\displaystyle q_i > 0$. Let $\displaystyle T(\omega) = \inf_{j \geq 1} T_j$. Provided $\displaystyle \sum_{i=1}^\infty q_i =q < \infty$ prove that for every $\displaystyle \omega \in \Omega$ (except for zero-probability set) exists exactly one index $\displaystyle j = J(\omega)$ such that $\displaystyle T(\omega) = T_j(\omega)$. What is more $\displaystyle T$ and $\displaystyle J$ are independent, and $\displaystyle P(J = j) = \frac{q_j}{q}$.
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  2. #2
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    I'm pretty sure, that "exponential with parameter $\displaystyle q_j$ " means that cumulative distribution function is
    $\displaystyle F_j(t) = 1 - e^{q_j t}$ (with support $\displaystyle [0, \infty)$)

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

    Because series is convergent, we know that $\displaystyle q_j \rightarrow 0$
    Hence $\displaystyle ET_j \rightarrow \infty$
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  3. #3
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    Solution!

    To solve the problem we shall use the following theorem

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

    ------------

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

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

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

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