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Math Help - 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 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}.
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  2. #2
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    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
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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 \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}

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

    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
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