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Thread: A complicated function of two random variables

  1. #1
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    May 2011
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    The CDF function of min {I/x, P} . y

    Hello everyone,

    I have faced a difficult function of two random variables. The problem is described as follows.

    Given two independent exponential random variables $\displaystyle x$ and $\displaystyle y$, $\displaystyle x \ge 0$ and $\displaystyle y \ge 0$, with the corresponding PDF functions as $\displaystyle {f_x}(x)$ and $\displaystyle {f_y}(y)$. The random variable $\displaystyle u = \min \{ \frac{I}{x},P\} \times y$, where $\displaystyle I$ and $\displaystyle P$ are constants, has the CDF function written as:

    $\displaystyle \Pr \{ u \le U\} = \int\limits_{x = I/P}^\infty \int\limits_{y = 0}^{x.U/I} {{f_x}(x){f_y}(y)dydx + } \int\limits_{x = 0}^{I/P} {} \int\limits_{y = 0}^{U/P} {{f_x}(x){f_y}(y)dydx} $

    I cannot understand how the CDF of $\displaystyle u$ is obtained as above. Can everyone give me some tips for solving the problem?. Thanks very much.
    Last edited by lptuyen; May 31st 2011 at 04:09 AM. Reason: SOLVED
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