2 lec 17 6 points assume and are independent

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Unformatted text preview: sume and are independent exponential random variables with the same mean value one. (a) Find the pdf of . (4 points) | | . (2 points) (b) Find the value of Solution: , (a) Note that First method: For 0, which means X and , X,Y Y, the cdf of . is given by , X,Y 1 dd 1 2 Therefore, we have d dz 1 2 For 0, which means X Y, the cdf of 2 is given by , X,Y dd 1 2 Therefore, we have d dz 1 2 2 Second method: For 0, using the conditioning method in Lec. 17, we have | | d d e 1 e 2 2 d 0, using the conditioning method in Lec. 17, we have For | | d d e d 1 e 2 2 (b) From the results in (a), we have || d 1 e 2 1 2 d 1 2 d ed 3. (Lec. 20, 5 points) The congregation of HKUST is scheduled in November each year. Assume that each student has the same probability of attending the congregation. HKUST wants to have a rough idea of how many students will attend the ceremony. Use the Chebyshev inequality to determine the probability of a student to attend the ceremony is within 0.04 of the actual probability with probability 0.95. Solution...
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This document was uploaded on 03/19/2014.

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