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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.
 Fall '14

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