Problem Set 1 - Solutions for Questions 2,3,4
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#2 (BCN 6.6)

   X = X1 + X2 is from an Erlang with k*theta=1.
   Since k=2, theta=0.5.

   The cdf is:
                 k-1   e^(-ktx) (ktx)^i
      F(X) = 1 - sum ( ---------------- )
                 i=0         i!
   (theta is shown as "t" in the formula above)


               1    e^(-2) (2)^i
   F(2) = 1 - sum ( ------------ )
              i=0        i!
        = 1 - ( e^(-2) + 2 e^(-2) )
        = 1 - 3 e^(-2)
        = .5960

   P(X1+X2>2) = 1 - F(2)
              = 1 - .5960
              = .4060

#3 (BCN 6.20)

   Let X be defined as the lifetime of the battery.  Then, X is
   exponential ( lamda = 1/48 ) with cumulative distribution function
      F(x) = 1 - e^(-x/48), x>0

   (a) The probability that the battery will fail within the next 12
       months, given that it has operated for 60 months is:

       P(X<=72 | X>60) = P(X<=12)
                       = F(12)
                       = 1 - e^(-12/48)
                       = .2212  due to the memoryless property.

   (b) Let Y be defined as the year in which the battery fails, then,

       P (Y=odd year) = P(die in 12 months) + P (die in 24-36 months) + ...
                      = (1-e^(-12/48)) + (1-e^(-36/48)) - (1-e^(-24/48)) + ... 
                      = (1-e^(-.25)) + (e^(-.50)-e^(-.75)) + ...

       P (Y=evn year) = P(die in 12-24 months) + P (die in 36-48 months) + ...
                      = (1-e^(-24/48)) - (1-e^(-12/48)) + 
                           (1-e^(-48/48)) - (1-e^(-36/48)) + ... 
                      = (e^(-.25)-e^(-.50)) + (e^(-.75)-e^(-1)) + ...

       Now multiply P(Y=odd year) by e^(-.25)
          P(Y=odd year)*e^(-.25) = (e^(-.25)-e^(-.50)) + (e^(-.75)-e^(-1)) + ...
                                 = P(Y=even year)

       We know that a battery fails in either a odd or even year, so
          P(Y=odd year) + P(Y=even year) = 1 

       And then substitute P(Y=even year) = P(Y=odd year)*e^(-.25) to get
          P(Y=odd year) + P(Y=odd year)*e^(-.25) = 1 
                                1
          P(Y=odd year) = ------------
                          e^(-.25) + 1 
                        = .5622

   (c) Due to the memoryless property of the exponential distribution,
       the remaining expected lifetime is 48 months.
       
#4 (BCN 6.36)

   (a) Normal (10,4)

          F(8)-F(6) = F((8-10)/2) - F((6-12)/2)
                    = F(-1) - F(-2)
                    = (1-.84134) - (1-.97725)
                    = .13591

   (b) Triangular (4,10,16)

                        (8-4)^2        (6-4)^2
          F(8)-F(6) = ------------ - ------------
                      (10-4)(16-4)   (10-4)(16-4)
                    = 1/6
                    = .1667

   (c) Uniform (4,16)

                       (8-4)   (6-4)  
          F(8)-F(6) = ------ - ------
                      (16-4)   (16-4)
                    = 1/6
                    = .1667