Applied probability model homework 12
1.
A viral linear DNA molecule of length, say, 1 is often known to contain a certain
“marked position,” with the exact location of this mark being unknown. One
approach to locating the marked position is to cut the molecule by agents that break
it at points chosen according to a Poisson process with rate  . It is then possible to
determine the fragment that contains the marked position. For instance, letting m
denote the location on the line of the marked position, then if L1 denotes the last
Poisson event time before m (or 0 if there are no Poisson events in [0, m]), and R1
denotes the first Poisson event time after m (or 1 if there are no Poisson events in [m,
1]), then it would be learned that the marked position lies between L1 and R1 .
Find
(a) P( L1  0),
(b) P( L1  x), 0  x  m,
(c) P( R1  1),
(d) P( R1  x), m  x  1.
By repeating the preceding process on identical copies of the DNA molecule, we are
able to zero in on the location of the marked position. If the cutting procedure is
utilized on n identical copies of the molecule, yielding the data Li , Ri , i  1,, n,
then it follows that the marked position lies between L and R, where
L  max Li ,
i
R  min Ri
i
(e) Find E[R-L ] , and in doing so, show that E[ R-L ]~
2
.
nλ
2.
Let S n denote the time of the nth event of the Poisson process {N(t), t  0} having
rate  . Show, for an arbitrary function g, that the random variable
the same distribution as the compound Poisson random variable

N (t )
i 1

N (t )
i 1
g ( S i ) has
g (U i ) ,
where U1 ,U 2 , is a sequence of independent and identically distributed uniform (0,
t) random variables that is independent of N, a Poisson random variable with mean
t. Consequently, conclude that
t
t
 N (t )

 N (t )

E   g ( S i )    g ( x)dx Var   g ( S i )     g 2 ( x)dx
0
0
 i 1

 i 1

3.
Some components of a two-component system fail after receiving a shock. Shocks of
three types arrive independently and in accordance with Poisson processes. Shocks
of the first type arrive at a Poisson rate 1 and cause the first component to fail.
Those of the second type arrive at a Poisson rate  2 and cause the second
component to fail. Let X 1 and X 2 denote the survival times for the two
components. Show that the joint distribution of X 1 and X 2 is given by
P{ X 1  s, X 2  t}  exp{1 s  2 t  3 max( s, t )}
This distribution is known as the bivariate exponential distribution.