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Statistics 116 - Fall 2004
Theory of Probability
Assignment # 6
Due Friday, November 05
Solution prepared by Xiaohu Zhang, [email protected]
Nov 5, 2004
Q. 1)
Q. 2)
(Ross #5.19)
Answer
Let X be a normal random variable with mean 12 and
variance 4. Find the value of c such that P (X > c) = 0:10.
: Letting Z = (x 12)=2 then Z is a standard normal. Now,
:10 = P fZ > (c 12)=2g. From table 5.1, we nd P fZ < 1:28g = :90 and
so
(c 12)=2 = 1:28 or c = 14:56
(Ross # 5.31)
(a) A re station is to be located along a road of length A, A < 1. If
res will occur at points uniformly chosen on (0; A), where should
the station be located so as to minimize the expected distance from
the re? That is, choose a so as to minimize
E (jX
aj):
(b) Now suppose that the road is of innite length { stretching from point
0 outward to 1. If the distance of a re from point 0 is exponentially
distributed with rate , where should the re now be located? That
is, we want to minimize E (jX aj) where X is now exponential with
rate :
Answer (a):
1
Z A
1
aj dx
A
0
Z a
Z A
1
(a x) dx +
(x
=
A
0
a
1 a2 A2
a2
=
( +
aA (
A 2
2
2
a2
A
=
a+
A
2
E jX
aj =
jx
1
a) dx
A
a2 ))
Let the derivative be zero, we get 2Aa 1 = 0, thus a = A2 . To check
this is the minimizer, one can check the second derivative which is always
positive, so it is indeed minimizer.
:
Answer (b)
E jX
Z A
aj =
jx
aje x dx
(a
x)e x dx +
0
Z a
=
0
= a
Z a
0
e
x dx
Z a
1 a 1
e
+ )
2 a 1
= a+ e
0
Z A
a
xe
(
= a(
(x
a)e x dx
x dx + a a
e
1
Z
a
xe
x dx
a
1
Z
a
1 a 1
a
1
e
+ 2 ) + ( e a + 2 e a )
2
Let the derivative be zero, we get 1 2e a = 0, thus a = (log 2)=. To
check this is the minimizer, one can check the second derivative which is
always positive, so it is indeed minimizer.
Q. 3)
(Ross #5.35)
The lung cancer hazard rate of a t-year-old male smoker,
(t) is such that
(t) = 0:027 + 0:00025 (t
40)2 ;
t 40:
Assuming that a 40-year-old male smoker survives all other hazards, what
is the probability that he survives to
(a) age 50 without contracting lung cancer;
(b) age 60 without contracting lung cancer.
Answer (a): This problem is very similar to Example 5f of the textbook
(pp216-217). The general result is as follows.
2
e x dx
1
a( e a )
Suppose the hazard rate function is (t), then probability for an A-year
person to survive until age B (B > A) is
P fA-year-old person reaches ageB g
= P fhis life time > B jhis life time > Ag
= P fhis life time > B g=P fhis life time > Ag
= (1 F (B ))=(1 F (A))
Z B
= expf
0
Z B
= expf
A
(t)dtg= expf
Z A
0
(t)dtg
(t)dtg
Applying to this problem, the probability
h R for him ito survive to age 50
50
:353 = 0:7026
without contracting lung cancer is exp
40 (t)dt = e
Answer (b): Similarly, the probability
h R for him ito survive to age 60 with60
out contracting lung cancer is exp
Q. 4)
(t)dt = e 1:207 = 0:2991
(Ross #5.5 { Theoretical) Use the fact that for any non-negative ran-
dom variable Y ,
E (Y ) =
to prove that
E (X n ) =
Answer:
E [X n ] =
=
=
Q. 5)
40
Z
Z0 1
Z0 1
0
0
1
Z
1
1
Z
0
P (Y > t) dt
ntn 1 P (X > t) dt:
P fX n > sgds
P fX n > tn gdtn by s = tn
P fX > tgntn 1 dt
(Ross # 5.28 Theoretical) Let X be a continuous random variable
having cumulative distribution function F . Dene a new random variable
Y by Y = F (X ). Show that Y is uniformly distributed over (0,1).
: Let FY () be the cumulative distribution function of Y . It
suces to show FY (y ) = y for y 2 (0; 1) since this is the cumulative distribution function for uniform random variable over (0,1). In the following
we use FX () to denote the cumulative distribution function of X .
Answer
3
P fY y g
P fFX (X ) y g
P fX FX 1 (y )g
FX (FX 1 (y ))
y
FY (y ) =
=
=
=
=
4