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Unformatted text preview: Statistics 265
Elements of Probability Theory
Fall Term 2000
Assignment 6
Due: Tuesday December 5, 2000 Solutions 6.2 Let Y be a random variable with a density function given by
(3 2
;1 y 1
f (y) = 2 y
0
otherwise.
a. Find the density function of U1 = 3Y:
b. Find the density function of U2 = 3 ; Y:
c. Find the density function of U3 = Y 2 :
Solution: The distribution function of Y is
F (y) = P (Y y) = Zy
;1 f (t) dt = Zy3
;
t2 dt = 1 1 + y3
2
2
;1 for ;1 y 1:
a. If U1 = 3Y then
FU1 (u) = P (3Y u) = P Y u = F u = 1 1 + u3
3
3
2
27 for ;3 u 3: Di erentiating, we have 82
<u
fU1 (u) = : 18
0 ;3 u 3
otherwise. b. If U2 = 3 ; Y then
FU2 (u) = P (3 ; Y u) = P (Y 1
3 ; u) = 1 ; P (Y < 3 ; u) = 1 ; F (3 ; u) = 2 1 ; (3 ; u)3 for ;1 3 ; u 1 that is, 2 u 4: Di erentiating, we have
(3
2
2u4
fU2 (u) = 2 (3 ; u)
0
otherwise.
c. If U3 = Y 2 then p p FU3 (u) = P (Y 2 u) = P (; u Y
u) = P (Y
p
p
1u3 + 1u2 = u3
2
= F ( u) ; F (; u) = 2 2 2 3 for 0 u 1: Di erentiating, we have fU3 (u) = u
0 31
2
2 0u1
otherwise. pu) ; P (Y < ;pu) 6.8 The total time from arrival to completion of service at a fast{food outlet, Y1 and the time spent waiting
in line before arriving at the service window, Y2 were given in Exercise 5.9 with joint density function
e;y1
0 f (y1 y2) = 0 y2 y1 < 1
otherwise. Another random variable of interest is U = Y1 ; Y2 the time spent at the service window.
a. Find the probability density function for U:
b. Find E (U ) and V (U ): Compare your answers with the results of Exercise 5.68.
Solution: a. The region for which y1 ; y2 u is shown in the gure, over the region for which y1 and y2 are positive
and y2 y1 :
y =y y 1 2 2 y =u+y
1 u 2 y 1 a. The distribution function of U = Y1 ; Y2 is given by
FU (u) = P (Y1 ; Y2 u) = Z 1 Z y2 +u
0 y2 e;y1 dy1dy2 = Z 1;
0 e;y2 ; e;(y2 +u) dy2 = 1 ; e;u for 0 u < 1: Di erentiating, we have
fU (u) = e;u
0 0 u<1
otherwise: b. The random variable U = Y1 ; Y2 has a gamma distribution with parameters
Therefore, E (U ) = = 1 and V (U ) = 2 = 1:
6.12 In Exercise 4.7, we determined that 8b
<
f (y) = : y2
0 = 1 and = 1: yb
otherwise is a bona de probability density function for a random variable, Y: Assuming b is a known constant and
U has a uniform distribution on the interval 0 1] transform U to obtain a random variable with the same
distribution as Y:
Solution: The distribution function of Y is given by
FY (y) = Zy ;1 f (t) dt = for y b:
2 Zyb
b
2 dt = 1 ; y
t
b Since U has a uniform distribution on 0 1] then
P (U u) = u b
for 0 u 1 and we want to nd a function G such that Y = G(U ) with FY (y) = 1 ; y for y b: Now, FY (y) = P (Y y) = P (G(U ) y) = P (U G;1(y)) = P (U u) = u
for 0 u 1 since U has a uniform distribution on 0 1]: Therefore, we want
b
u= 1; y or b
y = 1;u b
that is, Y = G(U ) = 1 ; U :
6.16 Let the random variable Y possess a uniform distribution on the interval 0 1]:
a. Derive the distribution of the random variable W = Y 2 :
p
b. Derive the distribution of the random variable W = Y :
Solution: a. If W = Y 2 then
FW (w) = P (W w) = P (Y pw ) = Z
ww p
w ) = P (; 2 pw 0 for 0 w 1: Di erentiating, the density function for W = Y 2 is
1
w; 2
0 1
2 fW (w) = 0<w<1
otherwise. p b. If W = Y then
FW (w) = P (W p w) = P ( Y w ) = P (Y w2 ) = p for 0 w 1: Di erentiating, the density function for W = Y is
fW (w) = 2w
0 3 0<w<1
otherwise. Z w2
0 p 1 dy = w 1 dy = w2 6.22 The Weibull density function is given by ( 1 m;1 ; ym
my e
f (y ) =
0 0<y<1
otherwise where and m are positive constants. This density function is often used as a model for the lengths of
life of physical systems. Suppose Y has the Weibull distribution just given.
a. Find the density function of U = Y m :
b. Find E (yk ) for any positive integer k:
Solution: 1
a. Let U = Y m then using the transformation approach, we have Y = U m and ;
dy = 1 u (1;m) = 1 u; (mm 1)
m
du m
m so that ;
1 m;1 ; u
1 u; (mm 1) = 1 e; u
gU (u) = 1 m u m
e
m for u > 0:
b. If Y has the Weibull distribution given above, and k is a positive integer, then Z1 k u
k
k
E (Y k ) = E U m = 1
u m e; du = 1 ; m + 1
0 k
m +1 k
=; m +1 k
m: Note that the integrand is the density (except for constants) of a gamma variable with parameters
k
m + 1 and so that the integration can be done by choosing the necessary constants.
6.24 Let Y have a uniform 0 1] distribution. Show that U = ;2 ln Y has an exponential distribution with
mean 2:
Solution: The density function for Y is given by f (y ) = 1 for 0
y 1 and since u = ;2 ln y then
2
y = e; u : Now
dy
1 ;u
2
du = ; 2 e
so that
dy
1
2
2
fU (u) = f (y) du = 1 ; 2 e; u = 1 e; u
2
for u > 0 which is the density function of an exponential distribution with = 2: 4 6.25 The speed of a molecule in a gas at equilibrium is a random variable V whose density function is given by
f (v) = av2 e;bv 2 v>0 where b = 2m where k T and m denote Boltzmann's constant, the absolute temperature, and the mass
kT
of the molecule, respectively.
a. Derive the distribution of W = 1 mV 2 the kinetic energy of the molecule.
2
b. Find E (W ):
Solution: r 2
W
a. If W = mV then V = 2m and
2 r dv = 1 2 w; 1 = p 1
2
dw 2 m
2mw
Therefore, ; p p a 2w
w
w
bw
2
fW (w) = p m e; 2m = a 2w e; kT = a 32 w 1 e; kT
3
2mw
m2
m2
and since the above density must integrate to 1 and since the variable part of the density is that of a
gamma variable with = 3 and = kT then a must be chosen so that
2 p a 2= ; 1
3
2
m 3 ; 3 (kT ) 2
2
and the density is 1
w
fW (w) = ; 3 1 3 w 2 e; kT
; 2 (kT ) 2 for w 0:
b. For a gamma{type random variable, we have E (W ) = 5 = 3 kT:
2 6.32 Suppose that Y1 and Y2 are independent, standard normal random variables. Find the density function
of U = Y12 + Y22 :
Solution: Since Y1 and Y2 are independent standard normal random variables, the moment{generating
functions for Y12 and Y22 can be written as
mY12 (t) =
and therefore, 1
1
(1 ; 2t) 2 and mY22 (t) = 1
2
(1 ; 2t) 1 1
mU (t) = mY12 (t) mY22 (t) = 1 ; 2t which is the moment generating function for a gamma random variable with = 1 and = 2: By
the uniqueness theorem, U has a gamma distribution with = 1 and = 2: Equivalently, U has a 2
distribution with 2 degrees of freedom.
6.45 Show that, if Y1 has a 2 distribution with 1 degrees of freedom and Y2 has a 2 distribution with 2
degrees of freedom, then U1 + U2 has a 2 distribution with 1 + 2 degrees of freedom, provided that Y1
and Y2 are independent.
Solution: From Example 4.13 with
= 2i and = 2 the generating functions are
mYi (t) = (1 ; 2t) 2i
for i = 1 2: Since Y1 and Y2 are independent, then
mU (t) = mY1 (t) mY2 (t) = (1 ; 2t);( 1 + 2 ) 2
which is the moment generating function for a 2 random variable with 1 + 2 degrees of freedom. The
result now follows by the uniqueness theorem for moment{generating functions.
6.46 Let Y1 and Y2 be independent normal random variables, each with mean 0 and variance 2 : De ne
U1 = Y1 + Y2 and U2 = Y1 ; Y2 : Show that U1 and U2 are independent normal random variables, each
with mean 0 and variance 2 2: Hint: If (U1 U2) has a joint moment{generating function m(t1 t2) then
U1 and U2 are independent if and only if m(t1 t2) = mU1 (t) mU2 (t):
Solution: We have i
h
ih
E et1 (Y1+Y2 )+t2 (Y1;Y2 ) = E eY1 (t1+t2 )+Y2 (t1;t2 )
h i = E e(t1 +t2 )Y1 e(t1 ;t2)Y2
= mY1 (t1 + t2) mY2 (t1 ; t2 )
2
2
2
2
= e 2 (t1 +t2 ) e 2 (t1 ;t2 )
1
2
= e 2 t2 e 2 t2
= mU1 (t1 ) mU2 (t2)
and since the joint moment{generating function is the product of the marginal moment{generating functions, then the random variables U1 and U2 are independent. 6 ...
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This note was uploaded on 10/19/2011 for the course MATH Statistics taught by Professor Issac during the Fall '00 term at University of Alberta.
 Fall '00
 Issac
 Statistics, Probability

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