Download Math 151 Homework 7 Solutions (Winter 2015)

Math 151 Homework 7 Solutions (Winter 2015)
Problem 6.16
Sn
(a) A = i=1 Ai
(b) Ai and Aj are disjoint because in Ai all points, in particular Pj , lie on the same semicircle
beginning from Pi and going clockwise. So assuming that Ai golds, Aj doesn’t hold. Thus
events Ai are mutually exclusive.
(b) Let Si denote the semicircle starting from Pi and going clockwise. Then for any 1 ≤ i ≤ n
P (Ai ) = P (Aj ∈ Si for all j 6= i) =
Y
P (Aj ∈ Si ) =
j6=i
So
P (A) = P
n
[
!
Ai
=
n
X
P (Ai ) =
i=1
i=1
Y1
j6=i
n
X
2
= 2−(n−1)
2−(n−1) = n2−(n−1)
i=1
Problem 6.18
4
Observe that f (x, y) = 2 if 0 < x < L/2, L/2 < y < L and f (x, y) = 0 otherwise is the joint
L
density function of X and Y . So
ZZ
P (Y − X > L/3) =
f (x, y)dydx
y−x>L/3
Z L/2 Z L
4
4
=
dydx +
dydx
2
2
0
L/2 L
L/6
L/3+x L
Z L/6
Z L/2 2
4
2L
=
dx + 2
− x dx
L
L L/6
3
0
L/2
1
4
2Lx x2 = + 2
−
3 L
3
2 L/6
Z
L/6
Z
L
1 4
+
3 9
7
=
9
=
1
Problem 6.20
Observe that if
xe−(x+y)
0
f (x, y) =
: x > 0, y > 0
: otherwise
then f (x, y) = g(x)h(y) where
xe−x
0
:x>0
: otherwise
e−y
0
:y>0
: otherwise
g(x) =
and
g(x) =
So X and Y are independent. Now if
2 : 0 < x < y, 0 < y < 1
0 : otherwise
f (x, y) =
then f (x, y) cannot be expressed as a product of g(x) and h(y). So in this case X and Y are not
independent.
Problem 6.22
(a) Since f (x, y) cannot be expressed as g(x)h(y) we deduce that X and Y are not independent.
(b) If 0 < x < 1 then
∞
Z
fX (x) =
1
Z
f (x, y)dy =
−∞
(x + y)dy = x +
0
So
(
x+
fX (x) =
0
1
2
1
2
:0<x<1
: otherwise
(c)
1
Z
Z
P (X + Y < 1) =
1−x
Z
1
(x + y)dydx =
0
0
1
Z
=
0
0
2
1−x
dx =
2
1−x
y 2 dx
xy +
2 0
1
x x 1
−
=
2
6 0
3
3
Problem 6.24
(a) Event {N = n} contains al outcomes where the first n − 1 trials are 0 and n’th trial isn’t. So
P (N = n) = pn−1
(1 − p0 )
0
2
(b) Even {X = j} is same as event that a trial results in j given that it is not 0. So
pj
P (X = j) =
1 − p0
(c) Event {N = n, X = j} contains all outcomes where the first n trials result in 0 and n + 1’th
trial results in j. So
pj
= P (N = n)P (X = j)
P (N = n, X = j) = pn0 pj = pn0 (1 − p0 )
1 − p0
Problem 6.25
For 1 ≤ i ≤ 106 let Xi = 1 if i’th person arrive in the first hour and let Xi = 0 otherwise. So Xi is
a Bernoulli random variable with success probability p = 1/106 . Then N = X1 + X2 + · · · + X106
is binomial random variable with parameters n = 106 and p = 1/106 . By approximating binomial
random variable with Poisson random variable we see that that N is approximately Poisson with
parameter λ = np = 1. Thus
e−1
1i
P (N = i) ≈ e−1 =
i!
i!
Problem 6.28
Let X and Y be the times that it takes to service for A.J.’s and M.J.’s car, respectively. Since X
and Y are independent exponential random variables with rate 1 we have fX,Y (x, y) = e−(x+y) .
So
(a) Observe that the event that M.J.’s car will finish before A.J.’s car is {X > Y + t}. So
Z ∞Z ∞
Z ∞
1
P (X > Y + t) =
e−(x+y) dxdy =
e−(t+2y) dy = e−t
2
0
t+y
0
(b) The event that M.J.’s car will finish before time 2 is {X + Y < 2}. So
Z 2 Z 2−y
Z 2
P (X + Y < 2) =
e−(x+y) dxdy =
e−y − e−2 dy = 1 − 3e−2
0
0
0
Problem 6.31
Let Xm and Xw denote the number of men and women who never eat breakfast. Then E[Xm ] =
200(0.252) = 50.4 and E[Xm ] = 200(0.236) = 47.2. Also Var(Xm ) = 200(0.252)(0.748) = 37.7 and
Var(Xw ) = 200(0.236)(0.764) = 36.1.
(a) Since X = Xm + Xw ∼ N (97.6, 73.8)
109.5 − 97.6
X − 97.6
√
√
≥
≈ P (Z ≥ 1.385) = 1 − P (Z ≤ 1.385) ≈ 0.0831
P (X ≥ 110) = P
73.8
73.8
3
(b) Since Y = Xm − Xw ∼ N (3.2, 73.8)
Y − 3.2
0.5 − 3.2
√
√
P (Y ≤ 0) = P
≤
≈ P (Z ≤ −0.314) ≈ 0.3764
73.8
73.8
Problem 6.T.8
(a) Since X and Y are independent
FW (s) = P (W ≤ s) = P (min(X, Y ) ≤ s)
= P ({X ≤ s} ∪ {Y ≤ s})
= P (X ≤ s) + P (Y ≤ s) − P (X ≤ s, Y ≤ s)
= P (X ≤ s) + P (Y ≤ s) − P (X ≤ s)P (Y ≤ s)
= FX (s) + FY (s) − FX (s)FY (s)
(b) Differentiate above equation to get
fW (s) = fX (s) + fY (s) − fX (s)FY (s) − FX (s)fY (s) = fX (s)(1 − FY (s)) + fY (s)(1 − FX (s))
Also observe that
1 − FW (s) = 1 − FX (s) − FY (s) + FX (s)FY (s) = (1 − FX (s))(1 − FY (s))
So
fW (s)
fX (s)(1 − FY (s)) + fY (s)(1 − FX (s))
=
1 − FW (s)
(1 − FX (s))(1 − FY (s))
fX (s)
fY (s)
=
+
1 − FX (s) 1 − FY (s)
= λX (s) + λY (s)
λW (s) =
Problem 6.T.11
Z ∞ Z ∞Z ∞Z ∞Z ∞
I=
f (x5 )f (x4 )f (x3 )f (x2 )f (x1 )dx5 dx4 dx3 dx2 dx1
−∞ x1
x2
x3
x4
Z ∞ Z ∞Z ∞Z ∞
∞
=
F (x5 ) f (x4 )f (x3 )f (x2 )f (x1 )dx4 dx3 dx2 dx1
x4
−∞ x1
x2
x3
Z ∞ Z ∞Z ∞Z ∞
=
(1 − F (x4 ))f (x4 )f (x3 )f (x2 )f (x1 )dx4 dx3 dx2 dx1
−∞
∞
Z
−∞
∞
Z
Z
x1
∞
Z
x2
∞
x1
∞
Z
=
Z
x2
∞
=
−∞
x1
x2
x
∞
3
1
f (x3 )f (x2 )f (x1 )dx3 dx2 dx1
− (1 − F (x4 ))2
2
x3
1
(1 − F (x3 ))2 f (x3 )f (x2 )f (x1 )dx3 dx2 dx1
2
4
∞
Z
−∞
∞
Z
Z
∞
=
Z
x1
∞
∞
1
3
− (1 − F (x3 ))
f (x2 )f (x1 )dx2 dx1
6
x2
1
(1 − F (x2 ))3 f (x2 )f (x1 )dx2 dx1
6
−∞ x1
∞
Z ∞
1
4
− (1 − F (x2 ))
=
f (x1 )dx1
24
−∞
x1
Z ∞
1
(1 − F (x1 ))4 f (x1 )dx1
=
24
−∞
∞
1
5
(1 − F (x1 ))
= −
120
x1
1
=
120
=
So I does not depend on F . Since Xi are independent and identically distributed by symmetry
P (Xσ(1) < Xσ(2) < Xσ(3) < Xσ(4) < Xσ(5) ) is constant for any permutation σ of {1, 2, 3, 4, 5}.
Since there are 5! = 120 possible permutations we can intuitively deduce that this probability must
be 1/120.
5