AMS 570 Spring 2021
Answers to Sample Exam I
1. Let be a -algebra and A1; A2; : : : 2 . Then A1C ; A2C ; : : : 2 , and thus
S1
i=1Ai
C 2 .
Hence
S1
i=1Ai
C
C
=
T1
i=1Ai 2 by DeMorgan's law.
2.
P
1[
n=1
An
B
!
= lim
k!1
P

k[
n=1
An
B
!
= lim
k!1
P
hSk
n=1An
iT
B

P (B)
= lim
k!1
P
Sk
n=1[An \B]

P (B)
= lim
k!1
Pk
n=1 P (An \B)
P (B)
=
1X
n=1
P (AnjB)
3. (a) Let A =fthe chip is redg and B =fthe chip came from bowl 2g. Then
P (A) = P (AjBc)P (Bc) + P (AjB)P (B) =

3
10

1
2

+

6
10

1
2

=
9
20
= 0:45:
(b)
P (BjA) = P (AjB)P (B)
P (A)
=
(6=10)(1=2)
9=20
=
2
3
(c) A and B are not independent because
P (B) =
1
2
6= P (BjA) = 2
3
:
4. (a)
EX =
1X
x=1
x(1 p)x1p = p
"

1X
x=1
d
dp
(1 p)x
#
= p d
dp
" 1X
x=1
(1 p)x
#
= p d
dp

1 p
p

= p

1
p2

=
1
p
(b) Let X be the number of children until the rst daughter is born. Then from part (a),
EX =
1
1=2
= 2:
5. (a)
M(t) = E

etX

=
1X
x=1
etx

1
2
x
=
1X
x=1

et
2
x
=
et=2
1 (et=2) =
et
2 et ; t < ln 2
because et=2 < 1 =) t < ln 2.
1
(b)
M 0(t) =
2et
(2 et)2 =) E(X) = M
0(0) = 2
(c)
M 00(t) =
2et(2 + et)
(2 et)3 =)M
00(0) = 6
Var(X) = M 00(0) [M 0(0)]2 = 6 4 = 2
6. (1) In tossing a fair coin n times, there are

n
k

ways to obtain k heads. (2) In tossing a
fair coin n times, there are

n
k

ways to obtain k tails. If we combine (1) and (2), there are
nX
k=0

n
k
2
ways for all k. (3) If we add the two trials in (1) and (2), we obtain n heads (or n
tails) in 2n tosses. This is

2n
n

.
7. Let X be the number of defective strawberries. Then
(a)
P (X 1) = 1 P (X = 0) = 1

48
5

50
5

= 1

48!
5!43!

50!
5!45!

= 1 45 44
50 49 =
47
245
(b) Find x such that
1

48
x

50
x

>
1
2
=)

48
x

50
x

=

48!
x!(48 x)!

50!
x!(50 x)!

=
(50 x)(49 x)
(50)(49)
<
1
2
=) x2 99x+ 1225 < 0 =) x > 14:5
He should examine at least 15 strawberries.
8. f(r) > f(r 1) and f(r) > f(r + 1)
f(x) =

n
x

px(1 p)nx; x = 0; ; n
f(r + 1)
f(r)
=

n
r+1

pr+1(1 p)nr1
n
r

pr(1 p)nr =
(n r)p
(r + 1)(1 p) < 1
=) r > (n+ 1)p 1 (1)
f(r 1)
f(r)
=

n
r1

pr1(1 p)nr+1
n
r

pr(1 p)nr =
(1 p)r
(n r + 1)p < 1
=) r < (n+ 1)p (2)
From (1) and (2), (n+ 1)p 1 < r < (n+ 1)p.
2
9. (a) Z 1
1
f(x)dx =
Z 4
2
(kx 1)dx =

k
2
x2 x
4
2
= 8k 4 (2k 2) = 6k 2 = 1) k = 1
2
(b) For 2 < x < 4,
F (x) =
Z x
1
f(x)dx =
Z x
2

t
2
1

dt =

t2
4
t
x
2
=

x2
4
x

(1 2) = 1
4
(x 2)2
Thus,
F (x) =
8<:
0; x < 2
1
4(x 2)2; 2 x 4
1; x > 4
(c)
P (2:5 < X < 3) = F (3) F (2:5) = 1
4
1
16
=
3
16
= 0:1875
(d)
E(X) =
Z 1
1
xf(x)dx =
Z 4
2

x2
2
x

dx =

x3
6
x
2
2
4
2
=
10
3
E(X2) =
Z 1
1
x2f(x)dx =
Z 4
2

x3
2
x2

dx =

x4
8
x
3
3
4
2
=
34
3
Var(X) = E(X2) [E(X)]2 = 34
3
100
9
=
2
9
(e)
MX(t) = Ee
tX =
Z 4
2
x
2
1

etxdx =
1
2
Z 4
2
xetxdx
Z 4
2
etxdx
=
1
2

x
t
etx
4
2
1
t
Z 4
2
etxdx


Z 4
2
etxdx
=
1
2t

4e4t 2e2t 1
2t
+ 1
Z 4
2
etxdx =
2
t
e4x 1
t
e2x

1
2t
+ 1

1
t
etx
4
2
=
2
t
e4t 1
t
e2x 1 + 2t
2t2

e4t e2t = 2t 1
2t2
e4t +
1
2t2
e2t
(f)
3
4
= F (x) =
1
4
(x 2)2 ) (x 2)2 = 3) x 2 =
p
3) x = 2 +
p
3 = 3:732
(g)
y =
p
x 2 =) x = y2 + 2 =) dx
dy
= 2y
fY (y) = 2yfX

y2 + 2

= 2y

y2 + 2
2
1

= y3; 0 < y <
p
2
3
10.
P

Z x+ ax

P (Z x) =
R1
x+ a
x
ez2=2dzR1
x e
z2=2dz
=
h
zez2=2
i1
x+ a
x
+
R1
x+ a
x
z2ez2=2dz
zez2=2
1
x
+
R1
x z
2ez2=2dz
=
x+ ax e 12x2+2a+ a2x2 + R1x+ a
x
z2ez2=2dz
xex22 + R1x z2ez2=2dz
) lim
x!1
P

Z x+ ax

P (Z x) = limx!1
1 + a
x2

e
1
2

x2+2a+ a
2
x2

+ 1x
R1
x+ a
x
z2ez2=2dz
ex22 + 1x
R1
x z
2ez2=2dz
=
e
x2
2
a
e
x2
2
= ea 8a > 0
11. For X Unif(0; 1), = 1=2 and 2 = 1=12, thus
P (jX j > k) = 1 P
X 12
kp12

= 1 P

1
2
kp
12
X 1
2
+
kp
12

=
(
1 kp
3
0 k < p3
0 k p3
For X Exponential(), = and 2 = 2, thus
P (jX j > k) = 1 P (jX j k) = 1 P ( k X + k)
=

1 + e(k+1) ek1 0 k < 1
e(k+1) k 1
Chebychev's Inequality gives the bound P (jX j > k) 1=k2. Chebychev's Inequality is
quite conservative compared to the exact probabilities.
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