UC Berkeley
Department of Statistics
STAT 210A: Introduction to Mathematical Statistics
Problem Set 1  Solutions
Fall 2009
Issued:
Tuesday, September 1, 2009
Due:
Tuesday, September 8, 2009
Problem 1.1
1.
Y
n
=
±
0
,
with probability 1

1
n
n,
with probability
1
n
.
For all
δ >
0, 0
≤
P
(

Y
n

0
 ≥
δ
)
≤
1
n
. Hence, for all
δ >
0, lim
n
→∞
P
(

Y
n

0
 ≥
δ
) = 0, which
implies that
Y
n
p
→
0. However,
E
(
Y
n
) = 1 for all
n
and thus lim
n
→∞
E
(
Y
n
) = 1
9
0.
2. Let
Y
n
=

√
n,
with probability
1
2
n
0
,
with probability 1

1
n
√
n,
with probability
1
2
n
.
Similarly as in part (a), for all
δ >
0, 0
≤
P
(

Y
n

0
 ≥
δ
)
≤
1
n
and thus lim
n
→∞
P
(

Y
n

0
 ≥
δ
) = 0,
so
Y
n
p
→
0. Also we have
μ
=
E
(
Y
n
) = 0. On the other hand,
E
(
Y
n

μ
)
2
= 2(
√
n
)
2 1
2
n
= 1 for all n,
so lim
n
→∞
E
(
Y
n

μ
)
2
= 1
9
0.
Problem 1.2
(a) It is enough to prove that lim
n
→∞
E

Y
n

μ

= 0. Notice that for any ﬁxed
± >
0:
0
≤
E

Y
n

μ

=
Z

Y
n

μ

1
(

Y
n

μ
 ≤
±
)d
P
(
Y
n
) +
Z

Y
n

μ

1
(

Y
n

μ

> ±
)d
P
(
Y
n
)
≤
±
·
1 +
s
Z

Y
n

μ

2
d
P
(
Y
n
)
Z
1
(

Y
n

μ

> ±
)d
P
(
Y
n
)
≤
±
+
p
(
M
+

μ

)
2
·
P
(

Y
n

μ

> ±
)
.
Let
n
→ ∞
, the second term above converges to 0, due to convergence in probability. It follows that for
any
± >
0,
0
≤
lim
n
→∞
E

Y
n

μ
 ≤
±
Hence, lim
n
→∞
E

Y
n

μ

= 0, which proves the conclusion.
(b) The same example in Problem 1.1 part(b) disproves the claim.
(c) For any
± >
0 and positive interger
k
, we have:
0
≤
E

Y
n

μ

k
=
Z

Y
n

μ

k
I
(

Y
n

μ
 ≤
±
)
dP
(
Y
n
) +
Z

Y
n

μ

k
I
(

Y
n

μ

> ±
)
dP
(
Y
n
)
≤
±
k
+ (

M

+

μ

)
k
·
P
(

Y