Homework # 2
20.3
n
log
1/n
1
=
Y j
n
j =1
n
X j
j =1
−→ −→ EX
a.s.
1
Hence, n
1/n
lim
Y j
n→∞
= e EX
j =1
20.7. X 1 + X 12
· · · + X n = X + · · · + X n + · · · + X n n 1
2
By the law of the large numbers, X 1 +
lim
X 12 +
X 12 +
· · · + X n = 1
2
· · · + X n n
2 1
Thus
= V ar (X 1 ) + EX 1
X 1 +
lim
n→∞
X 12
· · · + X n = 1 + · · · + X n 4 2
2
= 4 a.s.
a.s.
21.2. We have EX 12 = 2
EX 1 = 0,
Notice that
n
√ √ X j
n
j =1 n
n
1
=
n
X j2
X j
j =1
n
1
n
2
X j
j =1
j =1
By the law of the large numbers, 1
lim
n→∞
n
n
X j2 = 2 a.s.
j =1
By central limit theorem, 1
n
√ n
j =1
X j
−d→ U ∼ ∼ N (0, 2) 1
1
−
a.s.
2
· · · + X n = E X n
n→∞
n
n→∞
lim
a.s.
1
1
−
Therefore,
n
−→ √ X j
j =1 n
n
d
X j2
1 1 U ∼ N (0, ) 2 2
j =1
21.6. Let X λ and Z λ be independent Poisson random variable with means [λ] and d λ − [λ], respectively. Then Y λ = X λ + Z λ . It is sufficient to show that X λ + Z λ
d √ − λ −→ N (0, 1)
λ
By Exercise 21.5, X λ
[λ] [λ]
−
Therefore,
√ − [λ] =
X λ
λ
d −→ N (0, 1)
[λ] X λ − [λ] λ [λ]
By the decomposition,
d −→ N (0, 1)
X λ + Z λ
(λ − [λ]) √ − λ = X λ√ − [λ] + Z λ − √ λ
λ
all we need is to show that
Z λ
λ
− √ (λ − [λ]) P −→ 0 λ
In deed, by Chebyshev inequality
P
Z λ
− √ (λ − [λ]) ≥ ǫ ≤ ǫ 1λ V ar (Z λ) = λ ǫ−λ[λ] −→ 0
λ
2
2
(λ → ∞ )
21.7. Let X k be the i.i.d. random variables with common distribution Poisson (1). We have EX 1 = V ar(X 1 ) = 1. By CLT, n
√ 1
n
X j
j =1
− −→ n
d
N (0, 1)
In particular, n
P
j =1
X j
n
√ ≤ n
= P
1
n
j =1
X j
− ≤ −→ √ n
0
2
1 2π
0
−∞
x2 /2
−
e
dx =
1 2
(n → ∞)
On the other hand,
n
∼ ≤
P oisson(n)
X j
j =1
So
n
P
n
X j
n
=
n
j =1
P
X j = k
j =1
k=1
Hence,
n
lim e
n
n
−
n→∞
=
k nn
−
e
k!
k=1
nk
k=1
k!
=
1 2
20.12*. Let X k be independent copies of X . Inductively, one can show that
√ 1m 2
2
m
1 X k = √ 2 k=1
−1
m
2
√ 1
2m
1
−
m
X k +
k=1
2
1
√ m 2
1
−
X k
k=2
−1 +1
m
Take expectation on the both sides,
√ m
2 EX = E X
m = 1, 2,
· ··
Consequently, EX = 0. On the other hand, by CLT, m
√ 1m 2
2
k=1
X k
d −→ N (0, σ ) 2
So we have X ∼ N (0, σ 2 ).
3
d
=
√ 1 (X + Y ) =d X 2
