Math 151. Rumbos
Spring 2012
1
Solutions to Assignment #20
1. Let 𝑋1 , 𝑋2 , 𝑋3 , . . . denote a sequence of independent, identically distributed
random variables with mean 𝜇. Assume that the moment generating function
of 𝑋1 exists in some interval around 0. Use the mgf Convergence Theorem
to show that the sample means, 𝑋 𝑛 , converge in distribution to a limiting
distribution with pmf
{
1 if 𝑥 = 𝜇;
𝑝(𝑥) =
0 elsewhere.
Proof: The mgf of the sample mean, 𝑋 𝑛 , is given by
( )]𝑛
[
𝑡
𝜓𝑋 𝑛 (𝑡) = 𝜓𝑋1
,
𝑛
where 𝜓𝑋1 (0) = 1, and 𝜓𝑋′ (0) = 𝜇.
1
To find the limit of 𝜓𝑋 𝑛 (𝑡) as 𝑛 → ∞, we first consider
[
( )]
)
(
𝑡
lim ln 𝜓𝑋 𝑛 (𝑡) = lim 𝑛 ln 𝜓𝑋1
.
𝑛→∞
𝑛→∞
𝑛
Since 𝜓𝑋1 (0) = 1, the limit on the right–hand side can be computed by means
of L’Hospital’s Rule:
[
( )]
𝑡
[
( )]
ln 𝜓𝑋1
𝑡
𝑛
= lim
lim 𝑛 ln 𝜓𝑋1
1
𝑛→∞
𝑛→∞
𝑛
𝑛
( ) (
)
1
𝑡
𝑡
′
( ) 𝜓𝑋
⋅ − 2
1
𝑡
𝑛
𝑛
𝜓𝑋1
𝑛
= lim
.
1
𝑛→∞
− 2
𝑛
We then get that
( )
𝑡
′
𝜓𝑋
(
)
𝜓𝑋′ (0)
1
𝑛
( ) =𝑡⋅ 1
lim ln 𝜓𝑋 𝑛 (𝑡) = lim 𝑡 ⋅
= 𝜇𝑡.
𝑛→∞
𝑛→∞
𝑡
𝜓𝑋1 (0)
𝜓𝑋1
𝑛
It then follows that lim 𝜓𝑋 𝑛 (𝑡) = 𝑒𝜇𝑡 , which is the mgf of 𝑝(𝑥).
𝑛→∞
Math 151. Rumbos
Spring 2012
2
𝑌𝑛 − 𝑛𝑝
for
2. Let 𝑌𝑛 ∼ Binomial(𝑝, 𝑛), for 𝑛 = 1, 2, 3, . . ., and define 𝑍𝑛 = √
𝑛𝑝(1 − 𝑝)
𝑛 = 1, 2, 3, . . . Use the Central Limit Theorem to find the limiting distribution
of 𝑍𝑛 .
Suggestion: Recall that 𝑌𝑛 is the sum of 𝑛 independent Bernoulli(𝑝) trials.
Solution: Observe that 𝑌𝑛 =
𝑛
∑
𝑋𝑘 , where 𝑋1 , 𝑋2 , 𝑋3 , . . . are in-
𝑘=1
dependent Bernoulli(𝑝) random variables. Thus, by the Central Limit
Theorem,
1
𝑌𝑛 − 𝑝
𝐷
√ 𝑛
√ −→ 𝑍 ∼ Normal(0, 1) as 𝑛 → ∞.
𝑝(1 − 𝑝)/ 𝑛
Thus, multiplying the numerator and denominator in the fraction by
𝑛,
𝑌 − 𝑛𝑝
𝐷
√ 𝑛
−→ 𝑍 ∼ Normal(0, 1) as 𝑛 → ∞.
𝑛𝑝(1 − 𝑝)
Consequently, the limiting distribution of 𝑍𝑛 is 𝑍 ∼ Normal(0, 1). □
3. Suppose that 75% of the people in a certain metropolitan area live in the city
and 25% of the people live in the suburbs. If 1200 people attending certain
concert represent a random sample from the metropolitan area, what is the
probability that the number of people from the suburbs attending the concert
will be fewer than 270?
Solution: Let 𝑋 denote the number of people in the sample that
are from the suburbs. Then, 𝑋 ∼ Binomial(𝑝, 𝑛) where 𝑝 = 0.25 and
𝑛 = 1200. Using the approximation
)
(
𝑋 − 𝑛𝑝
Pr √
⩽ 𝑧 ≈ Pr(𝑍 ⩽ 𝑧),
𝑛𝑝(1 − 𝑝)
for large 𝑛, where ∼ Normal(0, 1), we have that
(
)
𝑋 − 300
270 − 300
Pr(𝑋 ⩽ 270) = Pr
⩽
15
15
≈ 𝑃 (𝑍 ⩽ −2) = 1 − 𝐹𝑍 (2)
≈ 0.0227.
Math 151. Rumbos
Spring 2012
3
Thus, the probability that 270 or fewer people in the sample are from
the suburbs is about 2.27%.
□
4. Suppose that a random sample of size 𝑛 is to be taken from a distribution for
which the mean is 𝜇 and the standard deviation is 3. Use the Central Limit
Theorem to determine approximately the smallest value of 𝑛 for which the
following relation will be satisfied:
Pr(∣𝑋 𝑛 − 𝜇∣ < 0.3) ⩾ 0.95.
Solution: By the Central Limit Theorem,
(
)
𝑋𝑛 − 𝜇
√ ⩽ 𝑧 ≈ Pr (𝑍 ⩽ 𝑧) ,
Pr
𝜎/ 𝑛
for all 𝑧 ∈ ℝ and large values of 𝑛, where 𝑍 ∼ Normal(0, 1). We then
get that, for 𝑧 > 0,
)
(
∣𝑋 𝑛 − 𝜇∣
√
⩽ 𝑧 ≈ Pr (∣𝑍∣ ⩽ 𝑧) ,
Pr
𝜎/ 𝑛
for large values of 𝑛, where
Pr(∣𝑍∣ ⩽ 𝑧) = Pr(−𝑧 < 𝑍 ⩽ 𝑧) = 𝐹𝑍 (𝑧) − 𝐹𝑍 (−𝑧) = 2𝐹𝑍 (𝑧) − 1.
We first find 𝑧 > 0 such that Pr(∣𝑍∣ ⩽ 𝑧) ⩾ 0.95 or 2𝐹𝑍 (𝑧) − 1 ⩽
0.95. This yields 𝑧 = 1.96. With this value of 𝑧, we then get that,
approximately,
)
(
∣𝑋 𝑛 − 𝜇∣
√
⩽ 1.96 ⩾ 0.95,
Pr
𝜎/ 𝑛
or
(
)
𝜎
Pr ∣𝑋 𝑛 − 𝜇∣ ⩽ 1.96 √
⩾ 0.95.
𝑛
We therefore choose 𝑛 so that
𝜎
1.96 √ ⩽ 0.3,
𝑛
from which we get that
(
𝑛⩾
1.96𝜎
0.3
We therefore take 𝑛 to be 385.
)2
≈ 384.16.
□
Math 151. Rumbos
Spring 2012
4
5. Let 𝑋𝑛 be a random variable having a binomial distribution with parameters 𝑛
and 𝑝𝑛 . Assume that lim 𝑛𝑝𝑛 = 𝜆. Prove that the mgf of 𝑋𝑛 converges to the
𝑛→∞
mgf of a Poisson distribution with parameter 𝜆 as 𝑛 → ∞.
Solution: Since 𝑋𝑛 ∼ Binomial(𝑝𝑛 , 𝑛) for all 𝑛 = 1, 2, 3, . . ., the mgf
of 𝑋𝑛 is
𝜓𝑋𝑛 (𝑡) = (𝑝𝑛 𝑒𝑡 + 1 − 𝑝𝑛 )𝑛 ,
for 𝑛 = 1, 2, 3, . . . We then have that
𝜓𝑋𝑛 (𝑡) = (1 + 𝑝𝑛 (𝑒𝑡 − 1))𝑛 ,
for 𝑛 = 1, 2, 3, . . .
𝜆
Now, for large values of 𝑛, 𝑝𝑛 ≈ , since lim 𝑛𝑝𝑛 = 𝜆. We therefore
𝑛→∞
𝑛
have that
(
)𝑛
𝜆 𝑡
𝜓𝑋𝑛 (𝑡) ≈ 1 + (𝑒 − 1) ,
𝑛
for large values of 𝑛. Consequently,
(
)𝑛
𝜆(𝑒𝑡 − 1)
𝑡
lim 𝜓𝑋𝑛 (𝑡) = lim 1 +
= 𝑒𝜆(𝑒 −1) ,
𝑛→∞
𝑛→∞
𝑛
which is the mgf of a Poisson(𝜆) random variable.
□