Math 472 Homework Assignment 6
Problem 5.2.1. Let X n denote the mean of a random sample of size n
from a distribution that is N (µ, σ 2 ). Find the limiting distribution of X n .
Solution 5.2.1. Since the random sample is taken from a distribution with
finite mean µ and finite variance σ 2 , we may apply the weak law of large
numbers to conclude that {X n } converges to µ in probability. Theorem
5.2.1 says that if a sequence of random variables {Yn } converges to the
random variable Y in probability, then {Yn } converges to Y in distribution.
Therefore, since {X n } converges to µ in probability we conclude that {X n }
converges to µ in distribution. Hence the limiting distribution for X n is the
distribution of the degenerate random variable µ,
(
0, x < µ
Fµ (x) =
.
1, x ≥ µ
Problem 5.2.2. Let Y1 denote the minimum of a random sample of size n
from a distribution that has pdf f (x) = e−(x−θ) , θ < x < ∞, zero elsewhere.
Let Zn = n(Y1 − θ). Investigate the limiting distribution of Zn .
Solution 5.2.2. A routine calculation shows that P (Y1 > y) = e−n(y−θ) for
y > θ, and P (Y1 > y) = 1 for y ≤ θ. Since the support of Y1 is the interval
(θ, ∞) it follows that the support of Zn = n(Y1 − θ) is the interval (0, ∞)
and therefore FZn (t) = P (Zn ≤ t) = 0 for all t ≤ 0. Let t > 0, then
FZn (t) = P (Zn ≤ t)
= P (n(Y1 − θ) ≤ t)
t
= P (Y1 ≤ + θ)
n
t
= 1 − P (Y1 > + θ)
n
= 1 − e−t .
We see that, for every natural number n, FZn is the cdf for the exponential
distribution with mean µ = 1. Therefore, {Zn } converges in distribution to
an exponential distribution with mean µ = 1.
Problem 5.2.5. Let the pmf of Yn be pn (y) = 1, y = n, zero elsewhere.
Show that Yn does not have a limiting distribution. (In this case, the probability has “escaped” to infinity.)
Solution 5.2.5. The cdf for the degenerate random variable Yn is
(
0, y < n
FYn (y) =
.
1, y ≥ n
For all y, limn→∞ FYn (y) = 0. To show that Yn does not have a limiting
distribution, we must show that there does not exist a distribution function
1
2
F with the property that F (y) = 0 for every y ∈ C(F ). Equivalently, we
must show that if F is a distribution function then there exists a point y
such that F is continuous at y and F (y) 6= 0.
Let F be a distribution function. Since limy→∞ F (y) = 1, there exists
a real number y0 such that F (y) > 1/2 for all y > y0 . Recall that the set
of discontinuous points for a distribution is always a countable set. Since
the interval (y0 , ∞) is an uncountable set, there must be at least one point
y ∈ (y0 , ∞) such that F is continuous at y. Since y > y0 , we also have that
F (y) > 1/2 > 0. Therefore every distribution function F has at least one
point y such that F is continuous at y and F (y) 6= 0. We conclude that Yn
does not have a limiting distribution.
Problem 5.2.7. Let Xn have a gamma distribution with parameter α = n
and β, where β is not a function of n. Let Yn = Xn /n. Find the limiting
distribution of Yn .
Solution 5.2.7. From Appendix D on page 667 we see that E[Xn ] = nβ
and Var[Xn ] = nβ 2 , and therefore E[Yn ] = E[Xn /n] = β and Var[Yn ] =
Var[Xn /n] = β 2 /n. Since the expected value of Yn does not depend on n
and since the variance of Yn is a constant divided by n, Exercise 5.1.3 on page
293 shows that {Yn } converges to β in probability. Applying Theorem 5.2.1
on page 298 shows that {Yn } converges to β in distribution. We conclude
that the limiting distribution of Yn is the distribution for the degenerate
random variable β,
(
0, t < β
Fβ (t) =
.
1, t ≥ β
Problem 5.2.12. Prove Theorem 5.2.3.
Solution 5.2.3.
Theorem. Suppose Xn converges to X in distribution and Yn converges in
probability to 0. Then Xn + Yn converges to X in distribution.
Proof. Let x be a continuous point for FX . We want to show that limn→∞ FXn +Yn (x) =
FX (x).
Let > 0 be given. Suppose that x+ and x− are also continuous points
for FX . Observe that Xn + Yn ≤ x and |Yn | ≤ imply that Xn ≤ x + .
Thus,
FXn +Yn (x) = P (Xn + Yn ≤ x)
= P (Xn + Yn ≤ x, |Yn | < ) + P (Xn + Yn ≤ x, |Yn | ≥ )
(1)
≤ P (Xn + Yn ≤ x, |Yn | ≤ ) + P (Xn + Yn ≤ x, |Yn | ≥ )
≤ P (Xn ≤ x + ) + P (|Yn | ≥ )
= FXn (x + ) + P (|Yn | ≥ ).
3
D
→ X and since x + is a continuity point for FX we have that
Since Xn −
lim FXn (x + ) = FX (x + ),
n→∞
P
→ 0 we have that
and since Yn −
lim P (|Yn | ≥ ) = 0.
n→∞
The inequality in (1), together with these last two limits, shows that
lim sup FXn +Yn (x) ≤ FX (x + ).
(2)
n→∞
Now, observe that Xn + Yn > x and |Yn | < imply that Xn > x − .
Thus,
1 − FXn +Yn (x) = P (Xn + Yn > x)
= P (Xn + Yn > x, |Yn | < ) + P (Xn + Yn > x, |Yn | ≥ )
≤ P (Xn > x − ) + P (|Yn | ≥ )
= 1 − FXn (x + ) + P (|Yn | ≥ ).
Rearranging the terms of this inequality yields
FXn (x − ) − P (|Yn | ≥ ) ≤ FXn +Yn (x).
(3)
D
Since Xn −
→ X and since x − is a continuity point for FX we have that
lim FXn (x − ) = FX (x − ),
n→∞
P
and since Yn −
→ 0 we have that
lim P (|Yn | ≥ ) = 0.
n→∞
The inequality in (3), together with these last two limits, shows that
(4)
FX (x − ) ≤ lim inf FXn +Yn (x).
n→∞
Combining inequalities (2) and (4) we have
(5)
FX (x − ) ≤ lim inf FXn +Yn (x) ≤ lim sup FXn +Yn (x) ≤ FX (x + ).
n→∞
n→∞
The inequalities in (5) require that x − and x + be continuity points
for FX . We claim that there exists a sequence of positive numbers {k } such
that limk→∞ k = 0 and for all natural numbers k the points x ± k are
continuity points for FX . This claim is an immediate consequence of the
fact that the set of discontinuous points of FX is countable.
Choose {k } as described in the previous paragraph. Since FX is continuous at x, limk→∞ FX (x ± k ) = FX (x). Since FX is continuous at the points
x ± k for every natural number k,
(6)
FX (x − k ) ≤ lim inf FXn +Yn (x) ≤ lim sup FXn +Yn (x) ≤ FX (x + k ).
n→∞
n→∞
Finally, letting k → ∞ in (7) shows that limn→∞ FXn +Yn (x) = FX (x), and
D
we conclude that Xn + Yn −
→ X.
4
Problem 5.2.20. Use Stirling’s formula, (5.2.2), to show that the first limit
in Example 5.2.3 is 1.
Solution 5.2.20. Let us introduce three sequences,
Γ n−1
2 +1
an = √
(7)
n/2
2π n−1
e−(n−1)/2
2
√
(n−1)/2 −(n−2)/2
2π n−2
e
2
bn =
(8)
n−2
Γ 2 +1
r
2 n − 1 n/2
−1/2
1−
(9)
cn = e
.
n n−2
Stirling’s formula states that
Γ(k + 1)
= 1.
lim √
1
k→∞
2πk k+ 2 e−k
We may apply Sterling’s formula to an with k = (n − 1)/2 to conclude that
limn→∞ an = 1. In a similar way we may apply Steling’s formula to bn with
k = (n − 2)/2 to conclude that limn→∞ bn = 1.
Notice that limn→∞ [(n − 1)/(n − 2)]n/2 is a 1∞ indeterminant form. Applying Calculus II methods that are taken from the section on L’Hôpital’s
Rule we are able to evaluate that limit. We leave it as an exercise to confirm that limn→∞ [(n − 1)/(n − 2)]n/2 = e1/2 . Knowing this, we see that
limn→∞ cn = 1.
After some simplification, which we leave to the student, we find that
Γ[(n + 1)/2]
.
an · bn · cn = p
n/2 Γ(n/2)
We therefore conclude that
Γ[(n + 1)/2]
lim p
= lim an · bn · cn = 1 · 1 · 1 = 1.
n→∞
n/2 Γ(n/2) n→∞