Homework # 1
18.6. Notice that for any m > 0
m n
P |X n | > m = 1 −
when n > m. Consequently, X n is not uniformly tight. Therefore, X n is not pre-compact in the sense of weak convergen convergence. ce. Thus, Thus, X n is not convergen convergentt in distribution. distribution. Since, Since, the conver convergence gence in distribut distribution ion is the weakes weakestt conver convergence gence,, so X n is not convergent in any sense. 18.8. By checki checking ng the characte characterist ristic ic function, function, the sample sample average average 1 n
(X 1 + · · · + X n )
has the Cauchy distribution with α = 0 and β = 1. So Y n converges in distribution in a triv trivail way way. In the followin followingg I give give two two differe different nt argumen arguments ts to show show that that Y n is not convergent in probability. p
Proof Proof 1. Assume that Y n −→ Y . Then Then Y must must be a Cauchy Cauchy random variable variable.. On the other hand, I will show that Y is almost surely a constant. a.s.
Indeed, by Theorem 17.3, there is a subsequence {nk } such that Y n −→ Y . By the independence of the Cauchy sequence { X j } and by 0-1 law, Y = C a.s. for some constant C . p p Proof 2. Assume that Y n −→ Y , or Y n − Y −→ 0. Write Y 2 n − Y n = (Y 2n − Y ) − p (Y n − Y ). We would would have have Y 2n − Y n −→ 0. On the othe otherr hand, hand, Y 2n − Y n has the same distribution as X 1 . k
18.9. By Chebyshev inequality, for any m > 0,
P |X n | > m
≤
1 m2
EX n2
Consequently,
sup P |X n | > m ≤ n≥1
1 m2
n = 1, 2, · · ·
sup EX n2
n≥1
Therefore,
lim sup P |X n | > m = 0
m→∞ n≥1
So { X n } is uniformly tight. 18.15 Let F n (x) and F (x) be distribution functions of X n and X , respectively respectively.. Fix the continuous point x of F (x). For any ǫ > 0, P {X n + Y n ≤ x} ≤ P {X n ≤ x + ǫ} + P {|Y n | ≥ ǫ} = F n (x + ǫ) + P {|Y n | ≥ ǫ}
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Similarly, P {X n + Y n ≤ x} ≥ F n (x − ǫ) − P {|Y n | ≥ ǫ}
When x + ǫ and x − ǫ are continuous points of F (x), lim F n (x + ǫ) = F (x + ǫ) and
n→∞
lim F n (x − ǫ) = F (x − ǫ)
n→∞
Thus, F (x − ǫ) ≤ lim inf P {X n + Y n ≤ x} ≤ limsup P {X n + Y n ≤ x} ≤ F (x + ǫ) n→∞
n→∞
We now pick a sequence {ǫm } of positive numbers such that x ± ǫm are continuous points of F (x) and ǫm → 0 as m → ∞. Take ǫ = ǫ m F (x−ǫm ) ≤ liminf P {X n +Y n ≤ x} ≤ lim sup P {X n +Y n ≤ x} ≤ F (x+ǫm ) n→∞
m = 1, 2, · · ·
n→∞
Letting m → ∞, lim P {X n + Y n ≤ x} = F (x)
n→∞
18.16. Write f (x) = |x| p1[−N,N ] (x). The problem is to show that Ef (X ) ≤ lim sup Ef (X n) < ∞ n→∞
Notice that f (x) is bounded function. So the finiteness of the right hand side is obvious. The difficult here is the discontinuity of f . For this we take a sequence f ǫ (x) of continuous functions sucht that 0 ≤ f ǫ (x) ↑ f (x) (ǫ → 0+ ) for every x. By the definition of convergence in distribution, for each ǫ > 0, Ef ǫ (X ) = lim Ef ǫ (X n ) ≤ lim sup Ef (X n) n→∞
n→∞
where the last step follows from the fact that f ǫ (x) ≤ f (x). Let ǫ → 0 + on the left hand side. By monotonic convergence lim Ef ǫ (X ) = E f (X )
ǫ→0+
So we have proved the requested conclusion.
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