Solutions Probability Essentials 18,19 [PDF]

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Homework # 1 18.6. Notice that for any m > 0 n o m P |Xn | > m = 1 − n when n > m. Consequently, Xn is not uniformly tight. Therefore, Xn is not pre-compact in the sense of weak convergence. Thus, Xn is not convergent in distribution. Since, the convergence in distribution is the weakest convergence, so Xn is not convergent in any sense. 18.8. By checking the characteristic function, the sample average 1 (X1 + · · · + Xn ) n has the Cauchy distribution with α = 0 and β = 1. So Yn converges in distribution in a trivail way. In the following I give two different arguments to show that Yn is not convergent in probability. p

Proof 1. Assume that Yn −→ Y . Then Y must be a Cauchy random 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 Ynk −→ Y . By the independence of the Cauchy sequence {Xj } and by 0-1 law, Y = C a.s. for some constant C. p p Proof 2. Assume that Yn −→ Y , or Yn − Y −→ 0. Write Y2 n − Yn = (Y2n − Y ) − p (Yn − Y ). We would have Y2n − Yn −→ 0. On the other hand, Y2n − Yn has the same distribution as X1 . 18.9. By Chebyshev inequality, for any m > 0, o 1 P |Xn | > m ≤ 2 EXn2 m n

Consequently,

n = 1, 2, · · ·

n o 1 sup P |Xn | > m ≤ 2 sup EXn2 m n≥1 n≥1

Therefore, n o lim sup P |Xn | > m = 0

m→∞ n≥1

So {Xn } is uniformly tight. 18.15 Let Fn (x) and F (x) be distribution functions of Xn and X, respectively. Fix the continuous point x of F (x). For any ǫ > 0, P {Xn + Yn ≤ x} ≤ P {Xn ≤ x + ǫ} + P {|Yn | ≥ ǫ} = Fn (x + ǫ) + P {|Yn | ≥ ǫ} 1

Similarly, P {Xn + Yn ≤ x} ≥ Fn (x − ǫ) − P {|Yn | ≥ ǫ} When x + ǫ and x − ǫ are continuous points of F (x), lim Fn (x + ǫ) = F (x + ǫ) and

n→∞

lim Fn (x − ǫ) = F (x − ǫ)

n→∞

Thus, F (x − ǫ) ≤ lim inf P {Xn + Yn ≤ x} ≤ lim sup P {Xn + Yn ≤ 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 = ǫm F (x−ǫm ) ≤ lim inf P {Xn +Yn ≤ x} ≤ lim sup P {Xn +Yn ≤ x} ≤ F (x+ǫm ) n→∞

m = 1, 2, · · ·

n→∞

Letting m → ∞, lim P {Xn + Yn ≤ x} = F (x)

n→∞

18.16. Write f (x) = |x|p 1[−N,N] (x). The problem is to show that Ef (X) ≤ lim sup Ef (Xn) < ∞ 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 fǫ (x) of continuous functions sucht that 0 ≤ fǫ (x) ↑ f (x) (ǫ → 0+ ) for every x. By the definition of convergence in distribution, for each ǫ > 0, Efǫ (X) = lim Efǫ (Xn ) ≤ lim sup Ef (Xn) n→∞

n→∞

where the last step follows from the fact that fǫ (x) ≤ f (x). Let ǫ → 0+ on the left hand side. By monotonic convergence lim Efǫ (X) = Ef (X)

ǫ→0+

So we have proved the requested conclusion.

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