A Probability Path.pdf

# Thus taking probabilities and using subadditivity we

• Test Prep
• 464

This preview shows pages 186–189. Sign up to view the full content.

Thus, taking probabilities and using subadditivity, we get from (6.2) E E P[IXr- Xsl >E)::; P[IXr- XI> 2J + P[IXs- XI> 2). If for n 2:; no(E, 0), then for r, s 2:: no. 0 P[IXn- XI >E)::; 2 P[IXr -Xsl > t:] :58 (ii) Next, we prove the following assertion: If {Xn} is Cauchy i.p., then there ex- ists a subsequence {X n i} such that {X n i} converges almost surely. Call the almost sure limit X. Then it is also true that also p To prove the assertion, define a sequence n i by n 1 = 1 and ni = inf{N > nj-1: P[IXr -X 5 I > 2-i] < Ti forallr,s 2:: N}. (In the definition (6.1) of what it means for a sequence to be Cauchy i.p., we let = 8 = 2- i .) Note, by construction n i > n j-1 so that n i oo. Consequently, we have and thus 00 LP[IXni+l -Xnil > 2-j] < 00. j=1

This preview has intentionally blurred sections. Sign up to view the full version.

6.3 Connections Between a.s . and i.p. Convergence 173 The Borel-Cantelli Lemma implies P(N) := P{limsup[IXni+l -Xnil > 2-j]} = 0. j-+00 (6.3) for all large j and thus {X n i ( w)} is a Cauchy sequence of real numbers. The Cauchy property follows since (6.3) implies for large l that L IXni+l (w)- Xn/W)I :S L2-j = 2 · 2- 1 , and therefore for any k > I large, we get IXnk(w)- Xnl(w)l::: L IXnj+l (w)- Xnj(w)l::: 2. T 1 Completeness of the real line implies exists; that is _ lim Xni(w) J-+00 w e Nc implies _ lim X n i (w) exists. J-+00 This means that {X n i} converges a.s. and we call the limit X . p To show X n --+ X note E E P[IXn- XI> E] :S P[IXn- Xnjl > 2] + P[IXnj- XI> 2] . Given any TJ, pick n j and n so large that the Cauchy i.p . property guarantees E 1/ P[IXn -Xnil > 2] < S . X a.s. X . X P X mce n i --+ tmp tes n i --+ , f TJ P[IXni -XI> 2] < 2 for large n j. This finishes the proof of part (a). We now focus on the proof of (b) : Suppose Xn .!:. X. Pick any subsequence {Xnd· Then it is also true that Xnk .!:. X. From (ii) above, there exists a further subsequence {X nkci)} converging a.s . Conversely: Suppose every subsequence has a further subsequence converging alomst surely to X. To show Xn .!:. X, we suppose this fails and get a contradic-
174 6. Convergence Concepts tion. If {Xn} fails to converge in probability, there exists a subsequence {Xnk} and a 8 > 0 and E > 0 such that But every subsequence, such as {Xnk} is assumed to have a further subsequence {X nk(iJ} which converges a.s. and hence i.p. But contradicts convergence i.p. 0 This result relates convergence in probability to point wise convergence and thus allows easy connections to continuous maps. Corollary 6.3.1 (i) If Xn and is continuous, then g(Xn) g(X). (ii) If Xn and is continuous, then p g(Xn) --+ g(X) . Thus, taking a continuous function of a sequence of random variables which con- verges either almost surely or in probability, preserves the convergence. Proof. (i) There exists a null event N E B with P(N) = 0, such that if w E Nc, then Xn(w)--+ X(w) in JR., and hence by continuity, if w E Nc, then g(Xn(w))--+ g(X(w)). This is almost sure convergence of {g(Xn)}. (ii) Let {g(X nk)} be some subsequence of {g(X n)}. It suffices to find an a.s. con- vergence subsequence {g(Xnk(i))}.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '08
• Staff

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern