Em relação à convergência em probabilidade

12

Seja { X n } n 1{Xn}n1 uma sequência de variáveis ​​aleatórias st X naXna em probabilidade, onde a > 0a>0 é uma constante fixa. Estou tentando mostrar o seguinte: X num

Xna
e umX n1
aXn1
ambos em probabilidade. Estou aqui para ver se minha lógica estava correta. Aqui está o meu trabalho

TENTATIVA

Para a primeira parte, temos | X n -a | <ϵ| X n - a | < £ | X n +a | =£| (X n -sqrta)+2a |

|Xna|<ϵ|Xna|<ϵ|Xn+a|=ϵ|(Xnsqrta)+2a|
£| X n -a | +2£um <ε2+2εa
ϵ|Xna|+2ϵa<ϵ2+2ϵa
Observe que ϵ2+2ϵa >ϵa
ϵ2+2ϵa>ϵa
Segue-se então que P(|X n -a | ϵ)P(|Xn-a|ϵa )1a sn
P(|Xna|ϵ)P(|Xna|ϵa)1asn
X numaeu np r o b a b i l i t y
Xnainprobability

Para a segunda parte, temos | umaX n -1| =| Xn-aX n | <ϵ| X n - a | < £ | X n |

|aXn1|=|XnaXn|<ϵ|Xna|<ϵ|Xn|
Agora, como X naXna como n n , temos que X nXn é uma sequência delimitada. Em outras palavras, existe um número real M < M< st | X n | M|Xn|M . Assim, | X n - a | < £ | X n ||Xna|<ϵM
|Xna|<ϵ|Xn||Xna|<ϵM
Looking at it in probability, we have P(|aXn1|>ϵ)=P(|Xna|>ϵ|Xn|)P(|Xna|>ϵM)0asn
P(|aXn1|>ϵ)=P(|Xna|>ϵ|Xn|)P(|Xna|>ϵM)0asn

I'm pretty confident in the first one, but am pretty iffy on the second. Was my logic sound?

Savage Henry
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6
Consider the sequence XnXn where Pr(Xn=a)=11/nPr(Xn=a)=11/n and Pr(Xn=n)=1/nPr(Xn=n)=1/n. It seems to me that since 11/n111/n1 this sequence converges to aa in probability, but clearly it is unbounded since sup(Xn)=max(a,n)sup(Xn)=max(a,n).
whuber
2
Continuous mapping theorem?
Christoph Hanck

Respostas:

13

The details of the proofs matter less than developing appropriate intuition and techniques. This answer focuses on an approach designed to help do that. It consists of three steps: a "setup" in which the assumption and definitions are introduced; the "body" (or a "crucial step") in which the assumptions are somehow related to what is to be proven, and the "denouement" in which the proof is completed. As in many cases with probability proofs, the crucial step here is a matter of working with numbers (the possible values of random variables) rather than dealing with the much more complicated random variables themselves.


Convergence in probability of a sequence of random variables YnYn to a constant aa means that no matter what neighborhood of 00 you pick, eventually each YnaYna lies in this neighborhood with a probability that is arbitrarily close to 11. (I won't spell out how to translate "eventually" and "arbitrarily close" into formal mathematics--anybody interested in this post already knows that.)

Recall that a neighborhood of 00 is any set of real numbers containing an open set of which 00 is a member.

The setup is routine. Consider the sequence Yn=a/XnYn=a/Xn and let OO be any neighborhood of 00. The objective is to show that eventually Yn1Yn1 will have an arbitrarily high chance of lying in OO. Since OO is a neighborhood, there must be an ϵ>0ϵ>0 for which the open interval (ϵ,ϵ)O(ϵ,ϵ)O. We may shrink ϵϵ if necessary to ensure ϵ<1ϵ<1, too. This will assure that subsequent manipulations are legitimate and useful.

The crucial step will be to connect YnYn with XnXn. That requires no knowledge of random variables at all. The algebra of numeric inequalities (exploiting the assumption a>0a>0) tells us that the set of numbers {Yn(ω)|Yn(ω)1(ϵ,ϵ)}{Yn(ω)|Yn(ω)1(ϵ,ϵ)}, for any ϵ>0ϵ>0, is in one-to-one correspondence with the set of all Xn(ω)Xn(ω) for which

a1+ϵ<Xn(ω)<a1ϵ.

a1+ϵ<Xn(ω)<a1ϵ.

Equivalently,

Xn(ω)a(aϵ1+ϵ,aϵ1ϵ)=U.

Xn(ω)a(aϵ1+ϵ,aϵ1ϵ)=U.

Since a0a0, the right hand side UU indeed is a neighborhood of 00. (This clearly shows what breaks down when a=0a=0.)

We are ready for the denouement.

Because XnaXna in probability, we know that eventually each XnaXna will lie within UU with arbitrarily high probability. Equivalently, Yn1Yn1 will eventually lie within (ϵ,ϵ)O(ϵ,ϵ)O with arbitrarily high probability, QED.

whuber
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I apologize for such a late best answer. It's been a busy week. Thank you so for much this!!!
Savage Henry
5

We are given that

limnP(|Xnα|>ϵ)=0

limnP(|Xnα|>ϵ)=0

and we want to show that

limnP(|αXn1|>ϵ)=0

limnP(αXn1>ϵ)=0

We have that

|αXn1|=|1Xn(αXn)|=|1Xn||Xnα|

αXn1=1Xn(αXn)=1Xn|Xnα|

So equivalently, we are examining the probability limit

limnP(|1Xn||Xnα|>ϵ)=?0

limnP(1Xn|Xnα|>ϵ)=?0

We can break the probability into two mutually exclusive joint probabilities

P(|1Xn||Xnα|>ϵ)=P(|1Xn||Xnα|>ϵ,|Xn|1)+P(|1Xn||Xnα|>ϵ,|Xn|<1)

P(1Xn|Xnα|>ϵ)=P(1Xn|Xnα|>ϵ,|Xn|1)+P(1Xn|Xnα|>ϵ,|Xn|<1)

For the first element we have the series of inequalities

P(|1Xn||Xnα|>ϵ,|Xn|1)P[|Xnα|>ϵ,|Xn|1]P[|Xnα|>ϵ]

P(1Xn|Xnα|>ϵ,|Xn|1)P[|Xnα|>ϵ,|Xn|1]P[|Xnα|>ϵ]

The first inequality comes from the fact that we are considering the region where |Xn||Xn| is higher than unity and so its reciprocal is smaller than unity. The second inequality because a joint probability of a set of events cannot be greater than the probability of a subset of these events.
The limit of the rightmost term is zero (this is the premise), so the limit of the leftmost term is also zero. So the first element of the probability that interests us is zero.

For the second element we have

P(|1Xn||Xnα|>ϵ,|Xn|<1)=P(|Xnα|>ϵ|Xn|,|Xn|<1)

P(1Xn|Xnα|>ϵ,|Xn|<1)=P(|Xnα|>ϵ|Xn|,|Xn|<1)

Define δϵmax|Xn|δϵmax|Xn|. Since here |Xn||Xn| is bounded, it follows that δδ can be made arnitrarily small or large, and so it is equivalent to ϵϵ. So we have the inequality

P[|Xnα|>δ,|Xn|<1]P[|Xnα|>δ]

P[|Xnα|>δ,|Xn|<1]P[|Xnα|>δ]

Again, the limit on the right side is zero by our premise, so the limit on the left side is also zero. Therefore the second element of the probability that interests us is also zero. QED.

Alecos Papadopoulos
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5

For the first part, take x,a,ϵ>0x,a,ϵ>0, and note that |xa|ϵ|xa|ϵaa|xa|ϵax+a|(xa)(x+a)|ϵa|xa|ϵa.

|xa|ϵ|xa|ϵaa|xa|ϵax+a|(xa)(x+a)|ϵa|xa|ϵa.
Hence, for any ϵ>0ϵ>0, defining δ=ϵaδ=ϵa, we have Pr(|Xna|ϵ)Pr(|Xna|δ)0,
Pr(|Xna|ϵ)Pr(|Xna|δ)0,
when nn, implying that XnPraXnPra.

For the second part, take again x,a,ϵ>0x,a,ϵ>0, and cheat from Hubber's answer (this is the key step ;-) to define δ=min{aϵ1+ϵ,aϵ1ϵ}.

δ=min{aϵ1+ϵ,aϵ1ϵ}.
Now, |xa|<δaδ<x<a+δaaϵ1+ϵ<x<a+aϵ1ϵa1+ϵ<x<a1ϵ1ϵ<ax<1+ϵ|ax1|<ϵ.
|xa|<δaδ<x<a+δaaϵ1+ϵ<x<a+aϵ1ϵa1+ϵ<x<a1ϵ1ϵ<ax<1+ϵax1<ϵ.
The contrapositive of this statement is |ax1|ϵ|xa|δ.
ax1ϵ|xa|δ.

Therefore, Pr(|aXn1|ϵ)Pr(|Xna|δ)0,

Pr(aXn1ϵ)Pr(|Xna|δ)0,
when nn, implying that aXnPr1aXnPr1.

Note: both items are consequences of a more general result. First of all remember this Lemma: XnPrXXnPrX if and only if for any subsequence {ni}N{ni}N there is a subsequence {nij}{ni}{nij}{ni} such that XnijXXnijX almost surely when jj. Also, remember from Real Analysis that g:ARg:AR is continuous at a limit point xx of AA if and only if for every sequence {xn}{xn} in AA it holds that xnxxnx implies g(xn)g(x)g(xn)g(x). Hence, if gg is continuous and XnXXnX almost surely, then Pr(limng(Xn)=g(X))Pr(limxXn=X)=1,

Pr(limng(Xn)=g(X))Pr(limxXn=X)=1,
and it follows that g(Xn)g(X)g(Xn)g(X) almost surely. Moreover, gg being continuous and XnPrXXnPrX, if we pick any subsequence {ni}N{ni}N, then, using the Lemma, there is a subsequence {nij}{ni}{nij}{ni} such that XnijXXnijX almost surely when jj. But then, as we have seen, it follows that g(Xnij)g(X)g(Xnij)g(X) almost surely when jj. Since this argument holds for every subsequence {ni}N{ni}N, using the Lemma in the other direction, we conclude that g(Xn)Prg(X). Hence, to answer your question you can just define continuous functions g(x)=x and h(x)=a/x, for x>0, and apply this result.
Zen
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Zen thank you for you answer. This was very clear!
Savage Henry