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This question is extended from Resnick's exercise 5.13 in his book A Probability Path.

Let the probability space be the Lebesgue interval:

$(\Omega=[0,1],\mathcal{B}([0,1]),\lambda)$ and define

$X_n:=\frac{n}{\log n}1_{(0,\frac 1n)}$

Show $X_n\to 0, E(X_n)\to 0$ even though DCT fails. And secondly, show

$\lim_{M\to\infty} \sup_{n\ge 2} E(X_n 1_{X_n>M})=0$ (uniform integrability)

Attempt/outline at a solution

:$X_{n} \rightarrow 0$. You should use that $X_{n}(w) \neq 0$ iif $w \in (0,1/n)$.

:For any $w \in (0,1)$, there exists $n^{*}$ such that $\frac{1}{n^{*}} < w$. Hence, for any $n > n^{*}$, $X_{n}(w) = 0$ and $\lim X_{n}(w) = 0$.

:$E[X_{n}] \rightarrow 0$. $X_{n}$ is constant and therefore,

:$E[X_{n}] = \frac{n}{\log(n)}P(Y \in (0,1/n)) = \frac{1}{\log(n)}$.

:Uniform integrability. $X_{n} \leq n$. Hence,

$$\sup_{n \geq 2} E[X_{n}I(X_{n} > M)] \geq \sup_{n \geq M} E[X_{n}I(X_{n} > M)] \geq \sup_{n \geq M} E[X_{n}] = \sup_{n \geq M}\frac{1}{\log(n)} = \frac{1}{\log(M)}$$

The result follows taking the limit in M.

Critique of the solution

Implicit in Resnick's instructions is that you explain why you cannot use DCT. (Hint: think about $Y = \sup_{n\ge2}X_n$ and its expectation.)

Your check of uniform integrability is wrong. To begin with, you have to bound the expression from above not from below. Second, since you have a supremum (over all n), the expression can only go to 0 because of the indicator for X_n>M; your argument appears to take no notice of it. Keep in mind that X_n is positive only on (0, 1/n) so no X_n dominates another.

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I'm confused; are you having a conversation with yourself? –  Nate Eldredge Jul 16 '12 at 17:06
    
Why in the critic it's written "you", not "I"? With this form, it looks like remarks of a teacher for example. –  Davide Giraudo Jul 16 '12 at 17:08
    
I am doing a self-study and trying to solve these myself. I then let a PhD friend of mine see them and those were his comments. Hope that helps (and believe me I wish I was taking a class on this and not trying to learn from scratch hah). –  Justin Jul 16 '12 at 17:40

1 Answer 1

up vote 1 down vote accepted
  • I think you mean $X_n\to 0$ almost everywhere. In fact, as you showed, it converges everywhere to $0$.
  • The fact that $E X_n\to 0$ is correct.
  • The hypothesis of the dominated convergence fail. Indeed, assume that we can find $X$ such that $|X_n(\omega)|\leq X(\omega)$ for all $n$ and almost every $\omega$. In particular, on $I_p:=\left[\frac 1{2^{p+1}},\frac 1{2^p}\right)$, we should have $$X(\omega)\geq \frac{2^p}{\log 2^p}=\frac{2^p}{p\ln 2}.$$ But such an $X$ is not integrable, since $$\int_{[0,1]}X(\omega)d\lambda(\omega)=\sum_{p=0}^{+\infty}\int_{I_p}X(\omega)d\lambda(\omega)\geq\sum_{p=0}^{+\infty}\frac 1{p\ln 2},$$ which is a divergent series.
  • But the family $\{X_n\}$ is uniformly integrable. Indeed, fix $M>0$. We can find an integer $n_0$ such that $\frac n{\ln n}\geq M$ if $n\geq n_0$. Necessarily, this integer $n_0$ is greater than $M$. The quantity $E[X_n\chi_{\{X_n\geq M\}}]$ is $0$ when $n\leq n_0$ (if $n_0$ is taken minimal, which is possible, since $n\mapsto \frac n{\ln n}$ is strictly increasing). For $n\geq n_0$, we have $$E[X_n\chi_{\{X_n\geq M\}}]=\frac 1{\ln n}\leq \frac 1{\log M},$$ hence $$\sup_{n\geq 2}E[X_n\chi_{\{X_n\geq M\}}]\leq \frac 1{\log M}.$$
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