I'm reading a paper concerning probability theory.

We have $X_i$ i.i.d random variables, such that $\mathbb{E}(|X_1|^t)<\infty$, where $t$ is some fixed number and $1\leq t< 2$, also $\mathbb{E}(X_1)=0$. Next we define a truncated random variable

$$X_{kn}=X_k\mathbb{I}_{(|X_k|<n^{1/t})}, k=1,2,...,n=1,2,...$$

Now the author said that by integration by parts and the fact that $\mathbb{E}(X_1)=0$, we can conclude that

$$n^{1-1/t}|\mathbb{E}(X_{1n})|\to 0,\quad \text{as }n\to \infty$$

I do not get the key to prove this and I don't see how integration by parts is used in proving the above. Any comment is really appreciated.

  • 1
    $\begingroup$ The question only seems to use $X_1$, not the sequence $\{X_i\}$. Is something missing? $\endgroup$
    – Math1000
    Jul 30 '16 at 2:04
  • $\begingroup$ @Math1000 Of course it holds for every $i$, i.e. $n^{1-1/t}|\mathbb{E}(X_{in})|\to0,\forall i=1,2,...$ $\endgroup$ Jul 30 '16 at 2:57
  • $\begingroup$ @Schrödinger'sCat I think what Math1000 is saying is that your question didn't need to mention the i.i.d. copies since you are only asking about one of them. $\endgroup$
    – angryavian
    Jul 30 '16 at 3:13
  • $\begingroup$ @angryavian hah, yes. actually the i.i.d. setting is not needed for the particular problem. :-) $\endgroup$ Jul 30 '16 at 3:21

First notice that since $X_1$ is centered, $\mathbb E\left[X_{1,n}\right]=\mathbb E\left[-X_{1}\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\}\right].$

To this aim, one can integrate the pointwise inequality $$n^{(t-1)/t} \left|X_{1} \right |\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\}\leqslant \left|X_{1} \right |^t\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\}.$$

$$n^{1-1/t}|\mathbb E(X_{1,n})| = n^{1-1/t}\mathbb E(-X_{1}\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\})\leq \mathbb E\bigg(\left|X_{1} \right |^t\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\}\bigg)\to 0$$

We are done.

  • $\begingroup$ Thanks for your answer. I made it a complete proof and corrected your typos. Seems that it does nothing to do with the integration by parts which is done by the author. $\endgroup$ Jul 30 '16 at 19:21
  • $\begingroup$ Indeed, there was a typo in the first line (which does not change the sequel). I don't know exactly what the author had in mind. Do you have a link to the paper? $\endgroup$ Jul 30 '16 at 20:39
  • 1
    $\begingroup$ Sure, the paper: Convergence rates in the law of large numbers; by Leonard E. Baum & Melvin Katz. The problem is in page 109, when they try to prove Theorem 1. $\endgroup$ Jul 31 '16 at 3:40

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