Rates of convergence in expectation of a random variable (probability theory)

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.

• The question only seems to use $X_1$, not the sequence $\{X_i\}$. Is something missing? Jul 30 '16 at 2:04
• @Math1000 Of course it holds for every $i$, i.e. $n^{1-1/t}|\mathbb{E}(X_{in})|\to0,\forall i=1,2,...$ Jul 30 '16 at 2:57
• @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. Jul 30 '16 at 3:13
• @angryavian hah, yes. actually the i.i.d. setting is not needed for the particular problem. :-) 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 $$n^{(t-1)/t} \left|X_{1} \right |\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\}\leqslant \left|X_{1} \right |^t\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\}.$$
$$n^{1-1/t}|\mathbb E(X_{1,n})| = n^{1-1/t}\mathbb E(-X_{1}\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\})\leq \mathbb E\bigg(\left|X_{1} \right |^t\mathbf 1\left\{ \left|X_{1 } \right | \geqslant n^ {1/t}\right\}\bigg)\to 0$$