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Let $X_1,...,X_n$ i.i.d random variables, square integrable, and with $E[X_1]=0$.

Let $Y_n = \frac{|X_1 +...+X_n|}{\sqrt{n}}$

I am trying to show that $(Y_n)$ is uniformly integrable, i.e

$\sup_n\mathbb E[Y_n\mathbb 1\{Y_n\gt K\} ] \to 0$ when $K \to +\infty$

I have tried to use Cauchy-Schwarz inequality:
$\mathbb E[Y_n\mathbb 1\{Y_n\gt K\} ]^2 \leq \mathbb E[Y_n^2] \mathbb E[1\{Y_n\gt K\}]=\text{Var}(X_1) \mathbb P(Y_n>K)$

By the CLT, $\mathbb P(Y_n>K) \to \mathbb P(Y>K)$ where $Y$ follows $|\mathcal{N}(0,1)|$.

Also $\mathbb P(Y>K) \to 0$ when $K \to +\infty$.

Ideally I would like to have $\mathbb E[Y_n\mathbb 1\{Y_n\gt K\} ] \leq \epsilon(K) $ where $\epsilon(K) \to 0$ but I don't see how to get it.

Thanks for your help.

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    $\begingroup$ Simply use $$P(Y_n>K)\leqslant K^{-2}E(Y_n^2)=K^{-2}\mathrm{Var}(X_1)$$ in your upper bound of $E(Y_n\mathbf 1_{Y_n>K})$. $\endgroup$ – Did Jun 29 '16 at 13:43
  • $\begingroup$ @Did Thanks, that was easy ! $\endgroup$ – Dark Jun 29 '16 at 13:46
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Note that $K\cdot Y_n\mathbb 1\{Y_n\gt K\}\leqslant Y_n^2 $, hence integrating, we get $$\mathbb E\left[ Y_n\mathbb 1\{Y_n\gt K\}\right]\leqslant\mathbb E\left[ Y_n^2\right] /K.$$ Since $(X_i)_{i\geqslant 1}$ is an uncorrelated sequence, we derive that $$\mathbb E\left[ Y_n\mathbb 1\{Y_n\gt K\}\right]\leqslant\mathbb E\left[X_1^2\right]/K.$$

Note that a similar reasoning yields the following result:

Let $\left(Y_n\right)_{n\geqslant 1}$ be a sequence such that $\sup_{n\geqslant 1}\mathbb E\left[Y_n^2\right]< +\infty$. Then $\left(Y_n\right)_{n\geqslant 1}$ is uniformly integrable.

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