Let $X_{n}$ and $Y_{n}$ be two sequences of non-negative random variables. Assume that $$ X_{n}\stackrel{d}{\to}X $$ and $$ E[X_{n}]\to E[X]. $$ Next, assume that $$ P(\underset{n\to\infty}{\liminf} \{Y_{n} \leq X_{n} \}) = 1. $$

Is the following correct then $$ \underset{n\to\infty}{\limsup}E[Y_{n}] \leq E[X] $$ ? Or we need more assumptions on the original sequences?


The conclusion does not hold under the mentioned assumption because

  1. first there is no guarantee that $Y_n$ is integrable. For example, if $Z$ is a non-integrable non-negative random variable, let $Y_n:=X_n+Z\cdot \mathbf 1\left\{Z\gt n\right\}$. For each $\omega$, if $n\geqslant Z\left(\omega\right)$, we have $X_n\left(\omega\right)=Y_n\left(\omega\right)$ but $\mathbb E\left[Y_n\right]$ is infinite for all $n$.

  2. Second, assuming $Y_n$ integrable for all $n$ is not enough. Indeed, let $Y_n:=X_n+n^2\cdot \mathbf 1\left\{Z\gt n\right\}$ where $Z$ is such that $n\mathbb P\left\{Z\gt n\right\}\gt 1$ for all $n$. Then for each $\omega$, if $n\geqslant Z\left(\omega\right)$, we have $X_n\left(\omega\right)=Y_n\left(\omega\right)$ but $\mathbb E\left[Y_n\right]=\mathbb E\left[X_n\right]+n^2\mathbb P\left\{Z\gt n\right\}$ which goes to infinity as $n$ goes to infinity.

However, if we assume that $\mathbb E\left[Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right]\to 0$, then the wanted conclusion holds as a consequence of the decomposition $$ \mathbb E\left[Y_n\right]=\mathbb E\left[Y_n\mathbf 1\left\{Y_n\leqslant X_n\right\}\right]+\mathbb E\left[Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right] $$ because the first term of the right hand side do not exceed $\mathbb E\left[X_n\right]$.

For example, if the sequence $\left(Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right)_{n\geqslant 1}$ is uniformly integrable, then $\mathbb E\left[Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right]\to 0$ (there is a pointwise convergence to $0$).

  • $\begingroup$ yes, let us assume that $Y_{n}$ is integrable. We can also say that both $X_{n}$ and $Y_{n}$ are asymptotically uniformly integrable. What would it be in this case? $\endgroup$
    – Mr.M
    Apr 11 '18 at 14:22
  • 1
    $\begingroup$ In this case, the conclusion you wrote in the opening post holds. $\endgroup$ Apr 11 '18 at 14:31
  • $\begingroup$ just integrability of $Y_{n}$ is enough? $\endgroup$
    – Mr.M
    Apr 11 '18 at 14:41
  • 1
    $\begingroup$ No (see the second point of the edit). $\endgroup$ Apr 11 '18 at 14:53
  • 1
    $\begingroup$ Yes. This is correct. $\endgroup$ Apr 11 '18 at 15:01

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