Bounded in probability and finite expectation Let $x_t = O_p(1)$, meaning that for all $\varepsilon > 0$ there exists $M_{\varepsilon} < \infty$ s.t. $P(|X_t| > M_{\varepsilon}) < \epsilon$ for all $t \in \mathbb{N}$. Does it imply that $\sup_{t} Ex_t < \infty$, namely, whether the expectation of a random variable which is bounded in probability is finite. If it is not the case, could you give me a  counter example?
 A: Here is a counter example. Let $X_{n}$, $n=1,2,\ldots$, be
a sequence of random variables, whose distributions are defined as
follows,
$$
X_{n}=\begin{cases}
0, & \mbox{with probability }\frac{n-1}{n},\\
n^{2}, & \mbox{with probability }\frac{1}{n}.
\end{cases}
$$
To see $X_{n}\sim O_{p}\left(1\right)$, let $\varepsilon>0$ be an
arbitrary positive number, and there exists $M_{\varepsilon}=\left\lfloor \varepsilon^{-1}+1\right\rfloor ^{2}$
($\left\lfloor \varepsilon^{-1}\right\rfloor $ is the integer part
of $\varepsilon^{-1}$) such that
\begin{eqnarray*}
\sup_{n}\Pr\left(\left|X_{n}\right|\geq M_{\varepsilon}\right) & = & \Pr\left(X_{\left\lfloor \varepsilon^{-1}+1\right\rfloor }=\left(\left\lfloor \varepsilon^{-1}+1\right\rfloor \right)^{2}\right)\\
 & = & \frac{1}{\left\lfloor \varepsilon^{-1}+1\right\rfloor }\\
 & < & \varepsilon.
\end{eqnarray*}
However, $\mathrm{{E}}\left(X_{n}\right)=n$ by the design, and $\sup_{n}\mathrm{{E}}X_{n}=\infty$.
Intuitively, 'bounded in probability' only restricts the probability placed on the extreme values; it does not say anything about the ratio between the extreme value and its associated probability. I hope this example could clarify something for your problem.
