Show that $(W_t)_{t \geq 0}$ and $(W_t^2-t)_{t \geq 0}$ are not uniformly integrable I'm considering the following martingale $M_t:=W_t^2-t,\ t\geq 1$, where the $W_t$ is a Brownian motion. I want to prove that this martingale and the Brownian motion are not uniformly integrable. I don't see how I should proceed in other to check that the following term goes not to $0$.
$\lim_{K\rightarrow \infty}\sup_{t\geq0}\mathbb{E}[\mid X_t\mid I_{\mid X\mid>K}]$ for $X=M$ and $X=W$
 A: Brownian motion, Solution I
Since $W_t \sim N(0,t)$, we have
$$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) = \frac{2}{\sqrt{2\pi t}} \int_K^{\infty} x \exp \left( -\frac{x^2}{2t} \right) \, dx = \sqrt{\frac{2}{\pi}} \sqrt{t} \exp \left(-\frac{K^2}{2t} \right)$$
and therefore
$$\sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})=\infty.$$
Brownian motion, Solution II
Fix $K \geq 1$. Then, using that $W_t \stackrel{d}{=} \sqrt{t} W_1$,
$$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) \geq \mathbb{P}(|W_t|>K) = \mathbb{P}\left(|W_1| > \frac{K}{\sqrt{t}} \right) \stackrel{t \to \infty}{\to} 1.$$
Consequently,
$$\lim_{K \to \infty} \sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})\geq 1.$$
(Non)Uniform integrability of $M_t := W_t^2-t$
For $K \geq 1$, we have by the triangle inequality
$$\mathbb{E}(|M_t| 1_{\{|M_t|>K\}}) \geq \mathbb{P}(|M_t|>K) \geq \mathbb{P}(|W_t|^2 > K-t).$$
Using again the scaling property, we find
$$\mathbb{E}(|M_t| 1_{\{|W_t|>K\}}) \geq \mathbb{P} \left( |W_1|^2 > \frac{K-t}{t} \right). \tag{1}$$
Choosing $t = K/2$, we get
$$\begin{align*}\lim_{K \to \infty} \sup_{t \geq 0} \mathbb{E}(|M_t| 1_{\{|M_t|>K\}}) &\geq \liminf_{K \to \infty} \mathbb{E}(|M_{K/2}| 1_{\{|M_{K/2}|>K\}}) \\ &\stackrel{(1)}{\geq} \mathbb{P}(|W_1|^2 > 1) >0. \end{align*}$$
This shows that $(M_t)_{t \geq 0}$ is not uniformly integrable.
