Hallo!

Is the following sequence of integrable random variables $\displaystyle (X_h)_{h \in [0,1]}$ also uniformly integrable?

$\displaystyle X_h=\prod\limits_{n=1}^{\frac{T}{h}}(1+\alpha_{(n-1)h}(\exp\{\mu_{(n-1)h}h+ \sigma (W_{nh}-W_{(n-1)h})\} -1))^\gamma$

whereas

$\displaystyle W$ is a standard Brownian motion,

$\displaystyle \sigma$ a constant $\displaystyle \in \matbb{R}_+, $

$\displaystyle \alpha_t $ is a random variable with values in [0,1],

$\displaystyle \mu_t$ is a standard-normal distributed random variable and $\displaystyle \mu_t$ is continous in $\displaystyle t$

[0,T] is the time interval and it is required that $\displaystyle N:=\frac{T}{h} \in \matbb{N}$ and

$\displaystyle \gamma $ is a constant $\displaystyle \in (0,1)$

I also know that $\displaystyle X_h$ converges in probability for $\displaystyle h \to 0$ to a integrable random variable $\displaystyle X$.

I've no idea how to show it.

Can anybody help me?

I found out that $\displaystyle \mathbb{E}[X_h]\leq \mathbb{E}[X]<\infty \; \forall h \in [0,1] \quad $ and $\displaystyle X_h$ converges in probability to this random variable $\displaystyle X.$

Can I now conclude that $\displaystyle (X_h)_{h \in [0,1]}$ is uniformly integrable?