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?
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?