# Thread: upper bound for a random variable

1. ## upper bound for a random variable

Hallo,
I've to find a constant $\displaystyle C \in \mathbb{R}_+$ such that $\displaystyle |Y_n|\leq C$ for a random variable $\displaystyle Y_n$ which is defined by

$\displaystyle Y_n:=\frac{2 \cdot (\pi e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h})})^3}{(1+\pi( e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h}))}-1))^3}-\frac{ (\pi e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h})})^2}{(1+\pi( e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h}))}-1))^2}$

whereas
$\displaystyle W$ is a standard Brownian motion
$\displaystyle \mu$ is a normal distributed random variable
$\displaystyle \pi$ and $\displaystyle \theta$ are random variables with values in the interval $\displaystyle [0,1]$
$\displaystyle 0<h<1$ is a parameter
$\displaystyle \sigma>0$ is a constant

Can anybody help me finding $\displaystyle C$ such that $\displaystyle |Y_n|\leq C$?
Thanks a lot!!

2. If it helps, the vast majority of the details you provided are completely irrelevant. If I did it correctly, it is very easy to show $\displaystyle |Y_n| \le 3$. Just call the exponential term $\displaystyle X_n$.

3. Thank you!
I've done the calculation again and now it is clear. For the first term it holds
$\displaystyle Y_n:=\frac{2 \cdot (\pi e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h})})^3}{(1+\pi( e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h}))}-1))^3}\leq \frac{2 \cdot (\pi e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h})})^3}{(\pi e^{\theta (\mu_n h+ \sigma (W_{nh}-W_{(n-1)h}))})^3}=2.$
For the second term I can do the same.