# Thread: upper bound for a random variable

1. ## upper bound for a random variable

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

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

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

Can anybody help me finding $C$ such that $|Y_n|\leq C$?
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 $|Y_n| \le 3$. Just call the exponential term $X_n$.

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