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Thread: upper bound for a random variable

  1. #1
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    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!!
    Last edited by Juju; Mar 9th 2011 at 04:22 AM.
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  2. #2
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    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 $\displaystyle X_n$.
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  3. #3
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    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.
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