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I have a pretty nasty looking function

$$\sigma (t,y) = \sqrt{\frac{\sum_{i=1}^N \lambda_i \sigma_i \exp \left (-\frac{1}{2 t \sigma_i^2}\left(\ln{\frac{y}{S_0}} - \left(r - \frac{\sigma_i^2}{2} \right)t \right)^2\right )}{\sum_{i=1}^N \frac{\lambda_i}{\sigma_i} \exp \left (-\frac{1}{2 t \sigma_i^2}\left(\ln{\frac{y}{S_0}} - \left(r - \frac{\sigma_i^2}{2} \right)t \right)^2\right ) }}$$

where $\lambda_i,\sigma_i,r,S_0$ all are strictly positive constants. I also know that $0< \alpha \leq \sigma(t,y) \leq \beta$. In other words $\sigma(t,y)$ is bounded from above by a constant and bounded away from zero by a constant. I am trying to prove that this function $\sigma(t,y)$ satisfies $$|\sigma(t,x)x-\sigma(t,y)y| \leq L|x-y|, \quad \forall x,y >0,$$ where $L$ is a constant. This follows immediately by the mean value theorem if I can prove that $$\left|\frac{\partial \sigma}{\partial y}(t,y)\right| \leq C,$$ but I cannot seem to prove this. Any help is appreciated!

Best regards,

Double Trouble

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  • 2
    $\begingroup$ L stands for Lipschitz perhaps! $\endgroup$ – Bravo Mar 26 '12 at 9:04
  • $\begingroup$ Well, your right in the sense that I'm trying to prove that sigma satisfies the Lipschitz regularity condition but the letter L was just arbitrarily chosen to symbolize a constant :) $\endgroup$ – DoubleTrouble Mar 27 '12 at 15:56

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