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How can I perform Euler discretization on this model where $\delta t=1$ and $\delta x_t = x_t-x_{t-1}$

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  • $\begingroup$ Hi Qurban Abbasov, welcome to Quant.SE! I've cleaned up the question, can you please verify? Also what is $\gamma$ in (4)? $\endgroup$ – Bob Jansen Oct 26 '15 at 9:45
  • $\begingroup$ Hi Bob Jansen, yes it is correct.thank you. $\gamma$ is just a parameter which is different from 0 $\endgroup$ – Qurban Abbasov Oct 26 '15 at 9:56
  • $\begingroup$ I think he meant whether it was to denote a given $x_t$ or is it used to mean $x_t$ raised to the power $\gamma$ $\endgroup$ – SRKX Oct 28 '15 at 2:38
  • $\begingroup$ @QurbanAbbasov: Can you please put more input to your question so that people can understand your intention or difficulty. $\endgroup$ – Gordon Oct 28 '15 at 12:50
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I would proceed as follows: \begin{align*} x_t &= x_{t-\delta t} + \alpha (\beta - x_{t-\delta t}) \delta t + \sigma x_{t-\delta t}^\gamma (w_t - w_{t-\delta t}),\\ x_t &= \max (x_t, \ 0), \end{align*} where $w_t - w_{t-\delta t}$ is a normal random variable with mean $0$ and variance $\delta t$, which can be obtained by an independent draw for each time step.

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  • $\begingroup$ Gordon, do you have to scale the diffusion term by sqrt(t)? $\endgroup$ – SmallChess Oct 30 '15 at 4:09
  • $\begingroup$ @StudentT: I purposely wrote my diffusion term as $w_t-w_{t-\delta t}$ to shorten the explanation. Yes, you can write it as $\xi \sqrt{\delta t}$, where $\xi$ is a standard normal random variable, which is an independent draw at each time step. $\endgroup$ – Gordon Oct 30 '15 at 15:06
  • $\begingroup$ @Gordon Thanks. It was my mistake. I misunderstood the term in the diffusion term was a standard normal. It was my fault. Thanks for the feedback. $\endgroup$ – SmallChess Oct 30 '15 at 15:09

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