Negativity in a CIR version discretized by Ito - Taylor development

Let $X = (X_t: t \in [0,T])$ be a stochastic procedure pleasing a CIR model $$ dX_t = \beta (X_t - \gamma) dt + \sigma\sqrt{X_t} dB_t, $$ where $B_t$ is a typical Brownian activity, $\beta$ is an adverse constant, $\gamma, \sigma$ declare constants. In order for the SDE to make good sense, think that $X_t > 0$ for all $t \in [0,T]$.

Take into consideration adhering to 2 means to imitate the version based upon discretization of $t$ with Ito - Taylor development :

  1. the Euler system : $$ X_{t + \Delta} \approx X_t + \beta(X_t - \gamma)\Delta + \sigma \sqrt{X_t} Z \Delta, $$ where $Z$ is $N(0, 1)$ Gaussian variable.
  2. the Milstein system $$ X_{t + \Delta} \approx X_t + \beta(X_t - \gamma)\Delta + \sigma \sqrt{X_t}Z\sqrt{\Delta} + \frac{1}{4}\sigma^2 \Delta (Z^2-1) $$ where $Z$ is $N(0, 1)$ Gaussian variable.

I was asking yourself why these 2 systems have a favorable probability of creating adverse values of $X_t$ and also consequently can not be made use of without ideal alterations?

Referrals (publication, tutorial and/or paper) will certainly be handy also!

Many thanks and also pertains to!

2019-12-02 03:12:07
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Answers: 1

Take the Euler system. Allow is think that you start with $X_0=\epsilon$. Allow is take $\beta=-1$, $\gamma =1$ and also $\sigma=1$ for definiteness. After that in the next action you'll have

$$X_{\Delta} = \epsilon + (1-\epsilon) \Delta + \sqrt{\epsilon} Z \Delta$$

What is the probability that this is smaller sized than absolutely no?

$$\mathbb{P}\left[X_{\Delta}<0\right]= \mathbb{P}\left[\epsilon + (1-\epsilon) \Delta + \sqrt{\epsilon} Z \Delta < 0\right]$$

Rearranging you'll get :

$$\mathbb{P}\left[X_{\Delta}<0\right]= \mathbb{P}\left[ Z < -\frac{\sqrt{\epsilon}}{\Delta}(1+(1-\frac{1}{\epsilon})\Delta)\right]$$

Using Chebyshev for a price quote this is smaller sized than

$$\mathbb{P}\left[X_{\Delta}<0\right] \leq \frac{\Delta^2}{2\epsilon}$$

So the closer you are to absolutely no, the greater the probability of going across that line. Yet requiring time actions completely tiny decreases the opportunity of going across quadratically. That need to offer you a suggestion of just how to show it for the various other system.

2019-12-03 01:29:33