# Negativity in a CIR model discretized by Ito-Taylor expansion

Let $X = (X_t: t \in [0,T])$ be a stochastic process satisfying a CIR model $$dX_t = \beta (X_t - \gamma) dt + \sigma\sqrt{X_t} dB_t,$$ where $B_t$ is a standard Brownian motion, $\beta$ is a negative constant, $\gamma, \sigma$ are positive constants. In order for the SDE to make sense, assume that $X_t > 0$ for all $t \in [0,T]$.

Consider following two ways to simulate the model based on discretization of $t$ with Ito-Taylor expansion:

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

I was wondering why these two schemes have a positive probability of generating negative values of $X_t$ and therefore cannot be used without suitable modifications?

References (book, tutorial and/or paper) will be helpful too!

Thanks and regards!

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I'd say that it's all in $Z$. Suppose your initial value for $X_t$ is very close to zero, if you happen to draw a large $|Z|$ with $Z<0$, you can get negative values. – Raskolnikov Dec 9 '10 at 15:07
Thanks! I think it is the right way to go. – steveO Dec 10 '10 at 17:31

Take the Euler scheme. Let's assume that you start with $X_0=\epsilon$. Let's take $\beta=-1$, $\gamma =1$ and $\sigma=1$ for the sake of definiteness. Then in the next step you'll have

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

What's the probability that this is smaller than zero?

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

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 an estimate this is smaller than

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

So the closer you are to zero, the higher the probability of crossing that line. But taking time steps sufficiently small lowers the chance of crossing quadratically. That should give you an idea of how to show it for the other scheme.

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Thanks! Nice to know more! – steveO Dec 10 '10 at 17:32