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Consider the Cox-Ingersoll-Ross (CIR) interest rate model: $\displaystyle d r_t = \kappa (\theta - r_t) \, d t + \sigma \sqrt{r_t} \,d W_t$ where $\kappa$, $\theta$, $\sigma$ are positive constants and $W_t$ is a standard Brownian motion.

A solution to the CIR stochastic differential equation is given by: $$ r_t = \theta + (r_0 - \theta) e^{-\kappa t} + \sigma e^{-\kappa t} \int_0^t e^{\kappa u} \sqrt{r_u} \,d W_u.$$

Since $\displaystyle d W_u \sim N(0, du)$ it is easy to derive that:

$$\mathrm{E}\left[r_t\right] = \theta + (r_0 - \theta) e^{-\kappa t}.$$

However, I am stuck on deriving the variance. I got as far as below. Anyone know how to solve for the variance?

Thanks.

My derivation so far:

$$ \operatorname{Var}\left[r_t\right] = \mathrm{E}\left[\left\{r_t - \mathrm{E}\left(r_t\right)\right\}^2\right] = \sigma^2 e^{-2 \kappa t} \mathrm{E}\left[\left\{\int_0^t e^{\kappa u} \sqrt{r_u} d W_u\right\}^2\right].$$

What to do with the square of that integral? The solution I found states:

$$ \sigma^2 e^{-2 \kappa t} \mathrm{E}\left[\left\{\int_0^t e^{\kappa u} \sqrt{r_u} \,d W_u\right\}^2\right] = \sigma^2 e^{-2 \kappa t} \int_0^t e^{2 \kappa u} \mathrm{E}\left(r_u\right) \,d u$$

and continues onwards from here. But I don't get how you can take the square inside the integral.

Michael Hardy
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Barbados
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  • Hint: if $M$ is a martingale, $M^2 - \langle M\rangle$ is amartingale. Therefore the expression inside the expectation equals to? – Lost1 Mar 20 '13 at 12:03
  • @Lost1 : Could you write $\langle M\rangle$ instead of $$? – Michael Hardy Mar 20 '13 at 12:06
  • @MichaelHardy teach me, I don't know how :( – Lost1 Mar 20 '13 at 12:06
  • @Lost1 : It's \langle M\rangle. – Michael Hardy Mar 20 '13 at 12:07
  • @MichaelHardy awesome – Lost1 Mar 20 '13 at 12:07
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    @Barbados : You'll notice that in four places where you used \displaystyle I changed it to an actual display (so display style is automatically used) and in one place I left it as \displaystyle since you had it in an inline setting. That one I might have set in display, but I've left it as you had it. I wonder why you use \displaystyle rather than an actual display when you have the whole thing on a line by itself? – Michael Hardy Mar 20 '13 at 12:09

1 Answers1

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The last equality $$ \sigma^2 \mathrm{e}^{-2 \kappa t} \mathrm{E}\left[\left\{\int_0^t \mathrm{e}^{\kappa u} \sqrt{r_u} \,d W_u\right\}^2\right] = \sigma^2 \mathrm{e}^{-2 \kappa t} \int_0^t \mathrm{e}^{2 \kappa u} \mathrm{E}\left(r_u\right) \,d u $$ follows by Ito isometry property of Ito integrals.

Sasha
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