1
$\begingroup$

Shreve II page 151, the Cox-Ingersoll-Ross model is defined as $$dr_t=(\alpha-\beta r_t)dt+\sigma\sqrt{r_t}dW_t$$ By applying Ito's Lemma, we obtain \begin{align} r_t&=r_0e^{-\beta t}+\frac{\alpha}{\beta}(1-e^{-\beta t})+\sigma e^{-\beta t}\int_0^te^{\beta u}\sqrt{r_u}dW_u\\ &=\frac{\alpha}{\beta}+\Big(r_0-\frac{\alpha}{\beta}\Big)e^{-\beta t}+\sigma e^{-\beta t}\int_0^te^{\beta u}\sqrt{r_u}dW_u \end{align} Now for the variance of $r_t$, Shreve suggests that we set $$X_t=e^{\beta t }r_t$$ and apply Ito's Lemma to obtain $dX_t$, after which $d(X_t^2)$ may be found. Then $d(X_t^2)$ is integrated to obtain $X_t^2$, from which $r_t^2$ is found. Finally, the variance is derived from $$Var(r_t)=E(r_t^2)-(E(r_t))^2$$ My question is, why not take the variance of $r_t$ immediately, that is \begin{align} Var(r_t)&=\sigma^2 e^{-2\beta t}\int_0^te^{2\beta u}E(r_u)du\\ &=\sigma^2 e^{-2\beta t}\int_0^te^{2\beta u}\Big(\frac{\alpha}{\beta}+\Big(r_0-\frac{\alpha}{\beta}\Big)e^{-\beta u}\Big)du \end{align} and from here the integration is simple. It yields the same result as Shreve's method.

One possibility is that I have assumed $E(r_u(dW_u)^2)$=$E(r_u)E((dW_u)^2)$, implying independence between $r_t$ and $(dW_t)^2$.

Any help is appreciated.

$\endgroup$
3
  • $\begingroup$ That is just a different approach. Did you verify that the results are the same? $\endgroup$
    – Gordon
    Oct 21, 2016 at 15:29
  • $\begingroup$ Yes, verified. I didn't type it here as the only possible concern was $E(r_u(dW_u)^2)$=$E(r_u)E((dW_u)^2)$. $\endgroup$
    – none
    Oct 21, 2016 at 15:53
  • 1
    $\begingroup$ The independence assumption is not needed; see answer below. $\endgroup$
    – Gordon
    Oct 21, 2016 at 16:10

1 Answer 1

3
$\begingroup$

The independence assumption is not needed. In fact, based on Ito's isometry and the Fubini theorem, \begin{align*} Var(r_t) &= E\left((r_t-E(r_t))^2 \right)\\ &=\sigma^2 e^{-2\beta t} E\left(\left(\int_0^te^{\beta u}\sqrt{r_u}dW_u\right)^2 \right)\\ &=\sigma^2 e^{-2\beta t} E\left(\int_0^te^{2\beta u} r_u du \right)\\ &=\sigma^2 e^{-2\beta t}\int_0^t e^{2\beta u}E(r_u) du. \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.