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CIR Short-Rate Variance from Itô’s Isometry

Article Quant Q&A · Author: none
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Summary

The document examines how to derive the variance of the short rate in the Cox–Ingersoll–Ross model, whose diffusion volatility depends on the square root of the rate. One proposed derivation transforms the rate, applies Itô’s lemma to its square, and then uses the variance identity based on the first and second moments. The question asks whether variance can instead be obtained directly from the stochastic-integral representation of the rate.

The answer uses Itô’s isometry to convert the expected square of the stochastic integral into an integral involving the expected rate, then applies Fubini’s theorem to exchange expectation and integration. This establishes the direct variance calculation without assuming the rate and Brownian increment are independent. The discussion is a focused derivation; it does not cover parameter restrictions, bond pricing, or broader model calibration and applications.

Key ideas

  • The CIR short rate has a diffusion term proportional to the square root of the current rate.
  • Its stochastic-integral representation permits a direct variance derivation.
  • Itô’s isometry converts the expected squared stochastic integral into an integral of the expected rate.
  • Fubini’s theorem justifies exchanging expectation and time integration under the relevant conditions.
  • Independence between the rate and the Brownian increment is not required for this argument.

Tags

Fixed incomeStatisticsVolatility
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# Variance of the Cox-Ingersoll-Ross short rate


# Variance of the Cox-Ingersoll-Ross short rate












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.

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/30665

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*}

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