Ito’s Isometry Links Expected Squared Returns to Integrated Variance

Summary

For a continuous-time martingale whose price changes are driven by volatility times Brownian motion, the document explains how to express the conditional expected squared one-day return. Ito’s isometry equates the expected square of the stochastic integral with the expected integral of instantaneous variance. Shifting the integration interval to run from zero to one gives the expression in terms of volatility at future offsets from the current time; conditional expectation can then be moved inside the time integral.

The answer also flags a notational problem in the original derivation: treating the squared return at each instant as an ordinary time-integrable quantity is imprecise. The more careful formulation uses the quadratic variation of the process. The result assumes the stated continuous-time martingale model and concerns conditional expected squared returns, rather than providing an empirical volatility estimator or a claim about realized returns in general.

Key ideas

• Ito’s isometry relates the expected square of a stochastic integral to integrated variance.
• The integration variable can be shifted to express the interval as offsets from the current time.
• Under suitable conditions, conditional expectation can be interchanged with the time integral.
• Quadratic variation provides a more precise formulation than integrating squared instantaneous returns.

Tags

# On an application of Ito's lemma


# On an application of Ito's lemma












Assume that instantaneous returns are generated by the continuous time martingale:

$$dp_t = \sigma_t dW_t$$

where $W_t$ denotes a standard Weiner process and One day returns are denoted by $r_{t+1} = p_{t+1} - p_t$. Then By Ito's lemma we have:

$$E_t (r_{t+1}^2) = E_t \Bigg( \int_0^1 r_{t + \tau}^2 d \tau \Bigg) = E_t \Bigg( \int_0^1 \sigma_{t + \tau}^2 d \tau \Bigg) = \int_0^1 E_{t} \Bigg( \sigma_{t + \tau}^2 \Bigg) d \tau $$

where $E_t$ denotes conditional expectation at time t.

I am very rusty with Ito's lemma applications and do not seem to recall where the $d \tau$ comes up from. Would anybody mind explaining these 3 equalities?

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/17270

Based on Ito's isometry, \begin{align*} E_t (r^2_{t+1}) &= E_t \bigg(\int_t^{t+1} \sigma_s dW_s \int_t^{t+1} \sigma_s dW_s\bigg)\\ &= E_t \bigg(\int_t^{t+1} \sigma_{\tau}^2 \,d\tau\bigg) \\ &= E_t\bigg(\int_0^1 \sigma_{\tau+t}^2 \,d\tau\bigg) \\ &=\int_0^1 E_t\big(\sigma_{\tau+t}^2\big) \,d\tau. \end{align*} The identity \begin{align*} E_t (r^2_{t+1}) &= E_t\bigg(\int_0^1 r_{\tau+t}^2 \,d\tau\bigg) \end{align*} is sloppy. It is better to write as \begin{align*} E_t (r^2_{t+1}) &= E_t\bigg(\int_0^1 d\langle r_{\tau+t}, r_{\tau+t}\rangle\bigg), \end{align*} where $\langle r_{\tau+t}, r_{\tau+t}\rangle$ is the quadratic variation.

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