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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?

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    $\begingroup$ This is basically Ito's isometry. $\endgroup$
    – Gordon
    Apr 6, 2015 at 17:42
  • $\begingroup$ @Gordon that is the first equality, right? $\endgroup$
    – Monolite
    Apr 6, 2015 at 18:04
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    $\begingroup$ That is correct. But the notation here is bit sloppy, the square $r^2_{t+\tau}$ within the first integral should be the quadratic variation. But in many book, such as that of John Hull, this sloppy notion is used. $\endgroup$
    – Gordon
    Apr 6, 2015 at 18:30
  • $\begingroup$ Thanks could you even tell me why we can exchange $r_{t+\tau}^2$ with $\sigma_{t+\tau}^2$ in the second equality? I know it is a popular approximation but why can we keep equality in this case? $\endgroup$
    – Monolite
    Apr 6, 2015 at 18:33

1 Answer 1

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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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  • $\begingroup$ Thanks now I understand the notation issue you where talking about. $\endgroup$
    – Monolite
    Apr 6, 2015 at 19:35

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