3
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ This is basically Ito's isometry. $\endgroup$
    – Gordon
    Apr 6 '15 at 17:42
  • $\begingroup$ @Gordon that is the first equality, right? $\endgroup$
    – Monolite
    Apr 6 '15 at 18:04
  • 2
    $\begingroup$ 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 '15 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 '15 at 18:33
4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks now I understand the notation issue you where talking about. $\endgroup$
    – Monolite
    Apr 6 '15 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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