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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 ...


Perhaps this works : http://en.wikipedia.org/wiki/Geometric_standard_deviation In particular, see under "Derivation"


For a random variable $\xi$, the variance is defined by $$mean\Big(\big(\xi -mean (\xi)\big)^2\Big).$$ Then the geometric variance should be defined by $$\prod_{i=1}^n\Bigg(1+ \bigg[x_i-\prod_{j=1}^n(1+x_j)^{1/n}\,\bigg]^2\, \Bigg)^{1/n}.$$ Addendum ---- The definition given in the link below is only a way of thinking. However, it does not provide a ...

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