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4

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 ...

0

The average of the exponentials is not the exponential of the average. It is always higher due to convexity (Jensen inequality). So there is no contradiction between the average of $X_T$ being negative and the average of $S_T$ being $S_0$. So the question is: are your results really significantly different from what you would expect? Have you tried ...

3

The second method you use is correct and, actually, is completely equivalent to the first one. The reason is that the proof of Ito's lemma relies on a Taylor expansion of the second order. Notice that Wikipedia's formulation of Ito's lemma is a bit misleading, as they write $$dX_t = \mu_t dt + \sigma_t dB_t$$ but, actually, the functions $\mu$ and ...

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