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Hope you will not mind if I place myself in continuous time. The discounted stock price at $T$ is $e^{-rT}S_T$. As you know that it is a martingale, you have that $\mathbf{E}^{\mathbf{P}}[e^{-rT}S_T | \mathscr{F}_t] = e^{-rt} S_t$ when $t\leq T$ which you can rewrite as $\mathbf{E}^{\mathbf{P}}\left[\frac{e^{-rT}S_T}{e^{-rt} S_t} | \mathscr{F}_t\right] = 1$ ...


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As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: \begin{equation} d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du}) \end{equation} For the first two terms on R.H.S, we have: \begin{equation} d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + ...


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$F=0$ seems like a good choice.


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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 ...


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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 ...



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