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3

The problem seems to be that you forgot the mean of the process. If $ds_t = rs_tdt + \sigma dW_t^\mathbb Q$, then the solution of the SDE is given by $$s_T = s_te^{r(T-t)} + \sigma\int_t^Te^{r(T-u)}dW^\mathbb Q_u.$$ Since the last integral is Gaussian, the distribution of the terminal price is given by $$s_T \sim\mathrm N\left(s_te^{r(T-t)}, \frac{\sigma^2}{...


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I think you have your argument slightly mixed up. Assuming the existence of a risk-neutral measure $Q$, you know from the risk-neutral pricing formula that the discounted bond price is a martingale, i.e. ($D(t)$ being the discount factor) $$D(t)p(t,T) = E_t^Q[D(T)p(T,T)] = E_t^Q[D(T)].$$ With $D(t) = \mathrm{exp}(-\int_0^t r(u) du)$, we immediately get $$p(t,...


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From my experience, in derivative trading space, fixing schedule is never continuous. The continuity distorsion of the product payoff (as proxy for fixing that occurs very frequently, daily, maybe hourly) was introduced by pricers hoping it would ease the PDE solving (making the underlying continuous average itself a third variable, along underlying and time)...


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No, $B$ is not a $Q$ martingale, neither is $1/B$, which you have assumed in your calculation (try using constant $r$, for an example of why this can be the case). The measure $Q$ is a risk neutral measure if the stock price processes that are discounted by $B$ are martingales.


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It would be an Asian strike call option, with the Asianing being computed over some period $[0;\tau]$. Not a big deal that the average is not computed from 0 to T. It even seems more natural to be done that way in practice. Before $\tau$, pricing would be similar to your option Asianing until T. For $t>=\tau$, you already know the value of the Asian ...


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