New answers tagged brownian-motion
1
Normality of returns follows from the fact that the Brownian increments are normal distributed (property 3), $W(t_i)-W(t_{i-1}) \sim N(0,t_i-t_{i-1})$, and furthermore that $t_i-t_{i-1}=T(\frac{i}{n}-\frac{i-1}{n})=\frac{T}{n}$. See that:
\begin{align}
r(t_i)&=p(t_i)-p(t_{i-1})\\
&=\left[p(0)+\mu\cdot t_i + \sigma W(t_i) \right] - \left[p(0)+\mu\cdot ...
3
That looks correct, but a bit complicated. We know that under Black-Scholes with no dividends, $E^Q(S_t) = Forward = S_0 e^{rt}$
$e^{-rT}E^Q(\int_0^TS_tdt) = e^{-rT}\int_0^TE^Q(S_t)dt
\\
= e^{-rT}\int_0^T S_0 e^{rt} dt = S_0 e^{-rT}\int_0^T e^{rt} dt
\\
= S_0 e^{-rT} \frac{1}{r}(e^{rT} - 1) = S_0\frac{1-e^{-rT}}{r}$
It is straightforward to generalize to the ...
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