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This is a special case of the question of why $$ \int_0^T f(t) dW_t $$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$ \sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point ...


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For simplicity, We assume that $\alpha$ is a positive constant. You need to show that, for any $t>0$, \begin{align*} \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed. Consider the process $\{X_t, t \geq 0\}$, where \begin{align*} X_t = \frac{1}{\sqrt{\frac{1}{t}\int_0^t e^{2\alpha u} du}}\int_0^t e^{\alpha u} dW_u, \end{align*} for ...



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