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I was wondering if Ito integral:

$\int_0^T B(t)dB(t) $

is Gaussian (in which B(t) is Brownian Motion)??

Thank you so much, I appreciate any help ^^

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$d(B_t^2) = 2B_t dB_t + dt$. Therefore your integral is $\frac12(B_T^2-T)$. $B_T$ is Gaussian $N(0,\sqrt T)$, therefore $B_T^2$ is $T$ times a $\chi^2_1$.

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As a simple and brief description, $\int_0^T B(t)dB(t)=\displaystyle\sum\limits_{j=0}^{n-1}B(t_j)\Delta B(t_j)$ you can expand sigma as below $$\sum\limits_{j=0}^{n-1}B(t_j)\Delta B(t_j)=\\ \sum\limits_{j=0}^{n-1}B(t_j)(B(t_{j+1})- B(t_j))=\\ \frac12\sum\limits_{j=0}^{n-1}2B(t_j)B(t_{j+1})- 2B^2(t_j)=\\ \frac12\sum\limits_{j=0}^{n-1}B^2(t_{j+1})-B^2(t_j)-B^2(t_{j+1})+2B(t_j)B(t_{j+1})- B^2(t_j)=\\ \frac12\sum\limits_{j=0}^{n-1}B^2(t_{j+1})-B^2(t_j)-(B^2(t_{j+1})-2B(t_j)B(t_{j+1})+B^2(t_j))=\\ \frac12\sum\limits_{j=0}^{n-1}B^2(t_{j+1})-B^2(t_j)-\frac12\sum\limits_{j=0}^{n-1}(B^2(t_{j+1})-2B(t_j)B(t_{j+1})+B^2(t_j))=\\ \frac12\sum\limits_{j=0}^{n-1}(B^2(t_{j+1})-B^2(t_j))-\frac12\sum\limits_{j=0}^{n-1}(B(t_{j+1})-B(t_j))^2=\\ \underbrace{\frac12\sum\limits_{j=0}^{n-1}(B^2(t_{j+1})-B^2(t_j))}_{telescopic}-\frac12\sum\limits_{j=0}^{n-1}(\underbrace{B(t_{j+1})-B(t_j)}_{\Delta B(t_j)})^2=\\ \frac12(B^2(T)-B^2(0))-\frac12\underbrace{(\Delta B(t_0)^2+\Delta B(t_1)^2+\cdots+\Delta B(t_{n-1})^2)}_{(\Delta t_0+\Delta t_1+...+\Delta t_{n-1})=T-0}=\\ \frac12(B^2(T)-B^2(0))-\frac12 T$$ if $B(0)=0$ then $$ \int_0^T B(t)dB(t)=\displaystyle\sum\limits_{j=0}^{n-1}B(t_j)\Delta B(t_j)=\frac12 B^2(T)-\frac12 T$$

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