Suppose we have a Brownian Motion $BM(\mu,\sigma)$ defined as
$X_t=X_0 + \mu ds + \sigma dW_t$
The quadratic variation of $X_t$ can be calculated as
$dX_t dX_t = \sigma^2 dW_tdW_t = \sigma^2 dt$
where all lower order terms have been dropped, therefore the quadratic variation (also the variance of $X_t$)
$[X_t,X_t]=\int_0^t \sigma^2 ds=\sigma^2 t$
I was trying to use the same tech solve the problem posted in Integral of Brownian Motion w.r.t Time
If I start as differential form $dX_t = W_tdt$ and calculate $dX_t dX_t$. After drop all lower order terms, I have $dX_tdX_t=0$. This means the quadratic variation is zero. Hence we have the variance is zero?
I understand this isn't correct. But I really want to know what, prevents me doing this problem as previous one?
I am pretty new to SDE and any help will be appreciated! Thanks a lot!