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!
Answer
Quadratic variation and variance are two different concepts.
Let $X $ be an Ito process and $t\geq 0$.
Variance of $X_t$ is a deterministic quantity where as quadratic variation at time $t $ that you denoted by $[X,X]_t $ is a random variable.
What is confusing you is the fact that when $X $ is a martingale then $X^2_t-[X,X]_t$ is a martingale thus you have
$$E (X_t^2)=E ([X,X]_t)+E (X^2_0) $$
In the case where $X_0=0$ ( and thus $E (X_t)=0$ because $X $ is a martingale) You have $$Variance (X_t)=E ([X,X]_t) $$
In the general case, it is not true. Your example is the case where $X $ is not a martingale and thus it is not true.
Please comment for any further explanation.
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