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!


1 Answer 1


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.

  • 1
    $\begingroup$ Thank you for great explanation! I have another question, for $X_t = X_0 + \mu dt + \sigma dW_t$ (BM with a drift term), which is apparently not a martingale. As to this case, we end up with quadratic variation is the same as variance, is this just a coincide? Or, something I have done in my calculation is incorrect. Thanks again! $\endgroup$ Jan 10, 2017 at 16:08
  • 2
    $\begingroup$ It is because your finite variation part is deterministic. Let $X_t = A_t + M_t$ where $A_t$ is deterministic and $M$ is a martingale starting from $0$, then you have again $Var(X_t)=Var(M_t)=E([M,M]_t)$ $\endgroup$
    – MJ73550
    Jan 10, 2017 at 16:23

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