$X_t = e^{B_t-\frac{1}{2}t^2}$

I need to find $[X]_t$, the quadratic variation process. I have tried to solve the problem and my main question is whether this approach is correct or not.

Given that

$$ \begin{equation} \begin{split} dX_t=&-tX_tdt+X_tdB_t+\frac{1}{2}X_td[B]_t \\ =&\left(\frac{1}{2}-t\right)X_tdt+X_tdB_t \end{split} \end{equation} $$.

$[X]$ is the unique process such that $X^2-[X]$ is a martingale.

$d( X^2_t)=2X_tdX_t+d[X]_t$

How do I compute $d[X]_t$ more rigorously than saying $dX_tdX_t=X_t^2dt$ because $dt\cdot dt=0$, $dt\cdot dB_t=0$, $dB_t\cdot dB_t=t$?

Now integrate from $0$ to $t$ and substitute in $dX_s$ (above)

$$ \begin{equation} \begin{split} X^2_t-X^2_0=&2\int_0^tX_sdX_s+\int_0^tX_s^2ds \\ =&\int_0^t X_s^2 \left(1-2s\right)ds+2\int_0^t X_s^2dB_s + \int_0^t X^2_s ds\\ &=2\int_0^t X_s^2 \left(1-s\right)ds + 2\int_0^t X_s^2 dB_s \end{split} \end{equation} $$

So it would seem that since the stochastic integral $2 \int_0^t X^2_s dB_s$ is a martingale, by uniqueness and since $X_0^2 = 1$, we have that $[X]_t=1+2\int_0^t X_s^2\left( 1-s \right) ds.$ Is this correct?

  • 1
    $\begingroup$ There are some errors in the computations, starting with $$X^2_t-X^2_0=2\int_0^tX_sdX_s+\int_0^tX_s^2dB_s,$$ which should read $$X^2_t-X^2_0=2\int_0^tX_sdX_s+\int_0^tX_s^2ds.$$ To compute $[X]$, indeed note that, if $$dX_t=a(X_t)dB_t+b(X_t)dt,$$ then $$d[X]_t=a(X_t)^2dt.$$ $\endgroup$
    – Did
    Jan 11, 2016 at 0:50
  • $\begingroup$ @Did Shouldn't it read $$X_t^2 - X_0^2 = 2 \int_0^t X_s \, dX_s + \int_0^t \, d\langle X \rangle_s$$....? How do you come up with the $\int_0^t X_s^2 \, ds$-term? (It might well be that I'm missing something obvious; I just got up ;).) $\endgroup$
    – saz
    Jan 11, 2016 at 7:14
  • $\begingroup$ @saz according to $d\langle X\rangle_s=a(X_s)^2 ds$ if we have $dX_s=a(X_s)dB_s+b(X_s)ds$ then with $dX_s=(1/2-s)ds+X_sdB_s$ we get $\int_0^tX_s^2ds$ for $\int_0^t \langle X\rangle_s$ right? $\endgroup$
    – shilov
    Jan 11, 2016 at 8:57
  • $\begingroup$ @shilov Ah, you do it this way... I see; thanks. $\endgroup$
    – saz
    Jan 11, 2016 at 9:31
  • $\begingroup$ @saz Oh perhaps you can get an expression for $\langle X \rangle_s$ from computing $$ \int_0^t d\langle X \rangle_s = X_t^2-X_0^2-2\int_0^tX_sdX_s $$ Is that what you thought I would have done? $\endgroup$
    – shilov
    Jan 11, 2016 at 9:35

1 Answer 1


The trouble is that there are two types of brackets, $\langle X \rangle_t$ (which I call angle bracket) and $[X]_t$ (sharp/square bracket). In this particular case they do not coincide because $(X_t)_{t \geq 0}$ is not a martingale (see also this answer).

$[X]$ is the unique process such that $X^2-[X]$ is a martingale.

$d(X_t^2) = 2X_t \, dX_t + d[X]_t$

These two characterizations are only equivalent if $(X_t)_{t \geq 0}$ is a martingale. For a semimartingale (with continuous sample paths) the sharp bracket $[X]_t$ is defined as

$$[X]_t := X_t^2-X_0^2 -2 \int_0^t X_s \, dX_s. \tag{1}$$

As long as $(X_t)_{t \geq 0}$ is not a martingale, we cannot expect that the process

$$X^2_t-[X]_t = X_0^2 + 2 \int_0^t X_s \, dX_s$$

is a martingale.

For your particular example, we obtain from Itô's formula that

$$X_t^2-X_0^2 = 2 \int_0^t X_s \, dX_s + \int_0^t X_s^2 \, ds,$$

so that, by $(1)$,

$$[X]_t = \int_0^t X_s^2 \, ds.$$ More generally, if $(X_t)_{t \geq 0}$ is an Itô process of the form

$$dX_t = b(t) \, dt + \sigma(t) \, dB_t,$$


$$[X]_t = \int_0^t \sigma^2(s) \, ds.$$


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