Ito Quadratic Variation

Question

In Ito's lemma for a Brownian motion $B_t$ the term in $dB_t^2$ is replaced with $dt$ without any averaging. It seems that higher moments are an order $dt$ smaller and that the term $dB_t^2$ is dominated by its expectation and it becomes deterministic and equal to $dt$.

I would like to demonstrate this by doing a numerical simulation. Now since $dB_t = g_t \sqrt{dt}$ where $g_t \simeq N(0,1)$ I could simply draw a value of $g_t$ and calculate $dB_t^2$. But such a term is random for all values of $dt$ and so does not demonstrate what I want to show.

Alternatively I can split the period $dt$ up into $N$ periods of length $dt/N$. Will this work ?

Answer 1

You seem to be confused by some terminology; I'll try to explain briefly.

Ito's lemma says that (in the context of a one-dimensional brownian motion; it is actually more general, see below):

$$f(B_t) = f(B_0) + \int_0^t f'(B_s) dB_s + \frac 12 \int_0^t f''(B_s) ds$$ for $f \in C^2(\mathbb R, \mathbb R)$. As you can see, the term $dB_s^2$ is nowhere to be seen.

It is useful now to define the quadratic variation of certain stochastic processes:

Let $X_t$ be a square integrable local martingale. Then the quadratic variation $\langle X\rangle_t$ is the unique, right-continuos, increasing adapted process starting at $0$ such that $X^2_t - \langle X\rangle_t$ is a local martingale. It can be shown that $\langle X\rangle_t$ arises as the quadratic variation of the trajectory of $X_\cdot$ along certain partitions, hence the name. One can additionally show that $\langle B \rangle_t = t$. Finally we also define $\langle M, N \rangle $ in a similar way; such that $M_tN_t - M_0 N_0 - \langle M, N \rangle_t $ is a local martingale. Note that $\langle X, X \rangle_t = \langle X \rangle _t$

Why is this important? Because the of the version of Ito's lemma for continuos semimartingales: Let $X_t$ be a continuos $d$-dimensional semimartingale, then

$$f(X_t) = f(X_0) +\sum_{i=1}^d\int_0^t f_{i}(X_{s})\,dX^i_s + \frac{1}{2}\sum_{i,j=1}^d \int_0^t f_{i,j}(X_{s})\,d\langle X^i,X^j\rangle_s $$

for $f \in C^2(\mathbb R^d, \mathbb R)$

If $d=1$ and $X_t = B_t$ a brownian motion, then using the fact that $\langle B\rangle_t = t$ one recovers the previous formula.

Another use of quadratic variation is the Ito isometry:

$$E\left[\left(\int_0^t Y_s dX_s\right)^2\right] = E\left[\int_0^t Y_s^2 d\langle X\rangle _s \right]$$. Note that for $X = B$ we get

$$E\left[\left(\int_0^t Y_s dB_s\right)^2\right] = E\left[\int_0^t Y_s^2 ds \right]$$

So as you can see $dB^2_s$ is not present in the above discussion.

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Why did you hear then that $dB_s^2 = ds$? This is a simplification and stems from different facts (namely, that $E[B^2_s] = s$ and $\langle B\rangle _s = s$). Let $X_t$ be a Ito diffusion process:

$$dX_t = b(t) dt + \sigma(t) dB_t$$ Using $dB^2_s = ds$ and $dB_s ds = 0$ you get

$$dX^2_t = \sigma^2(s) ds$$

And then you can "say" that the variance of $X^2_t$ is $\int_0^t \sigma^2(s) ds$, which is correct (as you can see properly from Ito's formula). So $dB_s^2 = ds$ can be useful but shouldn't be confused with the rigorous definition of quadratic variation.

Answer 2

Expanding a bit on Ant's last point so that your intuition might be satisfied:

We know that $E[dB_t^2] = dt$, but you can also easily show that $\text{var}[dB_t^2] = 2dt^2$ by noting that $\frac{B_t^2}{t}$~$\chi^2(1)$. Any RV with zero variance is a constant, so we have $dB_t^2$~$dt$, in exactly the same way that you have $dt^2$~$0$ (i.e. in the limit as $dt \to 0$).