In stochastic calculus, why do we have $(dt)^2=0$ and other results?

Question

I'm doing actuarial problems of Exam MFE and it covers some of the stochastic calculus (like Ito's Lemma). One of the frequently used results are the so-called "multiplication rules":

$(dt)^2=0$

$dZ(t)^2=dt$

$dZ(t) \, dt=0$

I tried to do some research online. There are tons of papers providing introduction to stochastic calculus, but strangely all of them seem to take these three rules as granted instead of proving them. Does anyone know the proof of these rules? Thanks.

Update: Someone mentioned that one needs to understand some analysis and measure theory to understand that. If someone could recommend relative textbooks on measure theory (I've taken analysis) and stochastic calculus, I would really appreciate it. Thanks.

Answer 1

Others already told you that a proper proof is more involved. However I want to point out, that there is a bit of intuition behind those rules. For me, having seen the way ordinary calculus is treated by physicists, the following is rather insightful:

Think of $dt$ as a really small increase in time $dB_t$ as the change the Brownian motion does in this small time increase and so on. Then the change of a quantity $X_t$ in some time interval is just the sum of all such small changes $dX_t$, or bending notation, the integral. In a way this is also what you do when defining the stochastic integral.

Using this, (dt)^2 is simple. If you separate your interval in about $\sim N$ steps, then $dt \sim \frac{1}{N}$. So $(dt)^2 \sim \frac{1}{N^2}$. Yet if you sum $N$ steps of size $\frac{1}{N^2}$, in the limit $N\to \infty$, you will end up with nothing. So $(dt)^2$ has no effect.

Now $dB_t$ is a strange beast. Comparatively it is of size $\sqrt{dt}$ (have a look at the scaling of the probability density in $x$ and $t$), so by the reasoning above, the integral should explode. Yet it changes sign all the time and those changes cancel each other quite nicely in the end.

On the other hand $(dB_t)^2 \sim \sqrt{dt}^2 = dt$ and has a fixed sign, so the calculation rule at least is not surprising. (That it is actually equal, however, is a not so trivial matter.)

The last one, $dB_t dt$ again is simple, as this is of order $\sim (dt)^{3/2}$, which allows us to use a similar argument to $(dt)^2$.

I should again stress, that this is only intuition, which cannot prove anything, only help in finding what to prove. For example using this, Ito's formula looks just like a Taylor expansion in $dB_t$ and $dt$, where you throw away the terms of order larger than $dt$.

Answer 2

For stochastic processes of the form $ dX_t = \theta_tdt+ K_tdB_t$ and $ dY_t = \gamma_tdt+ L_tdW_t$ where $B_t$ and $W_t$ are two correlated Brownian motions with correlation coefficient $\rho$ you have $d<X,Y>_t = \rho K_tL_tdt$ (for more details see convergence of quadratic variation and covariation of stochastic processes).

Examples you mentioned above are particular cases :

• $(dt)^2 = 0$ because your process have no diffusion i.e, $X(t)=t$ meaning $K_t=0$ and $\theta_t=1$ $\forall t$ and $d<X,X>_t=(dt)^2=0$.

• $dZ(t)^2 = dt$ you took $X(t)=B(t)$ where $B(t)$ is a B.M this gives $K_t=1$ and $\theta_t=0$ $\forall t$ and thus you obtain $d<X,X>_t=dt$ (notice that the correlation between a process and itself is 1).

• for the third example $X(t)=t$ and $Y(t)=W(t)$ this gives $K_t=O$ and $L_t=1$ and you get the result.

For all processes $X_t$ with bounded variation and any other stochastic process $Y_t$ you have $d<X,X>_t=d<X,Y>_t=0$

Answer 3

Other's have better answers, but in a simpler way:

$(dt)^2 ≈ 0$

Calculus looks at changes during small time steps $dt$ which when tiny, $(dt)^2 << dt$, so is approximated to be $0$. $[0.001 >> (0.001)^2 = 0.000001]$.

$dZ(t)^2 = dt$

To do with the 'scaling' chosen get a nice process. Out of three cases, $dZ(t)^2 \approx O(dt)$ provides a stable process, so we use $dZ(t)^2 = dt$.

[The two discarded cases are

• $dZ(t)^2 \rightarrow 0$ quicker than $dt$ does - process collapses to $0$,
• $dt \rightarrow 0$ quicker than $dZ(t)^2$ does - process grows indefinitely

.]

$dZ(t)dt=0$

uses the first and second points. Pretty much anything $< dt ≈ 0$, and we know $dZ(t) = \sqrt{dt}$ (from $dZ(t)^2 = dt$), therefore $dZ(t)dt < dt$.

EDIT: I think this is actually a botched answer more related to why we pick what we do when going from the BSE to the Kolmogorov equation.