Ito's lemma $f(t,W_t^2)$

Let $$f$$ be a function of $$t$$ and $$W_t^2$$.

a)Find a function $$f$$ such that $$f(t,W_t^2)$$ is a $$F_{t^-}$$ martingale, with $$F$$ the Brownian filtration.

b)Use Ito's lemma to show that $$f(t,W_t^2)$$ is a process with zero drift.

My attempt for first part, I got $$f(t,W_t^2)=W_t^2-t$$.

For the second part I know I'm supposed to use $$df(t,W_t)=(a\frac{\delta f}{\delta W_t}+\frac{1}{2}b^2\frac{\delta^2f}{\delta W_t^2}+\frac{\delta f}{\delta t})dt+b\frac{\delta f}{\delta W_t}dW_t$$

May I know how to determine the $$a$$ and $$b$$? From the marking scheme I see that it's $$a=0$$ and $$b=1$$. But how? Thank you in advance.

As stated here, for $$f = f(t, x) ∈ C^{1,2}(\mathbb{R}^2)$$ a deterministic function and Ito process $$X_t = W_t^2,$$ the stochastic process $$Y_t = f(t,X_t)$$ is an Ito process and we have $$df (t,X_t) = \partial_tf(t,X_t)\,dt + \partial_xf(t,X_t)\,dX_t + \frac{1}{2} \partial_{xx}^2f(t,X_t)(dX_t)^2.$$

Since $$dX_t = 2W_t dW_t + dt$$ and $$(dX_t)^2 = 4X_t dt,$$ we have

$$df (t,X_t) = \left(\partial_tf(t,X_t) + 2X_t \partial_{xx}^2f(t,X_t) +\partial_xf(t,X_t) \right)\,dt +2\partial_xf(t,X_t)W_t dW_t$$

So, to make $$f(t,X_t) = f(t,W_t^2)$$ martingale, all we need is deterministic functions $$f=f(t,x)$$ such that $$\partial_tf(t,x) + 2x\partial_{xx}^2f(t,x) +\partial_xf(t,x) = 0,$$

for all $$t$$ and $$x$$, which reduce the SDE to:

$$df (t,X_t) = 2\partial_xf(t,X_t)W_t dW_t$$

$$f(t,x)= x- t$$

and $$(\partial_xf)(t,x) = 1$$, so $$(\partial_xf)(t,X_t) = (\partial_xf)(t,W_t^2) = 1$$

Note 2: Another example (to bring in a non-zero second derivative in $$x$$) is:

$$f(t,x) = x^2-6xt +3t^2$$

Here, $$(\partial_xf)(t,x) = 2x-6t$$, so $$(\partial_xf)(t,X_t) = (\partial_xf)(t,W_t^2) = 2W_t^2 -6t$$.

(Example inspired by Hermite polynomials - fourth one, $$H_4(t,x) = x^4-6x^2t+3t^2$$ - which we know produce martingales.)

Let $$f(t,W_t)=W_t^2-t$$, then it is easier to derive the dynamics using the "general formula" for Itô's lemma (reference, see eq. 10):

$$df(t,W_t)=\frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial W_t} dW_t + \frac{1}{2}\frac{\partial^2f}{\partial W_t^2} dW_t^2$$

where,

$$\frac{\partial f}{\partial t} = -1, \qquad \frac{\partial f}{\partial W_t} =2W_t, \qquad \frac{\partial^2f}{\partial W_t^2} = 2.$$

Therefore we observe that:

\begin{align} df &= -1 \: dt + 2W_t \: dW_t + \frac{1}{2} \cdot 2 \: dW_t^2\\ &=- dt + 2W_t \: dW_t + dt\\ &=2W_t \: dW_t , \end{align} using that Brownian motions have finite quadratic variation equal to time-scale, ie. $$dW_t^2=dt$$. As seen above, the process has zero drift.

• Note that $f(t,W_t) \not= f(t,W_t^2)$.
– ir7
Jul 12, 2021 at 19:43