(I am an undergrad in Econ and new to this forum, so I'm sorry if this will be easy for you guys)

Im currently struggling with Stochastic Calculus, resp. Itô's Lemma. I understand that once we have an Itô Process $X_t$ that satisfies the Stochastic Differential Equation, we have to find a twice differentiable function. I posted an example, and wanted to ask if anyone knows how to get the:

$$Z_t := f(t, W_t) := t \cdot W_t$$

Thanks In advance, any form of help is very appreciated

  • $\begingroup$ Welcome to Math.SE! Please format your math using MathJax. Additionally, people will be more likely to help you if you show what you have tried. $\endgroup$ – csch2 Apr 11 at 13:22
  • $\begingroup$ $f(t,w)=tw$ is twice differentiable. It is not too farfetched that the integral $\int W_tdt$ or $\int tdW_t$ results in one term $tW_t$ plus others according to the Ito formula. $\endgroup$ – LutzL Apr 11 at 14:11

Itô's formula states that

$$f(t,W_t)-f(0,W_0) = \int_0^t \partial_x f(s,W_s) \, dW_s + \int_0^t \left( \frac{1}{2} \partial_x^2 f(s,W_s) + \frac{\partial}{\partial s} f(s,W_s) \right) \, ds \tag{1}$$

for any (nice) function $f$. Equivalently,

$$\int_0^t \partial_x f(s,W_s) \, dW_s = f(t,W_t)- f(0,W_0) - \int_0^t \left( \frac{1}{2} \partial_x^2 f(s,W_s) + \frac{\partial}{\partial s} f(s,W_s) \right) \, ds. \tag{2}$$

Now let's consider the stochastic integral $\int_0^t s \, dW_s$. If we can find a function $f$ such that

$$\partial_x f(s,x) = s \tag{3}$$

then the left-hand side of $(2)$ is nothing but $\int_0^t s \, dW_s$, i.e. exactly the stochastic integral which we are looking for. Integrating the equation $(3)$ with respect to $x$ yields

$$f(s,x) = sx + c$$

for some constant $c$; let's choose $c:=0$. Then $$\partial_x^2 f(s,x) = 0 \qquad \partial_s f(s,x) = x,$$ and therefore (2) gives $$\int_0^t s \, dW_s = t W_t - \int_0^t W_s \, ds.$$

See e.g. this question and this question for further examples.


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