# Apply Ito to $exp((\mu-0.5*\sigma^2)t+\sigma*W_t)$ for Brownian Motion $W_t$

Let us consider the model with only one risky asset with $$S_0^1=1$$. Omitting indices form $$S^1,W^i$$ and from the parameter $$\mu^1$$ and $$\sigma^{11}$$, the model of the risky asset becomes $$S_t=exp\biggl((\mu-\frac{1}{2}\vert \sigma \vert^2)t+\sigma W_t\biggl).$$ Now, we want to apply Ito's formula to the asset or stochastic process, and we know that we may get $$f(t,X_t)=f(0,X_0)+\int_0^t \partial_x f(s,X_s)dX_s+\int_0^t(\partial_t f(s,X_s)+\frac{1}{2}\sigma_s^2\partial_{x x}f(s,X_s))ds$$ using Ito's formula

First, we defining $$f(t,x)=exp\biggl((\mu-\frac{1}{2}\vert\sigma\vert^2)t+\sigma x\biggl)$$ we have $$S_t=f(t,W_t)$$.

After compute the required derivatives, we get $$\sigma f(t,x),(\mu-\frac{1}{2}\sigma^2)f(t,x)$$ and $$\sigma^2 f(t,x)$$ respectively, then in our lecture not, it follows that $$f(t,W_t)=S_t=f(0,0)+\int_0^t \partial_x f(s,W_s)dW_s+\int_0^t(\partial_t f(s,W_s)+\frac{1}{2}\partial_{x x}f(s,W_s))ds$$ $$=1+\int_0^t\sigma S_s dW_s+\int_0^t((\mu-\frac{1}{2}\vert \sigma \vert^2)S_s+\frac{1}{2}\sigma^2 S_s)ds$$

Problem: In the Ito's formula, we notice that there is a $$\sigma^2$$ in the last term, why is this term disappear when we apply Ito's formula to our specific case, where we only have$$\frac{1}{2}\partial_{x x}f(s,W_s)$$ in stead of $$\frac{1}{2}\sigma^2\partial_{x x}f(s,W_s)$$ in my opinion, it seems kind of weird to me.

It will be great if anyone could explain it in detail in the given case!

The problem here is how you have written Ito's formula, the way you have it written is actually incorrect if we are considering a general stochastic process $$X_t$$, the $$\sigma^2$$ term appears from the quadratic variation of the stock price $$S_t$$ when we apply Ito to $$S_t$$. See the following for details, (out of laziness I will use differential notation). Let $$X_t$$ be any Ito process, then for a fucntion $$f:\mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}:(t,x) \mapsto f(t,x)$$ such that $$f \in C^{1,2}$$, we have

\begin{align} 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)d \langle X \rangle_t, \end{align} where $$\langle \cdot \rangle$$ denotes quadratic variation. If we now define, $$S_t=e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t},$$ Then applying Ito we get, \begin{align} dS_t & = (\mu-\frac{\sigma^2}{2})e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t}dt+\sigma e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t}dW_t +\frac{\sigma^2}{2}e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t}dt \\ & = (\mu-\frac{\sigma^2}{2})S_tdt+\sigma S_tdW_t+\frac{\sigma^2}{2}S_tdt \\ &= \mu S_tdt+\sigma S_tdW_t, \end{align} Which is required SDE. Note here the quadratic variation of $$W_t$$ is just $$t$$.

• I believe his lecture limits itself to processes $(X_s)_{s\ge0}$ such that $d\langle X\rangle_s=\sigma_s^2\,ds$. This is rather common for lectures on introduction to stochastic calculus applied to quantitative finance, which seems to be the subject here.
– Will
Jul 12, 2023 at 21:29
• Yes, we have the same result, I am always confused by the integral form and differential form. Like how do we get$\frac{1}{2}\partial_{x x}f(s,X_s)= \frac{1}{2}\sigma^2 S_s$ in our case. Jul 12, 2023 at 21:32
• @Will Possibly, but if it was a basic course on SDE's for finance then you certainly would not have any processes with quadratic variation given as you have written. If they only use a geometric brownian motion then the required quadratic variation process would be $d\langle S \rangle_t=\sigma^2 S_t^2dt.$ Jul 12, 2023 at 21:34
• @Malik it is exactly what you have written, the second derivative of $S_t$ with respect to the brownian motion is exactly $\sigma^2 S_t$. Essentially we have an exponential function $f(t,x)= e^{at+bx}$, so that $\partial_{xx}^2f(t,x)=b^2e^{at+bx}=b^2f(t,x)$, I hope this helps. Jul 12, 2023 at 21:39
• Thanks for the detailed answer! Jul 12, 2023 at 21:46

The way you write Itô's formula, I suppose in your lecture you apply it to a so called Itô process $$X$$ which satisfies $$dX_s=\mu_s\,ds+\sigma_s\,dW_s$$ for some processes $$\mu$$ and $$\sigma$$.

In your application, $$\sigma_s$$ did not disappear. It is just that it is equal to $$1$$, because you apply Itô's formula with a Brownian motion. Do not confuse this $$\sigma_s$$ with the parameter $$\sigma$$ associated with $$S_t$$.

• Yes, we have used that expression for Ito process, do you mean that there is no $\sigma_s$ term in our $S_t$? Jul 12, 2023 at 21:02
• Yes there is. But you don't apply Itô's formula with $S_t$ here, you apply it with $W_t$, whose sigma process is $1$.
– Will
Jul 12, 2023 at 21:16
• Sorry, I still do not get the point why do we not apply Ito to $S_t$ but $W_t$? Jul 12, 2023 at 21:33
• You wrote the Itô formula for $f(t,X_t)$. Then did you apply it with $X_t=S_t$ or $X_t=W_t$?
– Will
Jul 12, 2023 at 21:37
• That is correct :)
– Will
Jul 12, 2023 at 21:43