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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!

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2 Answers 2

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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 $t$.

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    $\begingroup$ 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. $\endgroup$
    – Will
    Jul 12, 2023 at 21:29
  • $\begingroup$ 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. $\endgroup$
    – Malik
    Jul 12, 2023 at 21:32
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    $\begingroup$ @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 $d\langle S \rangle_t=\sigma^2 S_t^2dt.$ $\endgroup$
    – Emmet
    Jul 12, 2023 at 21:34
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    $\begingroup$ @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. $\endgroup$
    – Emmet
    Jul 12, 2023 at 21:39
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    $\begingroup$ Thanks for the detailed answer! $\endgroup$
    – Malik
    Jul 12, 2023 at 21:46
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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 $S_t$.

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

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