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My lecturer 'derived' Ito's lemma today, but there are a few steps that I'm unclear on;

Firstly, let the SDE be $dS = \mu S dt + \sigma S dW$, where $W$ is a Wiener process.

Now, consider a function of the stock price, $f(S)$, then from the Taylor series expansion we have $f(S + \delta S) = f(S) + \delta S \frac{\delta f}{\delta S} + \frac{1}{2} (\delta S)^2 \frac{ \delta^2 f}{\delta S^2} + ...$

He then states that $(dS)^2 = ... (dt)^2 + ... (dt)(dW) + \sigma^2 S^2 (dW)^2$. Then, informally states that $(dW)^2$ is roughly $dt$ as $dt$ tends to $0$, so the only term left will be the last term.

After this, he proceeds to state Ito's lemma; If $F = f(S,t)$, then $$dF = [\mu S \frac{\delta f}{\delta S} + \frac{\delta f}{\delta t} + \frac{1}{2} \sigma^2 S^2 \frac{ \delta^2 f}{\delta S^2}]dt + \sigma S \frac{\delta f}{\delta S} dW$$

How on earth did he derive this given the above information? What does the Taylor series expansion in $f$ have to do with it? It almost looks a little like he's substituted $dS$ in to the Taylor series expansion but I'm not sure. Can someone explain the steps for me please?

the man
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  • Your professor is using the classical calculus version of rules for Ito Calculus. All derivations above formally come from the theory of quadratic variation: https://math.stackexchange.com/questions/1896972/ito-quadratic-variation – Alex R. Apr 05 '18 at 21:23
  • Obviously the post is relevant but it doesn’t help with answering my question. I still don’t understand how the Taylor series in $f$ is used in deriving the lemma? – the man Apr 05 '18 at 21:34

1 Answers1

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The fact that $(dW_t)^2 = dt$ is key to derive Ito's lemma. Write $dW_t = W_{t+dt} - W_t$, then $dW_t \sim N(0,dt)$ by definition. Since $\mathbb{E}[(dW_t)^2] = dt$, it is reasonable to say that $(dW_t)^2 = dt$.

Given that $dS = \mu S dt + \sigma S dW$, it follows by Taylor's theorem: $$f(S+dS, t+dt) = f(S,t) + (dS)\frac{\partial f}{\partial S}(S,t) + dt \frac{\partial f}{\partial t}(S,t)+\frac{1}{2} (dS)^2 \frac{\partial^2 f}{\partial S^2}(S,t) + \ldots$$

Using that $(dS)^2 = \sigma^2 S^2 dt$ and substituting the expression for $dS$ into the above equation, one arrives to Ito's lemma: $$\underbrace{f(S+dS, t+dt) - f(S,t)}_{df(S,t)} = \left(\frac{\partial f}{\partial t}+\mu S \frac{\partial f}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 f}{\partial S^2}\right)dt + \sigma S \frac{\partial f}{\partial S}dW$$

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