Why Itô’s Formula Keeps Only First-Order Time Terms

Summary

The document concerns the Taylor expansion used when deriving the Black–Scholes equation for a function of a stochastic asset price and time. The question asks why the expansion retains a second derivative with respect to price, yet discards second-order time and mixed price-time terms. The response’s central point is that the derivation tracks terms at the order of the time increment: stochastic price changes have squared increments proportional to dt, so the price second-derivative term contributes at the same order as ordinary time terms.

By contrast, the time-squared and price-time products vanish at smaller orders as the time step shrinks, so they are omitted in Itô’s formula. This exchange gives only a brief intuition and does not lay out the full increment-ordering rules or derive the option pricing PDE. It is useful as a conceptual pointer, but readers may need a fuller treatment of stochastic calculus to see why the differentials scale differently.

Key ideas

• In a diffusion model, the squared asset-price increment contributes at the order of dt.
• The second derivative with respect to price therefore remains in the stochastic expansion.
• The time-squared and mixed price-time terms vanish at smaller orders as the time step shrinks.
• The response gives intuition but does not present a full derivation of the Black–Scholes equation.

Tags

# Black Scholes Stochastic Taylor expansion question


# Black Scholes Stochastic Taylor expansion question












I am currently deriving Black-Scholes formula, and i get the following equation when Im doing the Tayler expansion:

$dG=\frac{\partial G}{\partial S}dS+\frac{\partial G}{\partial t}dt+\frac{1}{2}\frac{\partial^{2}G}{\partial S^{2}}dS^{2}+\frac{1}{2}\frac{\partial^{2}G}{\partial t^{2}}dt^{2}+\frac{\partial^{2}G}{\partial S\partial t}dSdt$

Since it is a stochastic process, we need the to take into account the $\frac{1}{2} \frac{\partial^{2} G}{\partial S^{2}} d S^{2}$ part, since $d S^{2}=\sigma^{2} S^{2} d t$, which means that this term also has an effect on $dG$ (since expression is linear in $dt$).

Now, what I dont understand, is why we need $dG$ to be linear? Why can't we use second order terms?

I suspect the answer is quite straight forward, but I cant seem to wrap my head around it.

## Answer by mbih (score 2)

https://quant.stackexchange.com/a/63126

Oh i think i got it. We are only considering the time step $dt$, which is why, first of all we need $\frac{1}{2} \frac{\partial^{2} G}{\partial S^{2}} d S^{2}$, and also why we remove: $\frac{1}{2} \frac{\partial^{2} G}{\partial t^{2}} d t^{2}+\frac{\partial^{2} G}{\partial S \partial t} d S d t$

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