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Deriving the Discounted Option Differential with Itô's Product Rule

Article Quant Q&A · Author: Ria
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Summary

This explanation shows how the product rule in Itô calculus creates the discounting term when differentiating an option value divided by a bank account. Set the two factors to the option price and the inverse bank account, then apply the rule, including the cross-product of differentials. Since the bank account grows at the risk-free rate, its inverse has a negative drift proportional to that rate; multiplying by the option value produces the negative rate-times-price contribution.

Substituting the usual stochastic differential for the option price gives a discounted process whose drift includes the stock exposure, the time derivative, the gamma term, and the negative discounting term. The explanation addresses the algebra behind the term, though it assumes the Black–Scholes setup and does not develop the full derivation or discuss alternative market assumptions.

Key ideas

  • Itô's product rule includes a cross-product term when differentiating two stochastic processes.
  • The inverse bank account has drift equal to the negative risk-free rate times its value.
  • Multiplying the inverse bank account by the option price introduces a negative rate-times-option-price drift term.
  • Substitution of the option's stochastic differential yields the drift of the discounted option value.

Tags

OptionsDerivatives pricingStatistics
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# Black Scholes derivation and Ito's Product


# Black Scholes derivation and Ito's Product












In this derivation of the Black Scholes equation can someone please explain the last step where the author uses Ito's product rule? I do not understand where the "rC" term comes from.

## Answer by Kevin (score 4, accepted)

https://quant.stackexchange.com/a/77922

In general, Itô's product rule is $$\text d(XY)=X\text dY+Y\text dX+\text dX\text dY.$$

In your example, we have $X=\frac1B$ and $Y=C$ such that $$\text d\left(\frac{C}B\right) = \frac1B\text dC+C\text d\left(\frac1B\right) + \text d\left(\frac1B\right)\text dC.$$

Because $\text dB= rB\text dt$ and $\text d(1/x) = -1/x^2\text dx$, we have $\text d\left(\frac1B\right)=-\frac1{B^2} \text dB=-\frac{r}B\text dt$. Thus, $$\text d\left(\frac{C}B\right) = \frac1B\text dC-\frac{rC}{B}\text{d}t.$$

As is standard, $$\text{d}C = \left(C_t+rSC_S+\frac{1}{2}\sigma^2S^2C_{SS}\right)\text{d}t+\sigma SC_S\text{d}W.$$

Putting everything together gives the desired result, namely $$\text{d}C = \frac{1}{B}\left(C_t+rSC_S+\frac{1}{2}\sigma^2S^2C_{SS}-rC\right)\text{d}t+\sigma \frac{S}{B}C_S\text{d}W.$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.