Skip to content
All library documents

Spot Price and Quadratic Variation in an Informal Black–Scholes Derivation

Article Quant Q&A · Author: Shayan Slaesi
Read the original

Summary

The note clarifies that the symbol S in the discussed Black–Scholes derivation refers to the current spot price, written as S at time t. It connects that interpretation to geometric Brownian motion, where the asset price changes through a drift component and a volatility component driven by Brownian motion.

The answer informally squares the stochastic differential to show why the quadratic variation term is proportional to the square of the spot price, volatility squared, and elapsed time. Drift terms vanish from this calculation because finite-variation terms have zero quadratic variation, while Brownian motion contributes elapsed time. A second explanation confirms that the pricing equation uses the current spot price. The derivation is explicitly informal and relies on stochastic-calculus conventions; it is a conceptual explanation of notation and a term, rather than a full derivation of the Black–Scholes model or its assumptions.

Key ideas

  • The S in the equation denotes the asset's current spot price.
  • Geometric Brownian motion models price changes with drift and Brownian volatility terms.
  • The Brownian component contributes the quadratic variation proportional to volatility squared, spot squared, and time.
  • The explanation is informal and focuses on notation and one stochastic-calculus step.

Tags

OptionsDerivatives pricingStatistics
Full text
# Black Scholes informal derivation - question about a term in the equation


# Black Scholes informal derivation - question about a term in the equation












I am wondering what the term S means in the equation I have circled? I am not sure how to interpret it.

## Answer by Pleb (score 3)

https://quant.stackexchange.com/a/60462

From the GBM we have:

$$(S_{t}-S_{t-1}) \approxeq dS_t = \mu \cdot S_t \: dt + \sigma \cdot S_t \: dW_t,$$ where $W_t$ denotes a Brownian motion. Here, the first part, $(S_{t}-S_{t-1})$, can be seen as a discretization of $dS_t$. Now, squaring the GBM, $dS_t^2$, is an informal way of denoting the quadratic variation of the process, that is $[dS_t,dS_t]$. A slightly informal derivation gives us: \begin{align} dS_t^2 & = [dS_t,dS_t]\\ &= [\mu \cdot S_t \: dt + \sigma \cdot S_t \: dW_t, \: \mu \cdot S_t \: dt + \sigma \cdot S_t \: dW_t]\\ &= \mu^2 S_t^2 [d_t, d_t] +2\cdot \sigma\mu S_t^2 [dt, dW_t] +\sigma^2 S_t^2 [dW_t,dW_t]\\ &= \sigma^2 S_t^2 \: dt, \end{align} where $[dt,dW_t]=[dt,dt]=0$ since any finite/bounded variation process (read deterministic function) has zero quadratic variation (I'm talking about the $dt$ term), and $[dW_t,dW_t]=dt$, since quadratic variation of two Brownian motions equals the time difference (you can also look at Ito's multiplication table for these results). From your above formulation, $S = S_t$, and denotes the spot price, and probably that $dt\approx [t-(t-1)]=1$.

## Answer by D Stanley (score 1)

https://quant.stackexchange.com/a/60460

It's the current spot price, or $S_t$. The "next chapter" might show if/why they drop the subscript for this approximation, but in the end the variable in the black-scholes equation will be the current spot price.

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.