The Volatility Correction in the Risk-Neutral Stock Process

Summary

The document explains why a volatility-squared correction appears when expressing a stock price as an exponential Brownian process in the Black–Scholes framework. Under the risk-neutral measure, the discounted stock is modeled as a martingale. The answer starts with a stock following a geometric Brownian motion under the real-world measure, then describes changing measure so the discounted price has no drift.

Solving the resulting driftless stochastic differential equation gives an exponential with a negative one-half variance term in its exponent. Restoring the discount factor yields the familiar risk-neutral stock-price representation, with the correction arising from the stochastic calculus of the exponential process. This connects the term to solving the SDE, with Itô calculus underlying the correction. The response is a compact derivation sketch rather than a full step-by-step application of Itô’s lemma; it assumes familiarity with measure changes and Brownian motion.

Key ideas

• Under the risk-neutral measure, the discounted stock price is modeled as a martingale.
• Changing measure removes the drift from the discounted stock process.
• Solving the geometric Brownian motion produces a negative one-half variance term in the exponent.
• The correction term is connected to the stochastic calculus of the exponential solution.

Tags

# Black Scholes Formula, drift term


# Black Scholes Formula, drift term












In the formula, the stock return is modelled as a brownian motion that is a drift + a stochastic term, ok I get that. But the drift term is then modelled as r - volatility ^ 2 / 2. I am not sure how they derive this "volatility ^ 2 / 2". Is this derived out of the Ito Lemma??

## Answer by Slug Pue (score 4, accepted)

https://quant.stackexchange.com/a/17957

This drift comes from making the discounted stock a martingale in the risk-neutral measure $\mathbb Q$

You start with a stock in $\mathbb P$ having this form: $$ dS_t = \mu S_t dt + \sigma S_t dW_t $$ You also have a discount factor $e^{rt}$.

The idea is to remove the drift of the discounted process in $\mathbb Q$ so you get (after applying Girsanov's theorem) a martingale:

$$ d\hat S_t = \sigma \hat S_t d \tilde W_t $$ where $\hat S_t$ is the discounted stock and $\tilde W_t$ is a $\mathbb Q$-brownian motion.

If you solve this last SDE you get

$$ \hat S_t = \hat S_0\exp(\sigma W_t - \frac{1}{2}\sigma^2t) $$ Multiplying with $e^{rt}$ on both sides you get the un-discounted process and the drift you were asking about.

But the gist of why you get the correction term $\frac{1}{2}\sigma^2t$ is when solving the SDE $$ dX_t = \sigma X_t dW_t $$ you get $X_t = X_0 \exp(\sigma W_t - \frac{1}{2}\sigma^2t)$

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