Using Itô’s Formula to Verify a Black–Scholes Martingale

Summary

The document considers a no-dividend Black–Scholes stock and the power process Z(t)=(S(t)/H)^p, where H is positive and p is set by the interest rate and volatility. It shows how to apply Itô’s formula: differentiate the power function, include the quadratic variation term, and express the resulting dynamics as a drift component plus a Brownian shock. The apparent extra time term is expected during this calculation; the chosen exponent makes its coefficient zero.

With zero drift, the process has martingale dynamics under the model assumptions, and dividing by its initial value gives a positive process with initial value one. The answer gives an algebraic cancellation as evidence, but the exposition does not separately discuss the conditions that ensure a true martingale over the time horizon. The result depends on the specified Black–Scholes dynamics and parameter choice; it is not a general claim about arbitrary powers of stock prices.

Key ideas

• Itô’s formula for a power of the stock price includes a drift term from quadratic variation.
• The exponent p is chosen so the drift coefficient vanishes under the stated Black–Scholes model.
• The resulting process has Brownian shock dynamics with zero drift.
• Normalizing by the initial value gives a positive process that starts at one.
• The conclusion relies on the model and parameter assumptions in the question.

Tags

# Show that Z(t)/Z(0) is a positive mean-1 martingale


# Show that Z(t)/Z(0) is a positive mean-1 martingale












We look at a standard no dividends Black-Scholes model and here we have a process Z, which is defined by: Z(t)=(S(t)/H)^p , where H is a positive constant and p=1-2r/sigma^2

I am now asked to show that Z(t)/Z(0) is a positive mean-1 martingale.

My first intuition tells me to use Ito's formula to get dZ(t) and that shouldn't include a dt term, but I somehow keep seeing the dt term when I use Ito. I am getting increasingly frustrated. Can someone help me with this? From there on I would take the expectation and se that E(Z(t))=1.

- Mads

## Answer by vanna (score 1, accepted)

https://quant.stackexchange.com/a/16675

$$ Z_t = f(S_t) := \left( \frac{S_t}{H} \right)^p $$ $$ dZ_t = \partial_x f(S_t) dS_t + \frac{1}{2} \partial^2_{xx} f(S_t) d\langle S \rangle_t = p\frac{S_t^{p-1}}{H^p} dS_t + \frac{1}{2} p(p-1) \frac{S_t^{p-2}}{H^p} S_t^2 \sigma^2 dt $$ Thus $$ dZ_t = Z_t \left( p r + \frac{1}{2} p(p-1) \sigma^2 \right)dt + p Z_t \sigma dW_t $$ so that $$ \frac{dZ_t}{Z_t} = \mu dt + p\sigma dW_t $$

Now

$$ \mu = p r + (p\sigma^2 )\left( \frac{1}{2} (p-1)\right) = r - 2 \frac{r^2}{\sigma^2} - (\sigma^2-2r) \frac{r}{\sigma^2} = 0$$ which proves that $Z$ is a martingale

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