Pricing a Squared Stock Payoff Under Black–Scholes

Summary

The document asks whether a payoff equal to the square of the terminal stock price is worth the initial stock price squared grown at the risk-free rate. The response explains that taking a power does not commute with taking an expectation: convexity makes the expected squared price differ from the square of the expected price. Under a risk-neutral geometric Brownian motion, the second moment includes an extra term driven by volatility. Discounting that moment gives the price of the power payoff; the response’s general moment formula shows the effect for higher powers as well.

The discussion is a concise illustration of risk-neutral valuation and Jensen’s inequality, not a broad treatment of power options. Its displayed moment formula uses a discounted stock process with drift set to zero, so applying it to the question’s stock dynamics requires restoring the risk-free-rate factor. The result assumes the standard Black–Scholes model and does not address dividends, alternative payoff conventions, or other market dynamics.

Key ideas

• A convex power payoff depends on the terminal price’s distribution, not only its expected value.
• Under risk-neutral geometric Brownian motion, the second moment grows with volatility.
• The discounted expected payoff determines the derivative price under the risk-neutral measure.
• The stated general moment expression omits the risk-free-rate factor and must be adjusted for the question’s drift convention.

Tags

# Price of power option under Black Scholes Assumptions


# Price of power option under Black Scholes Assumptions












As in the title, is $$\pi_0(S^2_T)=S^2_0e^{rT}$$ the correct way to price power option with terminal payoff $$X=S^2_T$$ under Black Scholes assumptions of $S_T=S_0\exp((r-\sigma^2/2)T+\sigma W_T$? Here I'm using the risk neutral pricing formula of $\pi_0(X)=e^{-rT}E_Q(X_T)$ Many thanks

## Answer by Andrea (score 1)

https://quant.stackexchange.com/a/81224

It is definitely not correct, you have forgotten the whole role of convexity (aka Jensen inequality) (set $n\ge1$)

$E(X^n) \ge (E(X))^n$

For a BS process, you have

$S_T = S_0 e^{-\frac{1}{2} \sigma^2 T + \sigma W_T}$, and

$S_T^n = S_0^n e^{n \left (-\frac{1}{2} \sigma^2 T + \sigma W_T \right )}$

$E(S_T^n) = S_0^n e^{\frac{1}{2} \sigma^2 T (n^2 - n)}$

which is equal to $S_0^n$ only for $n=1$ or $\sigma=0$.

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