Pricing a Forward Contract on the Squared Stock Price

Summary

The document asks how to value a forward contract paying the square of a stock price at maturity under a Black–Scholes model. It changes to the risk-neutral measure, applies Itô’s lemma to the squared stock price, and uses the resulting drift to find its expected terminal value. Discounting that expectation gives the contract’s time-zero value, with the initial stock price of five and maturity of three years appearing in the stated expression.

The calculation illustrates why the square of a stock price does not itself grow at the risk-free rate: Itô’s lemma adds a volatility-dependent term to the drift. The no-arbitrage connection is the risk-neutral pricing principle, which values a traded claim by discounting its risk-neutral expected payoff. The document contains a likely transcription error in an intermediate expectation equation, where a coefficient is written inconsistently with the preceding stochastic differential equation; its final pricing expression follows the stated drift. The result assumes the model parameters and risk-neutral dynamics are valid and does not discuss practical hedging or market frictions.

Key ideas

• Under risk-neutral dynamics, apply Itô’s lemma to derive the evolution of the squared stock price.
• The squared price has a volatility-dependent drift in addition to twice the risk-free rate.
• Price the forward payoff by discounting its risk-neutral expected terminal value.
• The derivation assumes the Black–Scholes model and frictionless no-arbitrage pricing.

Tags

# Black-Scholes evaluating the squared of the stock price


# Black-Scholes evaluating the squared of the stock price












> Consider a Black-Scholes model $S_t = 5\exp{(\sigma W_t + \mu t)}$, $B_t = \exp{(rt)}$, where $W_t$ is Brownian motion with respect to a given measure $\mathbb{P}$. Suppose you hold a forward contract $X$, that pays at $T=3$, the value $X = (S_3)^2$ the square of the stock price at the terminal time. Compute the value of the contract $X$ at time $t=0$. Explain how the no arbitrage condition is related to your answer.

I asked a similar question like this before but I am confused now when $S_t$ is squared. Any suggestions is greatly appreciated.

## Answer by user16651 (score 4, accepted)

https://quant.stackexchange.com/a/31516

Let $$dS_t=r S_tdt+\sigma S_t dW^{\mathbb{Q}}_t\tag 1$$ where $S_0=5$. Set $X_t=S_t^2$. By application of Ito's lemma, we have $$dX_t=\left(2r+\sigma^2\right)X_tdt+2\sigma^2X_tdW^{\mathbb{Q}}_t\tag 2$$ in other words $$X_T=X_0+\left(2r+\sigma^2\right)\int_{0}^{T}X_t dt+2\sigma^2\int_{0}^{T}X_t dW^{\mathbb{Q}}_t\tag 3$$ thus $$\mathbb{E}^{\mathbb{Q}}_{0}[X_T]=\mathbb{E}^{\mathbb{Q}}[X_T]=X_0+\left(2+\sigma^2\right)\int_{0}^{T}\mathbb{E}^{\mathbb{Q}}[X_t] dt\tag 4$$ as a result $$d\,\mathbb{E}^{\mathbb{Q}}[X_T]=\left(2r+\sigma^2\right)\mathbb{E}^{\mathbb{Q}}[X_T]\tag 5$$ therefore $$\mathbb{E}^{\mathbb{Q}}[X_T]=25\,e^{(2r+\sigma^2)T}\tag 6$$ Finally, we have $$\Pi(0)=e^{-r(T-0)}\mathbb{E}^{\mathbb{Q}}[X_T]=25\,e^{(r+\sigma^2)T}\tag 7$$

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