# 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.

## 1 Answer

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 $$\Pi(0)=e^{-r(T-0)}\mathbb{E}^{\mathbb{Q}}[X_T]=25\,e^{(r+\sigma^2)T}\tag 7$$

• Is this the complete answer then? – Wolfy Dec 15 '16 at 18:05
• Yes it is complete answer – user16651 Dec 15 '16 at 18:06
• The question asked about S^2 not S^3 – dm63 Dec 16 '16 at 10:49
• @dm63 The solution holds for any $S_t^n$. It was edited. – user16651 Dec 16 '16 at 11:08
• Ok. Looks like the 125 should be 25. – dm63 Dec 16 '16 at 11:10