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 $$t=0$$.

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

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? Dec 15, 2016 at 18:05
• Yes it is complete answer
– user16651
Dec 15, 2016 at 18:06
• @dm63 The solution holds for any $S_t^n$. It was edited.
• @dm63 So thanks. Indeed $$\Pi(0,T)=S_0^n\exp\left[\left((n-1)r+\frac 12 n(n-1)\sigma^2\right)T\right]$$