# Price of financial assets at $t=0$ in Black-Scholes framework

Given the share price equation $$dS_t=rS_tdt+\sigma S_tdW_t$$ working in the framework of Black-Scholes model, find the price at $$t=0$$ of the following two financial assets:

(a) The asset pays at $$t=T$$ exactly $$S_T^2$$.

(b) The asset pays at $$t=T$$ exactly $$1$$ if $$S_T, where $$K$$ is constant specified in the contract.

My attempt at solution.

(a) Since the solution of share price equation is given by $$S_t=S_0\,\text{exp}\left[\left(r-\tfrac12\sigma^2\right)t+\sigma W_t\right]$$ we calculate using risk-neutral pricing \begin{align} C_0&=e^{-rT}\mathbb{E}_Q\left[C_T\right]\\ &=e^{-rT}\mathbb{E}_Q\left[S_T^2\right]\\ &=e^{(r-\sigma^2)T}S_0^2\,\mathbb{E}_Q\left[e^{2\sigma W_T}\right]\\ &=e^{(r-\sigma^2)T}S_0^2\,e^{2\sigma^2T} \\ &=e^{(r+\sigma^2)T}S_0^2\,\quad \text{(Answer)} \end{align}

(b) Denoting by $$\Theta(x)=\begin{cases}1, x\ge 0\\ 0, x<0\end{cases}$$ the unit step function we find \begin{align} C_0&=e^{-rT}\mathbb{E}_Q\left[C_T\right]\\ &=e^{-rT}\mathbb{E}_Q\left[\Theta(K-S_T)\right]\\ &=e^{-rT}\,\mathbb{Q}\left\{S_T Denote $$a=r-\tfrac12\sigma^2$$, $$b=\sigma\sqrt{T}$$, $$x=\frac{K}{S_0}$$. Then $$S_T/S_0=X$$, $$X\sim N(a,b^2)$$ and \begin{align} \mathbb{Q}\left\{e^X where $$\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-x^2/2}dx.$$ Thus $$C_0=e^{-rT}\Phi\left(-\frac{\ln(S_0/K)+r-\tfrac12\sigma^2}{\sigma \sqrt{T}}\right)\quad \text{(Answer)}$$

Question: Is this solution correct?

• Looks correct, except that in (a) $C_0=e^{(\color{red}{-}r+\sigma^2)T}S_0^2\,.$ Dec 8, 2022 at 19:38
• @KurtG. but isn't there $2rT$ that comes from raising $S_T$ to the second power. Thus $-rT+2rT=+rT$ ? Dec 8, 2022 at 20:02
• Looks correct. My mistake. Dec 8, 2022 at 20:04
• Thanks.$\phantom{.................}$ Dec 8, 2022 at 20:05
• Voting to reopen because the question is focused and specific enough, and OP also providers their own attempt at answering. Dec 9, 2022 at 12:55