# Risk-neutral pricing to determine no-arbitrage price

We are asked to consider a derivative with payoff $$C_t = S_{T}^{1/3}$$ at maturity $$T > 0$$ and to use risk neutral pricing to derve the no-arbitrage price process $$C_{t}$$.

Some context:

Let $$W$$ be a standard Browian motion. We are in a financial market consisting of a risky asset $$S$$ and a money-market account $$B$$ with:

$$dS_t = a(b - S_t)dt + \sigma S_tdW_t$$ $$dB_t = rB_tdt$$

where, $$B_0 = 1,\; S_0 = s_0, \;\sigma > 0 \; \text{and}\; a,b \; \text{are constants unequal to zero.}$$

I presume we have to either use the First Fundamental Theorem of Asset Pricing, Girsanov's theorem or both, however I have a hard time determining where to start. Could someone help me out?

$$\text{Quick note:}$$

The FFTAP tells us that under regularity conditions absence of arbitrage holds if and only if, for some numeraire $$N$$, there exists a probability measure $$\mathbb{Q} = \mathbb{Q}_N$$ such that:

1. $$\mathbb{Q} \sim \mathbb{P}$$
2. For any asset $$A$$ in the market, the discounted price process $$A/N$$ is a $$\mathbb{Q}$$-martingale, i.e. $$\frac{A_t}{N_t} = \mathbb{E_Q}\left[ \frac{A_T}{N_T} | \mathcal{F}_t \right]$$

To find the $$S$$-dynamics under $$\mathbb{Q}$$ we have to use Girsanov's theorem: $$dW_t^P=\varphi_t dt+dW_t^Q$$ Dynamics under $$\mathbb{Q}$$ is thus $$dS_t=a(b-S_t)dt+\sigma S_t(\varphi_t dt+dW_t^Q)=abdt-aS_tdt+\varphi_t\sigma S_tdt+\sigma S_t dW_t^Q$$ To avoid any arbitrage opportunities the (local) rate of return must be equal to the risk-free rate meaning that $$\mathbb{E}[dS_t]=rS_tdt$$ $$\iff$$ $$abdt-aS_tdt+\varphi_t\sigma S_tdt=rS_t dt$$ $$\iff$$ $$-\frac{ab}{S_t}+a+r=\varphi_t \sigma$$ $$\iff$$ $$\frac{-\frac{ab}{S_t}+a+r}{\sigma}=\varphi_t$$ We have thus found the Girsanov kernel. Plugging it into the dynamics $$dS_t=a(b-S_t)dt+\sigma S_t(\varphi_t dt+dW_t^Q)=abdt-aS_tdt+\frac{-\frac{ab}{S_t}+a+r}{\sigma}\sigma S_tdt+\sigma S_t dW_t^Q=rS_t dt+\sigma S_t dW_t^Q$$ The price of the derivative is the risk-neutral expectation discounted at the risk-free rate $$C_t=e^{-r(T-t)}\mathbb{E}^Q[S_T^{1/3}]$$ We can write $$S_T$$ as $$S_T=S_te^{(r-\frac{1}{2}\sigma^2)(T-t)+\sigma (W_T^Q-W_t^Q)}$$ which is equal in distribution with $$S_T=S_te^{(r-\frac{1}{2}\sigma^2)(T-t)+\sigma \sqrt{T-t}\varepsilon}$$ where $$\varepsilon$$ is a standard normal variable. This gives us $$S_T^{1/3}=S_t^{1/3}e^{\frac{1}{3}(r-\frac{1}{2}\sigma^2)(T-t)+\frac{1}{3}\sigma \sqrt{T-t}\varepsilon}$$ and $$\log(S_T^{1/3})=\log(S_t^{1/3})+\frac{1}{3}(r-\frac{1}{2}\sigma^2)(T-t)+\frac{1}{3}\sigma \sqrt{T-t}\varepsilon$$ So $$S_T^{1/3}$$ is log-normally distributed with mean $$\frac{1}{3}\left(\log(S_t)+(r-\frac{1}{2}\sigma^2)(T-t)\right)$$ and variance $$\frac{1}{3^2}\sigma^2(T-t)$$ The mean of a log-normal distribution is given by $$e^{\mu+\sigma^2/2}$$, so $$\mathbb{E}\left[S_T^{1/3}\right]=e^{\frac{1}{3}\left(\log(S_t)+(r-\frac{1}{2}\sigma^2)(T-t)\right)+\frac{\frac{1}{9}\sigma^2(T-t)}{2}}=S_t^{1/3}e^{\frac{1}{3}(r-\frac{1}{2}\sigma^2)(T-t)+\frac{1}{18}\sigma^2(T-t)}=S_t^{1/3}e^{\frac{1}{3}r(T-t)-\frac{1}{9}\sigma^2(T-t)}$$ It should now be easy to find the price of the derivative $$C_t=e^{-r(T-t)}S_t^{1/3}e^{\frac{1}{3}r(T-t)-\frac{1}{9}\sigma^2(T-t)}=S_t^{1/3}e^{-\frac{2}{3}r(T-t)-\frac{1}{9}\sigma^2(T-t)}$$