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. $$\frac{A_t}{N_t} = \mathbb{E_Q}\left[ \frac{A_T}{N_T} | \mathcal{F}_t \right]$$

1 Answer 1


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 $$C_t=e^{-r(T-t)}S_t^{1/3}e^{\frac{1}{3}r(T-t)-\frac{1}{9}\sigma^2(T-t)}=S_t^{1/3}e^{-\frac{2}{3}r(T-t)-\frac{1}{9}\sigma^2(T-t)}$$

  • $\begingroup$ Thanks for the elaborate answer mmencke! I'll have a look at it. $\endgroup$
    – George
    Commented Dec 12, 2021 at 12:03
  • $\begingroup$ quick question: could we just skip the first part of the answer and immediately conclude, using FFT, that the price of the derivative is the risk-neutral expectation discounted at the risk-free rate? $\endgroup$
    – George
    Commented Dec 12, 2021 at 12:59
  • $\begingroup$ To use the First Fundamental Theorem of Asset Pricing we would have to show either that there exists at least one equivalent probability measure or that the model is absent of arbitrage. I think the former is easier to show and that it is usually true, but I don't know of an easier way to prove that than doing the Girsanov transform of the process. Btw I should have checked the Novikov condition to make sure that the likelihood process is a martingale. $\endgroup$
    – mmencke
    Commented Dec 12, 2021 at 16:05
  • 1
    $\begingroup$ I figured it out, thank you. Don't think we have to know the Novikov condition, so you are fine there! $\endgroup$
    – George
    Commented Dec 13, 2021 at 11:57

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