No-Arbitrage Test for a Stock-Cubed Claim Price

# No-arbitrage Pricing


# No-arbitrage Pricing












We have a contract whose value is $A(S_t,t) = S_t^3$ at all times, not just at expiration. $S_t$, the underlying stock, follows a Geometric Brownian Motion, $\frac{dS}{S} = \mu dt + \sigma dB$. How would we go about showing that this is inconsistent with no-arbitrage pricing?

I thought a potential solution could be to show that it is not a Martingale under the Q-measure. Basically, we start by assuming that $A(S_t, t)$ is a Martingale, which implies that $e^{-rt}E^Q[A_t] = A_0 = S_0^3$. But, under the risk-neutral measure, we know that $S_t = S_0e^{(r-\frac{\sigma^2}{2})t + \sigma \sqrt{t} Z^Q}$ where $Z$ is standard normal. It follows that $A(S_t, t) = S_t^3 = S_0^3e^{3(r-\frac{\sigma^2}{2})t + 3\sigma \sqrt{t} Z^Q}$. Computing the expectation $e^{-rt}E^Q[S_t^3] = S_0^3 e^{-rt}\int_{z^*}^{\infty} \frac{dz}{\sqrt{2 \pi}} e^{\frac{-z^2}{2}}e^{3(r-\frac{\sigma^2}{2})t + 3\sigma \sqrt{t} Z^Q}$ we obtain $S_0^3 e^{2rt + 3\sigma^2t}$. Because $S_0^3 e^{2rt + 3\sigma^2t} \neq S_0^3$ we conclude that $A(S_t, t)$ is not a Martingale, so the fact that the contract has value $S_t^3$ at all times is inconsistent with no-abitrage pricing.

Would something like this work? Any help would be much appreciated. Thanks.

## Answer by user34971 (score 3, accepted)

https://quant.stackexchange.com/a/58731

Under the risk-neutral measure by application of Ito: $$ dS^3_t = 3 \left[ (r + \sigma^2)S^3_t dt + \sigma S^3_t dW_t \right] $$ The risk-neutral drift is not the risk-free rate and hence $S_t^3 \; \forall t$ cannot be the price of a claim or any other tradable asset.

So basically along the same lines as your proof, but without calculating expectations etc. Just need to look at the local drift.