# 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.

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.