Short story: the process for Stock price squared is not a martingale when discounted by the money-market numeraire under the risk-neutral measure. How can we then compute derivative prices on $S_t^2$ under the risk neutral measure? Wouldn't this lead to arbitrage?
Long story: I found some great posts on power options, for example Finding price of the power option. Whilst the maths is clear, I am still somewhat confused about the concept: starting with a simple option on Stock price squared, I do not fully comprehend how an optional claim can be priced within the regular B-S framework, when the price process for $S_t^2=S_0^2exp((2r-\sigma^2)t+2\sigma W_t)$ is not a martingale when discounted by $e^{rt}$ under the risk-neutral money-market numeraire.
I consider a single period model with zero rates. As outlined in the post What is the Risk Neutral Measure?, in the one-period model, the risk-neutral measure arises from no arbitrage assumption in the model. We assume that initially, the stock price is $S_0$ and after one period it can be either $S_u=S_0*u$ or $S_d=S_0*d$, with $u$ and $d$ being some multiplicative factors. Pricing a derivative claim with pay-off function $V(.)$ on the underlying stock $S_t$ via replication gives rise to:
$$V_0 = \left(V(S_u) \left( \frac{1 -d}{u-d} \right) + V(S_d) \left(\frac{u-1}{u-d} \right) \right)$$
Imposing $u \leq 1 \leq d$ will ensure that there is no arbitrage in the one-period model. Furthermore, as a consequence of the condition $u \leq 1 \leq d$, we get that $0 \leq \frac{1 -d}{u-d} \leq 1$ and $0 \leq \frac{u-1}{u-d} \leq 1$. Therefore, we can define $p_u:=\frac{1 -d}{u-d}$, $p_d:=\frac{u-1}{u-d}$ and we can call $p_u$ and $p_d$ "probabilities": indeed, in the one-period model, $p_u$ & $p_d$ form the discrete (risk-neutral) probability measure.
Now, the interesting point is that pricing the claim $V(.)$ on $S_t^2$ via replication in the one-period model actually leads to a different probability measure:
(i) Upper state: $S_{t_1}^2=S_0^2u^2$, denoting risk-free bond as $B$ we have $B_{t_1}=B_{t_0}=1$ since rates are zero and the option pay-off is $V_u=V((S_0u)^2)=[S_0^2u^2-K]^+$.
(ii) Lower state: $S_{t_1}^2=S_0^2d^2$, $B_{t_1}=B_{t_0}=1$, $V_d=V((S_0d)^2)=[S_0^2d^2-K]^+$.
Trying to replicate the payoff $V(S_{t_1}^2)$ in both states via the underlying stock and the risk-free bond, we get two equations with two unknowns (x = number of stocks, y = number of bonds I wanna hold to replicate option pay-off):
$$(i) V_u=x*S_0^2u^2+y*1$$
$$(ii) V_d=x*S_0^2d^2+y*1$$
Solving the system of equations yields:
$$ x=\frac{V_u-V_d}{S_0^2(u^2-d^2)}, y=\frac{u^2V_d-d^2V_u}{u^2-d^2}$$
Which then gives the claim price as (after some basic algebraic simplifications):
$$V_0=x*S_0^2+y*1=V_u*\frac{1-d^2}{u^2-d^2}+V_d\frac{u^2-1}{u^2-d^2}$$
Setting $p_u^*:=\frac{1-d^2}{u^2-d^2}$ and $p_d^*:=\frac{u^2-1}{u^2-d^2}$, the above can be re-written as:
$$V_0=V_up_u^*+V_dp_d^*=\mathbb{E}^{Q_2}[V_{t_1}]$$
In other words, the replication argument gives rise to some new probability measure where $p_u^*=\frac{1-d^2}{u^2-d^2}\neq p_u=\frac{1-d}{u-d}$ and $p_d^*=\frac{u^2-1}{u^2-d^2}\neq p_d=\frac{1-d}{u-d}$.
Instead, we actually have that $p_u^*=p_u \frac{1+d}{u+d}$ and $p_d^*=p_d \frac{1+u}{u+d}$.
Question: So going back to the the start and considering the thread Finding price of the power option, how come we can price power options under the B-S classical risk-neutral measure? That would be equivalent to saying that under the one period model (with rates being zero), the price of the claim $V(S^2_t)$ could be computed as $V_0=\mathbb{E}^Q[V_t(S_t^2)]=p_uV_u(S_t^2) + p_dV_d(S_t^2)$, which does not produce the correct result (indeed, above we instead get that $V_0=\mathbb{E}^{Q_2}[V_t(S_t^2)]=p_u^*V_u(S_t^2) + p_d^*V_d(S_t^2)$).