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I have an assignment in which $S_t$ is a stock price following a geometric Brownian motion. The task is now to show that at time t the risk-neutral price of a derivative on $S_t$ that pays $S_T^3$ at time T is
$$S_t^3 e^{(2r + 3\sigma^2)(T-t)}$$
Next, I need to work out the risk-neutral price of a derivative that pays $(S_T^2 - K)^+$ at time T. This is also in a continuos time setting.
Any help much appreciated.
Hint: under the risk neutral probability, the prices of the securities having an $L^2$ payoff are martingales. So you probably need to compute the Ito-differential of $$ (t,\omega)\to S^3_t e^{(2r+3σ^2)(T−t)} $$
For an integer $n>1$, let $D_n(t,T)$ be the price at $t\leq T$ of the derivative that pays $S_T^n$ at $T$.
The risk-Neutral distribution with respect of the stock numeraire is given by $\ln\mathcal{N}\left(\ln S_t+\left(r+\frac{1}{2}\sigma^2\right)(T-t),\sigma^2(T-t)\right)$. So now apply the martingale for $\frac{D_n(t,T)}{S_t}$.
$$E(\frac{D_n(T,T)}{S_T}|S_t)=\frac{D_n(t,T)}{S_t}\implies D_n(t,T)=S_tE(S_T^{n-1}|S_t)$$
Use the formula for $E(X^k)=e^{k\mu+\frac{k^2}{2}\sigma^2}$ for the moments of $X\sim \ln\mathcal{N}(\mu,\sigma^2)$.
$D_n(t,T)=S_t^ne^{(n-1)\left(r+\frac{1}{2}\sigma^2\right)(T-t)+\frac{(n-1)^2}{2}\sigma^2(T-t)}=S_t^ne^{(n-1)(r+\frac{n}{2}\sigma^2)(T-t)}$
