# Derivatives pricing w/ squared and cubed stock prices

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.

• For the first part I have tried using that $S_T$ can be written as $S_T = S_te^{\left(r - \frac{\sigma^2}{2}\right) (T-t) + \sigma(W_T-W_t)}$ where $W_t$ is a Brownian Motion. I seem to run into problems though.. Nov 10, 2014 at 23:25
• Here is something I've tried: Call the value of the option $v_t$. Under the risk-neutral measure it must be the case that $v_t = e^{-r(T-t)}E[S_T^3]$ Using the MGF of a lognormal variable I can get $E[S_T^3] = e^{3E[\ln(S_t)]+\frac{8\sigma^2}{2} = 3(\ln(E[S_t]) - \frac{\sigma^2}{2}) + \frac{8\sigma^2}{2} = S_t^3e^{\frac{5}{2}\sigma^2}$ This would make the option price $v_t = S_t^3e^{-r(T-t)+\frac{5}{2}\sigma^2}$ Any suggestions as to why this doesn't work? Nov 12, 2014 at 2:50

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 $$X\sim \ln\mathcal{N}(\mu,\sigma^2)$$.
$$D_n(t,T)=S_t^ne^{(n-1)\left(r+\frac{1}{2}\sigma^2\right)(T-t)+\frac{(n-1)^2}{2}\sigma^2(T-t)}=S_t^ne^{(n-1)(r+\frac{n}{2}\sigma^2)(T-t)}$$