# Finding price of the power option

Let's assume a market with $$d=1$$ and $$X=X^1$$ satisfying

$$dX_t=\sigma X_t\,dW_t,\: \: X_0=1,$$

where $$(W_t)$$ is a standard Brownian motion. Assume that $$\mathbb{F}$$ is the natural filtration of $$X$$ and $$\mathcal{F}=\mathcal{F}_T$$.

I would like to find the price of the contingent claim $$H=X_T^2$$.

The stock price process $$(X_t)$$ is a geometric Brownian motion with drift $$\mu=0$$. Thus, $$X_t=X_0\exp\left(-\frac{1}{2}\sigma^2t+\sigma W_t\right).$$ Assume you have constant interest rates $$r_t\equiv r$$ and are interested in a European-style claim, then, using risk-neutral pricing, the time zero price of a claim paying $$H=X_T^2$$ equals $$V_0 = e^{-rT}\mathbb{E}^\mathbb{Q}[X_T^2].$$

The time $$t$$ price would simply read as $$V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[X_T^2\mid\mathcal{F}_t]$$.

So, you need the moments of $$(X_T^2)$$ under the risk-neutral measure $$\mathbb{Q}$$. Regardless what mean $$(X_t)$$ has in the real-world, its drift in the risk-neutral world is $$r$$, (potentially $$r-q$$ where $$q$$ is the (constant) dividend yield). Thus, under $$\mathbb{Q}$$, $$\ln(X_t) \overset{\mathrm{Law}}{=} \ln(X_0)+\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t.$$ As you see, $$(X_t)$$ is log-normally distributed. In general, if $$\ln(Y)\sim N(m,s^2)$$, then $$\mathbb{E}[Y^k]=e^{km+\frac{1}{2}k^2s^2}$$ for all $$k\geq1$$, see here. Thus, putting everything together, and using that $$\ln(X_0)=0$$, you obtain as price of your power contract

\begin{align*} V_0 &= e^{-rT}\mathbb{E}^\mathbb{Q}[X_T^2] \\ &= \exp\left(-rT+2\left(r-\frac{1}{2}\sigma^2\right)T+2\sigma^2T\right) \\ &= \exp\left(\left(r+\sigma^2\right)T\right). \end{align*}

As a matter of fact, if $$\gamma>0$$, you can show that $$(X_t^\gamma)$$ is again a geometric Brownian motion using Ito's Lemma. Furthermore, you get that the time $$t$$ price of a European-style claim paying $$X_T^\gamma$$ is given by \begin{align*} V_t = (X_t)^\gamma\cdot\exp\left((\gamma-1)\left(r+\frac{1}{2}\gamma\sigma^2\right)(T-t)\right). \end{align*} Indeed, setting $$t=0$$, $$X_0=1$$ and $$\gamma=2$$ recovers the above solution.

• Thank you. Seems answer slightly different then the answer above, forgot to mention that interest rates r=0, also this is generic contigent claim ( not really European , or American) also what would change if contingent claim expressed like this $\frac{1}{H}$ Nov 4 '19 at 11:12
• @EdwardMoor The case $r=0$ does not matter, just use this value it in the formulae above. It does however matter whether the claim is European-style or American-style. When do you get the payoff? Only at maturity $t=T$ or can the buyer terminate the contract early and receive the payoff $H$ at any time $t\leq T$? This changes the pricing completely! Regarding a (European-style) claim paying $X_T^{-2}$ you need to find the distribution of $(X_t^{-2})$ under the risk-neutral measure $\mathbb{Q}$. Nov 4 '19 at 14:16

By applying the Ito's lemma on $$X_t^2$$, you find easily the process $$(X_t^2)$$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 = X_t^2 ( \sigma^2 dt +2\sigma dW_t)$$

so $$(X_t^2)$$ is a geometric Brownian motion with drift $$\mu = \sigma^2$$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +2\sigma dW_t$$

we can deduce that ( $$r = 0$$) $$V_0 = E^Q(X_T^2) = E^Q(x_0^2 e^{\sigma^2 T} e^{-\frac{1}{2}(2\sigma)^2 T+2\sigma W_T}) = x_0^2e^{\sigma^2 T} E^Q(e^{-\frac{1}{2}(2\sigma)^2 T+2\sigma W_T}) = x_0^2e^{\sigma^2 T} = e^{\sigma^2 T}$$

• Thank you , forgot to mention that interest rates r=0, also what would change if contingent claim expressed like this $\frac{1}{H}$ Nov 4 '19 at 11:11
• No, as you see, the price $V_t$ of $X_T^2$ doesn't depend on $r$.
– NN2
Nov 4 '19 at 12:57
• I made a wrong calculation at the last step. I just make a correction.
– NN2
Nov 4 '19 at 15:31