Pricing a Squared-Stock Payoff Under Geometric Brownian Motion

# Finding price of the power option


# 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 $H=X_T^2$.

## Answer by Kevin (score 7)

https://quant.stackexchange.com/a/49508

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.

## Answer by NN2 (score 3)

https://quant.stackexchange.com/a/49510

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

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