4

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 $H=X_T^2$.

Kevin
  • 15,879
  • 4
  • 31
  • 64
Edward Moor
  • 53
  • 4

2 Answers2

7

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.

Kevin
  • 15,879
  • 4
  • 31
  • 64
  • 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 $\frac{1}{H}$ – Edward Moor Nov 04 '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 $\mathbb{Q}$. – Kevin Nov 04 '19 at 14:16
3

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

NN2
  • 1,008
  • 6
  • 10