3
$\begingroup$

The following question is taken from Mark Joshi's Concepts and Practice of Mathematical Finance, second edition, Exercise $6.6$

Suppose a stock follows geometric Brownian motion in a Black-Scholes world. Develop an expression for the price of an option that pays $S_t^2-K$ if $S_t^2>K $ and zero otherwise. What PDE will the option price satisfy?

I am pretty confused by hints given for the first part of the question. I retype the hint below.

$$F_T(t) = e^{r(T-t)}S_t,$$ $$F_T(T) = F_T(0) e^{-\frac{1}{2}\sigma^2T + \sigma\sqrt{T} N(0,1)},$$ $$F_T(T)^2 = F_T(0)^2 e^{-\sigma^2T + 2\sigma\sqrt{T} N(0,1)} = F_T(0)^2 e^{\sigma^2 T} e^{-\frac{1}{2}(2\sigma \sqrt{T})^2 + 2\sigma \sqrt{T} N(0,1)}.$$ So to price, we just use the Black formula with forward equal to $$F_T(0)^2 e^{\sigma^2 T}$$ and volatility equal to $2\sigma.$

Why would the calculations above work for pricing an option? I thought we need to use the risk-neutral pricing formula plus the fact that stock price follows geometric Brownian motion to deduce a price for the option. Of course, the latter is quite tedious and prone to error. If the former works, then it would be good as its calculation is quite short. But I do not understand its logic behind.

$\endgroup$
  • 1
    $\begingroup$ Well, what the BS formula is really giving you is the price of a call option written on an asset which follows a geometric Brownian Motion with parameters $r$ and $\sigma$. Hence, for a call option written on another asset, if that asset also follows a GBM, then you can try to replace the values in the standard BS formula ($r$, $\sigma$) with those of your new asset. $\endgroup$ – Daneel Olivaw Dec 16 '19 at 16:04
  • $\begingroup$ @DaneelOlivaw But I do not see why would Mark Joshi uses forward price to price such option. $\endgroup$ – Idonknow Dec 16 '19 at 23:12
4
$\begingroup$

The price of the option is $$C_0=E\left[e^{-\int_{0}^{T}{r_sds}} \left(S_T^2-K\right)^+\right]$$ where $r$ is the risk free rate, $E$ is the expectation under the risk-neutral measure.

Define the process $U_t=S_t^2$ and apply Ito's lemma $$dU_t=2S_tdS_t+d<S_t,S_t>$$

The assumption of Black-scholes implies that the process $r$ is constant and the process $S_t$ follows : $$dS_t=rS_tdt+\sigma S_t dW_t$$

Therefore, $$dU_t=2rS_t^2dt+2\sigma S_t^2 dW_t+\sigma^2S_t^2dt$$ $$dU_t=(2r+\sigma^2)U_tdt+2\sigma U_t dW_t$$

We define $$\bar{r}=2r+\sigma$$ $$\bar{\sigma}=2 \sigma$$

Finally, we have

$$dU_t=\bar{r}U_tdt+\bar{\sigma} U_t dW_t$$

and the option price is $$C_0=E\left[e^{-\int_{0}^{T}{r_sds}} \left(U_T-K\right)^+\right]$$

You can use your Black-Scholes formula for the process $U_t$, and get the option formula.

EDIT :: Why do we use the forward ? The reason is that the forward underlying is usually defined in a way that it is a martingale under its natural measure, therefore we just need to define its volatility. Let start from the first equation, and let assume that the rate is stochastic.

The price of the option is $$C_0=E\left[e^{-\int_{0}^{T}{r_sds}} \left(S_T^2-K\right)^+\right]$$

Introducing the zero-coupon bond $$P(t,T)=E\left[e^{-\int_{t}^{T}{r_sds}} |\mathcal{F_t}\right]$$. $\mathcal{F}$ is the market filtration.

We can price the option as $$C_0=P(0,T)E^T\left[ \left(S_T^2-K\right)^+\right]$$

where $E^T$ is the expectation under the measure associated to the bond (called the T-forward measure).

Let's define the forward function as $$F_T(t)=E^T\left[S_T | \mathcal{F_t}\right]$$,

Obviously $F_T(T)=S_T$ and $F_T(t)$ is a martingale under the $T$-forward measure by construction( it is a conditional expectation).

The price of the option becomes : $$C_0=P(0,T)E^T\left[ \left(F_T(T)^2-K\right)^+\right]$$

So far, as you can see , I have not introduced any model , and inside the expectation terms, we only have one random variable, and it is a martingale. This is why using the forward underlying is very powerful. We can keep going , let assume that the forward underlying is a geometrical Brownian motion $$dF_T(t)=\sigma F_T(t)dW_t$$ where $W$ is a Brownian motion under the $T$-forward measure. or $$F_T(T)=F_T(0)e^{-\frac{1}{2}\sigma^2T+\sigma W_T}$$ You can finish the proof using Black-Scholes with $r=0$

We have the price of the option without assuming that the rate is deterministic/constant. In the rate world, where rates are stochastic, we must work with forward underlying to avoid having complex payoff. In the Black-Scholes world, I can understand that introducing the forward seems irrelevant. Btw, Mark Joshi (RIP) was a rate quant, that may explain why he did that.

Now, we need to calculate $F_T(0)$

$$F_T(0)=E^T\left[S_T \right]=E^T\left[\frac{S_T}{P(T,T)}\right]=\frac{S_0}{P(0,T)}$$

Finally, if you assume that rate is constant, we have $$P(0,T)=e^{-rT}$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ I understand that to price an option, one can use risk-neutral pricing formula, which is the first equality in your answer. But what I do not understand is that, why would forward price be used in calculating the option? I do not see where would forward price comes in. $\endgroup$ – Idonknow Dec 16 '19 at 23:05
  • $\begingroup$ Since $\overline{\sigma} = 2\sigma,$ I assume that the new Black-Scholes PDE is $\frac{\partial U}{\partial t} + rS_t \frac{\partial U}{\partial S_t}+ 2\sigma^2 \frac{\partial ^2 U}{\partial S_t^2} = rU?$ $\endgroup$ – Idonknow Dec 16 '19 at 23:23
  • $\begingroup$ What numeraire are you using? $\endgroup$ – UBM Dec 17 '19 at 9:30
  • $\begingroup$ I edited the answer $\endgroup$ – Canardini Dec 17 '19 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.