Definition 4.1

A power call option gives the holder the right (but not obligation) to buy 1 share of stock for $K$ and then immediately sell for market value raised to the power $\gamma$.

Pricing the power call option

It turns out that we can find an explicit formula for the risk-neutral G.B.M. valuation (fair price) of such options. This will be useful for comparing against estimation strategies later on.

Let $C_{\gamma }(S_{0},T,K,\sigma ^{2},r)$ denote the price of the European call option with power $\gamma$. As usual, $S_{0}$ is the initial stock price, $T$ is the maturity, $K$ is the strike price, $\sigma ^2$ is the volatility (under the assumption of G.B.M.) and $r$ is the interest rate. Hence, $C_{1}(S_{0},T,K,\sigma ^{2},r)=C(S_{0},T,K,\sigma ^{2},r)$ denotes the Black-Scholes price of the vanilla call option, that is, with power parameter 1. In particular, note that the Black-Scholes price of one European call option is the discounted expected payoff at maturity

where $S_T / S_0 \sim LN([r-\sigma ^2/2]T,\sigma ^2 T)$ under the assumption of risk-neutral G.B.M. We know that this expression can be evaluated to give the Black-Scholes formula of Section 2.

Now, the fair price of the power call option is

Now, under the assumption that stock prices follow the risk neutral G.B.M. we know that $S_{T}/S_{0}$ has a log-normal distribution as above. Note that

We can force this distribution to have the same form as G.B.M. by defining new parameters

Hence,

Now, we have $C_{\gamma }(S_{0},T,K,\sigma ^{2},r)$ as

where the last line follows by using the form of the Black-Scholes price (as a discounted expected payoff with respect to a G.B.M., where in this case $S_0$ is replaced by $S_0^\gamma$, $r$ by $r_\gamma$ and $\sigma$ by $\sigma _\gamma$). Hence we have found an explicit formula for the fair price of a power call option with parameter $\gamma$. Let’s see how this works with an example.

Example 4.1

Find the fair (risk-neutral) price of a power call option with maturity $T=2$ years, strike price $\pounds 3000$, initial stock price $S_{0}=\pounds 50$ and payoff, max$(S^{2}_{T}-K,0)$ (a power parameter of 2). Suppose further that the risk-free interest rate is $r=0.05$ and stock price follows a geometric Brownian motion with volatility parameter $\sigma ^{2}=0.01$.

Solution

The fair price of the option is $$C_{\gamma }(S_{0},T,K,\sigma ^{2},r)=e^{-rT}e^{r_{\gamma }T}C_{1}(S^{\gamma }_{0},T,K,\sigma ^{2}_{\gamma },r_{\gamma })$$ where $S_{0}=50$, $T=2$, $K=3000$, $\sigma ^{2}=0.01$, $r=0.05$, $\gamma =2$, $S_{0}^{2}=2500$, $\sigma ^{2}_{\gamma }=\gamma ^{2}\sigma ^{2}=0.04$ and $$\begin{eqnarray*} r_{\gamma }-\sigma _{\gamma }^{2}/2& =& 0.09\\ \Rightarrow r_{\gamma }& =& 0.09+0.02=0.11\, . \end{eqnarray*}$$ We compute $C_{1}(S^{\gamma }_{0},T,K,\sigma ^{2}_{\gamma },r_{\gamma })$ using the Black-Scholes formula of Section 2, $$C_{1}(S^{\gamma }_{0},T,K,\sigma ^{2}_{\gamma },r_{\gamma })=2500\Phi (\omega )-3000e^{-0.11\times 2}\Phi (\omega -0.2\sqrt{2})$$ where $$\omega =\frac{0.11\times 2+0.2^{2}\times 2/2-\log (3000/2500)}{0.2\sqrt{2}}=0.2746$$ and recall that $\log$ is base $e$. Hence we obtain $C_{1}(S^{\gamma }_{0},T,K,\sigma ^{2}_{\gamma },r_{\gamma })=324.59$. Therefore the fair price of the power call option is $$\begin{eqnarray*} C_{\gamma }(S_{0},T,K,\sigma ^{2},r)& =& e^{(0.05+2\times 0.1^{2}/2)2}\times 324.59\\ & =& 365.97. \end{eqnarray*}$$ We will revisit this example later on, when looking at Monte Carlo pricing.

Definition 4.2

There are several types of barrier option:

• A down-and-in barrier option gives the holder the right to exercise the option at time $T$ provided that the stock price goes below $\nu$ at some time before $T$ i.e. the option becomes alive only if the security’s price goes below $\nu$ before $T$.

• A down-and-out barrier option is killed if the security’s price goes below $\nu$ before $T$. Note that in both the down-and out and down-and-in options, $\nu$ is a value less than the initial stock price $S_{0}$.

• An up-and-in barrier option becomes alive only if the security’s price goes above $\nu$ before $T$.

• An up-and-out barrier option is killed if the security’s price goes above $\nu$ before $T$. Note that in both the up-and out and up-and-in options, $\nu$ is a value greater than the initial stock price $S_{0}$.

Illustration

Figure 2 shows an example of an up-and-out barrier call that either expires worthless (barrier breached) or not (barrier not breached and above strike at maturity).

Comments

• Provided the option remains alive, the payoff at time $T$ (for the European call) is max$(S_{T}-K,0)$. If the option is killed at any time, the payoff is 0.

• The same definitions exist for the European barrier put, the only difference being that if the option remains alive, the payoff at maturity is max$(K-S_{T},0)$.

• If you own both a down-and-out and a down-and-in call option both the same strike price $K$ and maturity $T$ then this is equivalent to owning one vanilla call option (with parameters $K$ and $T$). This is true since only one option can be in play at any time $t$ (the down-and-in option if the barrier is breached and the down-and-out otherwise). Consequently, if we denote by $C_{di}$ and $C_{do}$ the respective risk neutral present values of owning the down-and-in and down-and-out call options, then

  $$C_{di}+C_{do}=C$$

  where $C$ is the Black-Scholes valuation of the vanilla European call option given in Section 2. A similar argument follows for the up-and-in and up-and-out options.

• We typically observe stock price on a daily basis. Therefore, let

  $$S_{i}^{d}=S_{i/252}$$

  denote the price on day $i$ at some arbitrary time. Hence, the down-and-in barrier call option has payoff

  $$\left\{ \begin{array}{cc} (S_{T}-K)^{+} & \text{if \(S^{d}_{i}\leq \nu \) for some \(i=1,\ldots ,252T\)} \\ 0 & \text{if \(S^{d}_{i}> \nu \) for all \(i=1,\ldots ,252T\)} \end{array} \right.$$

  Similarly, the down-and-out call option has payoff

  $$\left\{ \begin{array}{cc} 0 & \text{if \(S^{d}_{i}\leq \nu \) for some \(i=1,\ldots ,252T\)} \\ (S_{T}-K)^{+} & \text{if \(S^{d}_{i}> \nu \) for all \(i=1,\ldots ,252T\)} \end{array} \right.$$

Definition 4.3

The holder of the Asian call option has the right (but not obligation) to buy 1 share for $K$ and sell for the average price realised over $(0,T]$.

Hence, assuming daily prices, the payoff is

A common Asian put option with strike price $K$ and maturity $T$ has payoff

Definition 4.4

The holder of the lookback call option has the right (but not obligation) to buy 1 share for

at time $T$.

Hence, the lookback call option with maturity $T$ has strike price given by the minimum end-of-day price up to the maturity time. The payoff at time $T$ is

The lookback put has strike price given by the maximum end-of-day price up to the maturity time. Hence, the payoff is

Note that because the payoffs of both the lookback and Asian type options depend on the price path followed, there are no known exact formulas for the risk-neutral valuations of these options. However, approximate valuations are possible by using Monte Carlo simulation methods.

Example 4.1 revisited

Find the fair (risk-neutral) price of a power call option with maturity $T=2$ years, strike price $\pounds 3000$, initial stock price $S_{0}=\pounds 50$ and payoff, max$(S^{2}_{T}-K,0)$ (a power parameter of 2). Suppose further that the risk-free interest rate is $r=0.05$ and stock price follows a geometric Brownian motion with volatility parameter $\sigma ^{2}=0.01$. Compare the analytic fair price to estimates obtained via Monte Carlo.

Solution

Recall that the exact fair price is $365.97$. We can also estimate the fair price of the option via Monte Carlo. A key step is the simulation of the stock price process. Recall from Section 2.3.1 that for an equally spaced partition of $[0,T]$ with time step $\Delta t$ $$S_{t_i}=S_{t_{i-1}}\exp \left\{ (r-\sigma ^2/2)\Delta t + \sigma (W_{t_{i}} - W_{t_{i-1}}) \right\}$$ where $W_{t_{i}}-W_{t_{i-1}} \sim N(0, \Delta t)$. To simulate on a daily basis, we set $\Delta t=1/252$ and, starting with $S_{0}^{d}$ as a known value, simulate $S^{d}_{1},\ldots S^{d}_{252T}$ via the recursion $$S_{i}^{d}=S_{i-1}^{d}\exp \left\{ (r-\sigma ^{2}/2)\frac{1}{252}+\sigma (W_{i}^{d}-W_{i-1}^{d})\right\} , \quad i=1,2,\ldots ,252T$$ where $252T$ is the maturity time in days. We then calculate a sample value of the expected payoff at time $T$, via $$X_{1}=\textrm{max}\left((S^{d}_{252T})^{2}-3000,0\right)\, .$$ We repeat these steps a further $N-1$ times to give $N$ sample payoffs $X_{1},\ldots ,X_{N}$. We take the average and discount at the risk free interest rate to give an estimate of the fair price of the power option as $$e^{-rT}\frac{1}{N}\sum _{i=1}^{N}X_{i}\, .$$ The following $\mathsf{R\; }$function takes as arguments $S_{0}$, $K$, $T$, $\gamma$, $\sigma$, $r$ and $N$, and returns the Monte Carlo estimate of the fair price of the power option.

A single execution of this function (with $N=10000$) gave an estimate of 372.10 which is in reasonable agreement with the actual price of 365.97. Note that the payoff is not dependent on the whole path, only the price of the stock at maturity. Since we know the distribution of $S_{T}$ (log-normal), it is far more efficient to simulate $N$ values of $S_{T}$ and then set $$X_{i}=\textrm{max}\left((S_{T})^{2}-3000,0\right)\, i=1,\ldots ,N$$ before taking the discounted average as an estimate of the fair price. The required $\mathsf{R\; }$function is

Finally, note that every time we execute the function, we get a different estimate, since we’re generating a realisation of the estimator, which is a random variable.

Example 4.2

Suppose that stock prices follow a G.B.M. with volatility parameter $\sigma ^{2}=0.01$, the initial stock price is $\pounds 50$, the risk free interest rate is $r=0.05$. Describe a detailed Monte Carlo algorithm to find the risk-neutral (fair) price of a down-and-in barrier call option with strike price $K=\pounds 51$, maturity $T=1$ year and barrier $\nu =49$.

Solution

Let us assume that the stock price is observed daily. The fair price of the option is the discounted expected payoff at time $T$ given by $$e^{-rT}\textrm{E}\left(I(S^{d}_{252}-K)^{+}\right)$$ where $I$ is an indicator function defined by $$I=\left\{ \begin{array}{cc} 1 & \text{if \(S^{d}_{i}\leq \nu \) for some \(i=1,\ldots ,252\)} \\ 0 & \text{if \(S^{d}_{i}> \nu \) for all \(i=1,\ldots ,252\)} \end{array} \right.$$ That is, the option becomes alive (and remains alive) if the end-of-day stock price falls below $\nu$ at any time before maturity. We estimate the fair price $C_{di}$ of the option by implementing the following sequence of steps:

1. Simulate a random path $S^{d}_{0},S^{d}_{1},\ldots ,S^{d}_{252T}$ in the risk-neutral world, via the recursion

   $$S_{i}^{d}=S_{i-1}^{d}\exp \left\{ (r-\sigma ^{2}/2)\frac{1}{252}+\sigma (W_{i}^{d}-W_{i-1}^{d})\right\} , \quad i=1,2,\ldots ,252$$

   where $W_{i}^{d}-W_{i-1}^{d}\sim \textrm{N}(0,1/252)$.

2. Calculate a sample payoff at time $T$ with

   $$X_{1}=I\times \textrm{max}(0,S_{252}^{d}-51).$$

3. Repeat steps 1 and 2 to get say $N$ sample values of the payoff,

   $$X_{1},X_{2},\ldots ,X_{N}$$

4. Calculate the mean of these sample payoffs to get an estimate of the expected payoff.

5. Discount the expected payoff at the risk-free interest rate to get an estimate of the value of the option, that is, calculate

   $$e^{-rT}\bar{X}.$$

In practice, this is achieved in $\mathsf{R\; }$with (for example) the following function.

A single call of this function with $N=10000$ gave an estimate of $C_{di}$ as 1.24. We can use this value to estimate the fair price $C_{do}$ of the down-and-out barrier call option (with the same parameters) by using the relation in the comments on page 54. We calculate the fair price of the vanilla call option (with the same parameters) via the Black-Scholes formula. Performing the desired calculation gives $C=2.80$. Hence an estimate of $C_{do}$ is 1.56.

Comments

• Note that between any two time instants $t_{i}$ and $t_{i+1}$ at which we observe the stock, $S_{t}$ could fall below $\nu$ but then increase sufficiently to become greater than $\nu$ before $t_{i+1}$. This breaking of the barrier would go undetected as we only have the sample path at discrete time intervals. However, provided we use a sufficiently fine discretisation, the probability of this occurring is very small. Hence, the error introduced from discrete sampling of the path is small.

Using Antithetic Variables

Recall that a realisation of the daily stock price process can be generated via the recursion

where $W_{i}^d-W_{i-1}^d \sim \textrm{N}(0,1/252)$. It will be helpful for us to re-write this equivalently as

Hence, in step 1 of the Monte Carlo algorithm we generate $Y_{1},\ldots ,Y_{252T}$ and use these values to compute $S_{1}^{d},\ldots ,S_{252T}^{d}$. In step 2, we calculate a realisation of the payoff $X_{1}=g(S_{1}^{d},\ldots ,S_{252T}^{d})$.

The anithetic technique re-uses / recycles the $Y_i$ in a clever way to calculate a second stock price realisation, and in turn, a second payoff $X_2$ that is negatively correlated with $X_1$. To this end, the anithetic technique sets

and uses the $Y_{i}^{*}$ to generate a new realisation of the price process, $S_{1}^{d*},\ldots ,S_{252T}^{d*}$, before finally computing $X_{2}=g(S_{1}^{d*},\ldots ,S_{252T}^{d*})$. The process then repeats to calculate pairs of negatively correlated payoffs $X_{3}$ and $X_{4}$, etc.

Is this a valid Monte Carlo strategy? Yes. Note that the $Y_{i}^{*}$ have the same distribution as the $Y_{i}$ but are negatively correlated with the $Y_{i}$. To see this, note that $Y_i^*$ is just a linear transformation of $Y_i$ and

Finally,

This negative covariance induces a negative covariance between $X_1$ and $X_2$, $X_3$ and $X_4$ etc. Consequently, for the antithetic scheme,

and we note that the first term is the variance of the estimator used in the standard Monte Carlo scheme and the second term is negative. Hence, comapared to the standard scheme, the antithetic scheme gives an estimator with a smaller variance.