6.2. The Greeks

All Greeks are first-order or second-order partial derivatives of the option price with respect to different underlying factors, which are used to measure the risks—as well as potential returns—of the financial derivative. The following Greeks for a derivative $f$ are routinely used by financial institutions:

Delta: $\Delta=\frac{\partial f}{\partial S}$; Gamma: $\Gamma=\frac{\partial^2f}{\partial S^2}$; Theta: $\Theta=\frac{\partial f}{\partial t}$; Vega: $\upsilon=\frac{\partial f}{\partial\sigma}$; Rho: $\rho=\frac{\partial f}{\partial r}$

Delta

For a European call with dividend yield $y$: $\Delta=e^{-y\tau}N(d_1)$

For a European put with dividend yield $y$: $\Delta=-e^{-y\tau}[1-N(d_1)]$

A. What is the delta of a European call option on a non-dividend paying stock? How do you derive the delta?

Solution: The delta of a European call on a non-dividend paying stock has a clean expression: $\Delta=N(d_1)$. For the derivation, though, many make the mistake by treating 150 $N(d_1)$ and $N(d_2)$ as constants in the call pricing formula $c=SN(d_1)-Ke^{-r\tau}N(d_2)$ and simply taking the partial derivative on $S$ to yield $N(d_1)$. The derivation step is actually more complex than that since both $N(d_1)$ and $N(d_2)$ are functions of $S$ through $d_1$ and $d_2$. So the correct partial derivative is $\frac{\partial c}{\partial S}=N(d_1)+S\times\frac{\partial}{\partial S}N(d_1)-Ke^{-r\tau}\frac{\partial}{\partial S}N(d_2)$.

Take the partial derivative with respect to $S$ for $N(d_1)$ and $N(d_2)$⁹:

So we have $\frac{\partial}{\partial S}N(d_2)=\frac SK e^{r\tau}N(d_1)\Rightarrow S\times\frac{\partial}{\partial S}N(d_1)-Ke^{-r\tau}\frac{\partial}{\partial S}N(d_2)=0$. Hence, the last two components of $\frac{\partial c}{\partial S}$ cancel out and $\frac{\partial c}{\partial S}=N(d_1)$.

B. What is your estimate of the delta of an at-the-money call on a stock without dividend? What will happen to delta as the at-the-money option approaches maturity date?

Solution: For an at-the-money European call, the stock price equals the strike price. $S=K\Rightarrow d_1=\frac{(r+\sigma^2/2)\tau}{\sigma\sqrt\tau}=(\frac r\sigma+\frac\sigma2)\sqrt\tau>0$ and $\Delta=N(d_1)>0.5$. As shown in Figure 6.2, all at-the-money call options indeed have $\Delta>0.5$ and the longer the maturity, the higher the $\Delta$. As $T-t\to0$, $(\frac r\sigma+\frac\sigma2)\sqrt\tau\to0\Rightarrow N(d_1)=N(0)=0.5$, which is also shown in Figure 6.2 ($T=10$ days). The same argument is true for calls on stock with continuous dividend rate $y$ if $r>y$.

Figure 6.2 also shows that when $S$ is large ($S\gg K$), $\Delta$ approaches 1. Furthermore, the shorter the maturity, the faster the delta approaches 1. On the other hand, if $S$ is small ($S\ll K$), $\Delta$ approaches 0 and the shorter the maturity, the faster the delta approaches 0.

C. You just entered a long position for a European call option on GM stock and decide to dynamically hedge the position to eliminate the risk from the fluctuation of GM stock price. How will you hedge the call option? If after your hedge, the price of GM has a sudden increase, how will you rebalance your hedging position?

Solution: Since $d_1=\frac{\ln(S/K)+(r-y+\sigma^2/2)\tau}{\sigma\sqrt\tau}$ and $\Delta=e^{-y\tau}N(d_1)$ is a monotonously increasing function of $d_1$, we have $S\uparrow\Rightarrow d_1\uparrow\Rightarrow\Delta\uparrow$.

One hedging method is delta hedging, for which we short $\Delta=e^{-y\tau}N(d_1)$ shares of stock for each unit of call option to make the portfolio delta-neutral. Since $\Delta$ shares of GM stock costs more than one unit of GM option, we also need to invest cash (if the option price exactly follows the Black-Scholes formula, we need to lend $$Ke^{-r\tau}N(d_2)$ for each 152 unit of option) in the money market. If there is a sudden increase in $S$, $d_1$ increases and $\Delta$ increases as well. That means we need to short more stock and lend more cash ($Ke^{-r\tau}N(d_2)$ also increases).

The delta hedge only replicates the value and the slope of the option. To hedge the curvature of the option, we will need to hedge gamma as well.

D. Can you estimate the value of an at-the-money call on a non-dividend paying stock? Assume the interest rate is low and the call has short maturity.

Solution: When $S=K$, we have $c=S(N(d_1)-e^{-r\tau}N(d_2))$. In a low-interest environment, $r\approx0$ and $e^{-r\tau}\approx1$, so $c\approx S(N(d_1)-N(d_2))$.

We also have $\displaystyle N(d_1)-N(d_2)=\int_{d_2}^{d_1}\frac1{\sqrt{2\pi}}e^{-\frac12x^2}dx$,

where $d_2=(\frac r\sigma-\frac\sigma2)\sqrt\tau$ and $d_1=(\frac r\sigma+\frac\sigma2)\sqrt\tau$.

For a small $r$, a typical $\sigma$ for stocks (< 40% per year) and a short maturity (< 3 months), both $d_2$ and $d_1$ are close to 0. For example, if $r=0.03$, $\sigma=0.3$, and $\tau=1/6$ year, then $d_2=-0.02$ and $e^{-\frac12d_2^2}=0.98$.

In practice, this approximation is used by some volatility traders to estimate the implied volatility of an at-the-money option.

(The approximation $e^{-\frac12x^2}\approx1$ causes a small overestimation since $e^{-\frac12x^2}<1$; but the approximation $-e^{-r\tau}K\approx-K$ causes a small underestimation. To some extent, the two opposite effects cancel out and the overall approximation is fairly accurate.)

Gamma

For a European call/put with dividend yield $y$: $\Gamma=\frac{N'(d_1)e^{-y\tau}}{S_0\sigma\sqrt\tau}$

What happens to the gamma of an at-the-money European option when it approaches its maturity?

Solution: From the put-call parity, it is obvious that a call and a put with identical characteristics have the same gamma (since $\Gamma=0$ for both the cash position and the underlying stock). Taking the partial derivative of the $\Delta$ of a call option with respect to $S$, we have $\Gamma=\frac{N'(d_1)e^{-y\tau}}{S\sigma\sqrt\tau}$, where $N'(d_1)=\frac1{\sqrt{2\pi}}e^{-\frac12d_1^2}$.

So for plain vanilla call and put options, gamma is always positive.

Figure 6.3 shows that gamma is high when options are at the money, which is the stock price region that $\Delta$ changes rapidly with $S$. If $S\ll K$ or $S\gg K$ (deep in the money or out of the money), gamma approaches 0 since $\Delta$ stays constant at 1 or 0.

The gamma of options with shorter maturities approaches 0 much faster than options with longer maturities as $S$ moves away from $K$. So for deep in-the-money or deep out-of-the-money options, longer maturity means higher gamma. In contrast, if the stock prices are close to the strike price (at the money) as the maturity nears, the slope of delta for an at-the-money call becomes steeper and steeper. So for options close to the strike price, shorter-term options have higher gammas.

As $\tau\to0$, an at-the-money call/put has $\Gamma\to\infty$ ($\Delta$ becomes a step function). This can be shown from the formula of gamma for a European call/put with no dividend, $\Gamma=\frac{N'(d_1)}{S\sigma\sqrt\tau}$:

When $S=K$, $d_1=\lim_{\tau\to0}(\frac r\sigma+\frac\sigma2)\sqrt\tau\to0\Rightarrow\lim_{\tau\to0}N'(d_1)\to\frac1{\sqrt{2\pi}}$. The numerator is $1/\sqrt{2\pi}$; yet the denominator has a limit $\lim_{\tau\to0}S\sigma\sqrt\tau\to0$, so $\Gamma\to\infty$. In other words, When $t=T$, delta becomes a step function. This phenomenon makes hedging at-the-money options difficult when $t\to T$ since delta is extremely sensitive to changes in $S$.

Theta

For a European call option:

For a European put option:

When there is no dividend, the theta for a European call option is simplified to $\Theta=-\frac{SN'(d_1)}{2\sqrt\tau}-rKe^{-r\tau}N(d_2)$, which is always negative. As shown in Figure 6.4, when $S\ll K$, $N(d_2)\approx0$ and $N'(d_1)\approx0$. Hence, $\Theta\to0$. When $S\gg K$, $N(d_2)\approx1$ and 155 $N'(d_1)\approx0$. Hence, $\Theta\to-rKe^{-r\tau}$. When $S\approx K$, $\Theta$ has large negative value and the smaller the $\tau$, the more negative the $\Theta$.

A. When will a European option have positive theta?

Solution: For American options as well as European calls on non-dividend paying assets, theta is always negative. But for deep in-the-money European puts, their values may increase as $t$ approaches $T$ if all other factors remain the same, so they may have positive theta.

A put option on a non-dividend paying asset has $\Theta=-\frac{SN'(d_1)\sigma}{2\sqrt\tau}+rKe^{-r\tau}N(-d_2)$. If the put option is deep in-the-money ($S\ll K$), then $N'(d_1)\approx0$ and $N(-d_2)\approx1$. Hence, 156 $\Theta\approx rKe^{-r\tau}>0$. That’s also the reason why it can be optimal to exercise a deep in-the-money American put before maturity.

For deep in-the-money European call options with high dividend yield, the theta can be positive as well. If a call option with high dividend yield is deep in-the-money ($S\gg K$), $N(d_1)\approx N(d_2)\approx1$, $N'(d_1)\approx0$, so the component $ySe^{-y\tau}N(d_1)$ can make $\Theta$ positive.

B. You just entered a long position for a call option on GM and hedged the position by shorting GM shares to make the portfolio delta neutral. If there is an immediate increase or decrease in GM’s stock price, what will happen to the value of your portfolio? Is it an arbitrage opportunity? Assume that GM does not pay dividends.

Solution: A position in the underlying asset has zero gamma. So the portfolio is delta-neutral and long gamma. Therefore, either an immediate increase or decrease in the GM stock price will increase the portfolio value. The convexity (positive gamma) enhances returns when there is a large move in the stock price in either direction.

Nevertheless, it is not an arbitrage opportunity. It is a trade-off between gamma and theta instead. From the Black-Scholes-Merton differential equation, the portfolio $V$ satisfies the equation $\frac{\partial V}{\partial t}+rS\frac{\partial V}{\partial S}+\frac12\sigma^2S^2\frac{\partial^2V}{\partial S^2}=\Theta+rS\Delta+\frac12\sigma^2S^2\Gamma=rV$. For a delta-neutral portfolio, we have $\Theta+\frac12\sigma^2S^2\Gamma=rV$. This indicates that gamma and theta often have opposite signs. For example, when an at-the-money call approaches maturity, gamma is large and positive, so theta is large and negative. Our delta neutral portfolio has positive gamma and negative theta. That means if the price does not move, the passage of time will result in a lower portfolio value unless we rebalance. So the portfolio does not provide an arbitrage opportunity.

Vega

For European options: $\upsilon=\frac{\partial c}{\partial\sigma}=\frac{\partial p}{\partial\sigma}=Se^{-y\tau}\sqrt\tau N'(d_1)$

At-the-money options are most sensitive to volatility change, so they have higher vegas than either in-the-money or out-of-the-money options. The vegas of all options decrease as time to expiration becomes shorter ($\sqrt\tau\to0$) since a long-term option is more sensitive to change in volatility.

A. Explain implied volatility and volatility smile. What is the implication of volatility 157 smile for the Black-Scholes pricing model?

Solution: Implied volatility is the volatility that makes the model option price equal to the market option price. Volatility smile describes the relationship between the implied volatility of the options and the strike prices for a given asset. For currency options, implied volatilities tend to be higher for in-the-money and out-of-the-money options than for at-the-money options. For equity, volatility often decreases as the strike price increases (also called volatility skew). The Black-Scholes model assumes that the asset price follows a lognormal distribution with constant volatility. In reality, volatilities are neither constant nor deterministic. In fact, the volatility is a stochastic process itself. Furthermore, there may be jumps in asset prices.

B. You have to price a European call option either with a constant volatility 30% or by drawing volatility from a random distribution with a mean of 30%. Which option would be more expensive?

Solution: Many would simply argue that stochastic volatility makes the stock price more volatile, so the call price is more valuable when the volatility is drawn from a random distribution. Mathematically, the underlying argument is that the price of a European call option is a convex function of volatility and as a result $c(E[\sigma])\le E[c(\sigma)]$, where $\sigma$ is the random variable representing volatility and $c$ is the call option price. Is the underlying argument correct? It’s correct in most, but not all, cases. If the call price $c$ is always a convex function of $\sigma$, then $\frac{\partial^2c}{\partial\sigma^2}\ge0$. $\frac{\partial c}{\partial\sigma}$ is the Vega of the option. For a European call option,

The secondary partial derivative $\frac{\partial^2c}{\partial\sigma^2}$ is called Volga. For a European call option,

$\upsilon$ is always positive. For most out-of-the-money call options, both $d_1$ and $d_2$ are negative; for most in-the-money call options, both $d_1$ and $d_2$ are positive. So $d_1d_2>0$ in most cases and $c$ is a convex function of $\sigma$ when $d_1d_2>0$. But theoretically, we can have conditions that $d_1>0$ and $d_2<0$ and $\frac{\partial^2c}{\partial\sigma^2}<0$ when the option is close to being 158 at-the-money. So the function is not always convex. In those cases, the option with constant volatility may have a higher value.

C. The Black-Scholes formula for non-dividend paying stocks assumes that the stock follows a geometric Brownian motion. Now assume that you don’t know the stochastic process followed by the stock price, but you have the European call prices for all (continuous) strike prices $K$. Can you determine the risk-neutral probability density function of the stock price at time $T$?

Solution: The payoff a European call at its maturity date is $Max(S_T-K,0)$. Therefore under risk-neutral measure, we have $\displaystyle c=e^{-r\tau}\int_K^{\infty}(s-K)f_{S_T}(s)ds$, where $f_{S_T}(s)$ is the probability density function of $S_T$ under the risk-neutral probability measure. Taking the first and second derivatives of $c$ with respect to $K$,¹⁰ we have

and $\displaystyle\frac{\partial^2c}{\partial K^2}=\frac{\partial}{\partial K}(\frac{\partial c}{\partial K})=e^{-r\tau}\frac{\partial}{\partial K}\int_K^{\infty}-f_{S_T}(s)ds=e^{-r\tau}f_{S_T}(K)$.

Hence the risk-neutral probability density function is $\displaystyle f_{S_T}(K)=e^{r\tau}\frac{\partial^2c}{\partial K^2}$.