Delta and Gamma Hedging

Motivation. Imagine you are an investment bank, selling (which is equivalent to short-selling) a call option. At the end, you have to pay to the holder $C_T=\max \{S_T-K,0\}$. When the option is on the money at expiration time $T$, you have to pay $S_T-K$. You want to pay less, though; so you try to hedge against small changes in underlyer price.

Delta Hedging for Call/Put Options §

At time $t=0$ (day 0) §

• Sell the option $-C_t^E$, and get the cash $C_t^E$
• Borrow ${\mathbb{L}}_t$ amount of money to buy ${\Delta }_t$ units of stock. (this is the hedging part)
  • & this ${\Delta }_t$ is the same as the delta of the call option.

$$V_t=\underset{\text{ short-sell }}{\underbrace{-C_t^E}}+\underset{\text{ cash }}{\underbrace{C_t^E}}-\underset{\text{ loan }}{\underbrace{{\mathbb{L}}_t}}+\underset{\text{ underlyer }}{\underbrace{{\Delta }_t{S}_t}}$$

It is obvious that the value of this portfolio is $V_t=0$, at time $t=0$, but to show that this portfolio at any time $t$ always has value $0$, we observe the fact that the last three terms (let $C_t^{\text{syn}}:=+C_t^E-{\mathbb{L}}_t+{\Delta }_t{S}_t$) is a self-financing portflio that tracks the call price exactly. Observe from BSM derivation that a portfolio $W_t=n_t{S}_t^C+b_t{B}_t$ (we already used $V$) is self-financing and replicates when:

• $n_t=\frac{S_t}{S_t^c}{\Delta }_f$
• $b_t=\frac{f(S_t,t)-n_t{S}_t^C}{B_t}$ In our case, we have:
• $n_t=\frac{S_t}{S_t^C}{\Delta }_t$
• $b_t=\frac{C_t^E-S_t}{B_t}$ And thus $\underset{=b_t{B}_t}{\underbrace{C_t^E-{\mathbb{L}}_t}}+\underset{=n_t{S}_t}{\underbrace{{\Delta }_t{S}_t}}$ by the above equation.

At time $t=0+dt$ (day 1) §

• Option value changes to $-C_{t+dt}$
• Cash holding grows to $C_t^E{e}^{r\cdot dt}$
• We must “Re-hedge” or “balance” the loan and underlier such that we maintain ${\Delta }_{t+dt}$ units of underlyer. We break it out into three cases:
  • ${\Delta }_{t+dt}={\Delta }_t$ (delta stays same): time value of money, and use dividend to pay back loan
    • ${\mathbb{L}}_{t+dt}={\mathbb{L}}_t\underset{\text{ time value }}{\underbrace{e^{(r-q)\cdot dt}}}$
  • ${\Delta }_t<{\Delta }_{t+dt}$ (delta increases): Borrow more money to buy more stock:
    • ${\mathbb{L}}_{t+dt}={\mathbb{L}}_t{e}^{(r-q)dt}+|d{\Delta }_t|S_{t+dt}$
  • ${\Delta }_{t+dt}>{\Delta }_t$ (delta decreases): Sell stock to reimburse loan.
    • ${\mathbb{L}}_{t+dt}={\mathbb{L}}_t{e}^{(r-q)dt}-|d{\Delta }_t|S_{t+dt}$
  • In all three cases, we can say:

$${\mathbb{L}}_{t+dt}={\mathbb{L}}_t{e}^{(r-q)dt}+(d{\Delta }_t)S_{t+dt}$$

• We then purchase or sell $d{\Delta }_t$ units of the security to get ${\Delta }_{t+dt}{S}_{t+dt}$ As we show above, $C_{t+dt}^{\text{syn}}:=+C_{t+dt}^E-{\mathbb{L}}_t+{\Delta }_{t+dt}{S}_{t+dt}$ is self-financing. This means that total value of the portfolio is:

$$\begin{array}{rl}{V}_{t+dt}& =-C_{t+dt}^E+C_t^E{e}^{r\cdot dt}-{\mathbb{L}}_{t+dt}\\ & =-C_{t+dt}^E+\underset{=C_{t+dt}^E}{\underbrace{C_t^E{e}^{r\cdot dt}-\stackrel{\text{ loan }}{\overbrace{{\mathbb{L}}_t{e}^{(r-q)dt}+(d{\Delta }_t)S_{t+dt}}}+{\Delta }_{t+dt}{S}_{t+dt}}}\\ & =0\end{array}$$

At time $t=T$, (expiration date) §

On expiration and before the call is cashed out, the value of the portfolio is:

$$V_T=0=-\max \{S_T-K,0\}+C_0^E{e}^{r\tau }-\underset{{\mathbb{L}}_T}{\underbrace{{\mathbb{L}}_0{e}^{(r-q)\tau }+(d{\Delta }_{T-dt})S_T}}+{\Delta }_T{S}_T$$

We then consider our customer cashing out their loan:

1. If $S_T>K$ in the money,
   1. $-C_T^E$ becomes $-S_T+K$
   2. ${\Delta }_T=1$
   3. Thus $V_T=-S_T+K+C_0^E{e}^{r\tau }-{\mathbb{L}}_T+S_T=0$ since synthetic call still holds
      1. i.e., liquidate the stock to pay $S_T$
   4. Thus ${\mathbb{L}}_T=K+C_0^E{e}^{r\tau }$
      1. i.e., pay back the loan with remaining $K$ and cash invested
2. If $S_T\le K$ out of the money:
   1. $-C_t^E$ becomes $0$
   2. Thus $V_T=+C_0^E{e}^{r\tau }-{\mathbb{L}}_T+{\Delta }_T{S}_T=0$ since synthetic call still holds
   3. Thus ${\mathbb{L}}_T=C_0^E{e}^{r\tau }+{\Delta }_T{S}_T$
      1. i.e. pay back the loan with cash invested and liquidating the stock

Greek Neutral Portfolio of Derivatives §

Any derivative we can modify to make delta-neutral. For a security modeled by geometric brownian motion (with no dividends, $q=0$):

$$d{S}_t=m{S}_{t}dt+\sigma S_{t}d{\mathfrak{B}}_t$$

Notation

Since there are lots of notaions here, let’s outline them first:

• $S_t$ is the underlying security
• $C_t^E$ is the call option
• $\tilde{V}$ is the delta-neutral portfolio when parameter $N_s={\Delta }_C$
• $\hat{V}$ is the gamma-neutral portflio when parameters $N_s={\Delta }_C$, $N_C=-\frac{\tilde{\Gamma }}{{\Gamma }_C}$
• $V^{\prime }$ is the delta-gamma neutral portflio when parameters

let us create a portfolio of a derivative of this stock by having one call option and $N_s$ units of the underlier. The value of this portflio is:

$$\tilde{V}(S_t,t)=C_t^E+N_s{S}_t$$

• $C_t^E$ have greeks ${\Delta }_C,{\Gamma }_C,{\Theta }_C$
• which have greeks $\tilde{\Delta }={\Delta }_C+N_S$, $\tilde{\Gamma }={\Gamma }_C$, $\tilde{\Theta }={\Theta }_C$. Using ito’s formula, we get:

$$\begin{array}{rl}d\tilde{V}& =\left(\underset{\tilde{\Theta }}{\underbrace{\frac{\partial \tilde{V}}{\partial t}}}+m{S}_t\underset{\tilde{\Delta }}{\underbrace{\frac{\partial \tilde{V}}{\partial S_t}}}+\frac{1}{2}{\sigma }^2{S}_t^2\underset{\tilde{\Gamma }}{\underbrace{\frac{{\partial }^2\tilde{V}}{\partial S_t^2}}}\right)dt+\sigma S_t\underset{\tilde{\Delta }}{\underbrace{\frac{\partial \tilde{V}}{\partial S_t}}}d{\mathfrak{B}}_t\\ & =\tilde{\Gamma }\frac{1}{2}\underset{\text{ See below }}{\underbrace{{\sigma }^2{S}_t^{2}dt}}+\tilde{\Theta }dt+\tilde{\Delta }(\underset{=d{S}_t}{\underbrace{m{S}_{t}dt+\sigma S_{t}d{\mathfrak{B}}_t}})\\ & =\frac{1}{2}(d{S}_t{)}^2\tilde{\Gamma }+\tilde{\Theta }dt+\tilde{\Delta }d{S}_t\end{array}$$

Where from Ito Isometries

(dS_{t})^{2}&=m^{2}S_{t}^{2}\underbrace{ (dt)^{2} }_{ =0 }+\sigma^{2}S_{t}^{2}\underbrace{ d\mathfrak{B}_{t}^{2} }_{ = dt}+2m\sigma S_{t}^{2}\underbrace{ dtd\mathfrak{B}_{t} }_{ =0 } \\ &= \sigma^{2}S_{t}^{2}dt \end{align}

Now, from the above formula we see that there are three components that cause changes in $d\tilde{V}$

• $|\tilde{\Delta }|>0$ i.e. non-zero delta
• $|\tilde{\Gamma }|>0$, i.e. non-zero gamma
• $|\tilde{\Theta }|>0$, but considering $\Theta :=\frac{\partial S}{\partial t}$ it’s impossible for this to be zero.

Delta hedging. §

Lets make $\tilde{\Delta }=0$. We can do this simply by setting $N_s\equiv {\Delta }_C$:

$$\begin{array}{rl}\tilde{V}& =-C_t^E+{\Delta }_S{S}_t\\ \frac{\partial \tilde{V}}{\partial S_t}& =-\frac{\partial V}{\partial S_t}+{\Delta }_C=-{\Delta }_C+{\Delta }_C=0\end{array}$$

Delta-Gamma Hedging §

After the delta hedging we still have a positive gamma. Lets make $\tilde{\Gamma }=0$. We create a new portfolio of the delta-neutral portfolio, and $N_C$ units of the call:

$$\begin{array}{rl}\hat{V}& =\tilde{V}+N_C{C}_t^E\\ \hat{\Gamma }& =\frac{{\partial }^2\hat{V}}{\partial S_t^2}=\tilde{\Gamma }+N_C{\Gamma }_C\end{array}$$

Setting $N_C\equiv -\frac{\tilde{\Gamma }}{{\Gamma }_C}$ will make $\hat{\Gamma }=0$. But this incurs a new problem: $\hat{\Delta }$ is non-zero again. We add $N_{\text{more}}$ more units of $S_t$ to combat this:

$$V^{\prime }=\tilde{V}+N_C{C}_t^E+N_{more}$$