• Use the one-period binomial model to price European call and put options on a non-dividend paying stock
• No-arbitrage arguments:
  • Define and compute delta of an option
• Risk-neutral valuation:
  • Define and compute risk-neutral probabilities
  • Compare real-world (actual) and risk-neutral probabilities
• Compute the parameters of a binomial model from observed market data
• Extensions:
  • Multi-period binomial trees
  • Pricing of American options
  • Adjust the binomial model to cater for dividend yield

• Consider a portfolio consisting of:
  • a long position in $\mathrm{\Delta }$ units of the asset, and
  • a short position in 1 call option
• The portfolio value at maturity is $(\mathrm{\Delta }\times S_T)-f_T$ (where $S_T$ is the stock price and $f_T$ the option payoff at maturity $T$):
  • If the stock rises: $(\mathrm{\Delta }\times \mathrm{\$}120)-\mathrm{\$}20$
  • If the stock falls: $(\mathrm{\Delta }\times \mathrm{\$}80)-\mathrm{\$}0$
• The portfolio is riskless when the two state payoffs are equal:
  $$(\mathrm{\Delta }\times \mathrm{\$}120)-\mathrm{\$}20=(\mathrm{\Delta }\times \mathrm{\$}80)-\mathrm{\$}0$$
  which gives:
  $$\mathrm{\Delta }=\frac{\mathrm{\$}20-\mathrm{\$}0}{\mathrm{\$}120-\mathrm{\$}80}=\frac{20}{40}=0.5$$
• $\mathrm{\Delta }$ is the change in the option price over the change in the stock price (the hedge ratio).

• Substituting $\mathrm{\Delta }=0.5$, the terminal portfolio value is:
  • If the stock rises: $(0.5\times \mathrm{\$}120)-\mathrm{\$}20=\mathrm{\$}60-\mathrm{\$}20=\mathrm{\$}40$
  • If the stock falls: $(0.5\times \mathrm{\$}80)-\mathrm{\$}0=\mathrm{\$}40$
• Identical in both states — the portfolio is completely risk-free. With rate 0 ($e^{-rT}=e^0=1$) there is no discounting:
  $$\mathrm{\$}40e^{-0\times 1}=\mathrm{\$}40$$
• The initial portfolio value is:
  $$(\mathrm{\Delta }\times S_0)-f_0=\mathrm{\$}40$$
  from which it follows that:
  $$f_0=(0.5\times \mathrm{\$}100)-\mathrm{\$}40=\mathrm{\$}50-\mathrm{\$}40=\mathrm{\$}10$$

• Consider the portfolio consisting of:
  • a long position in $\mathrm{\Delta }$ units of the asset, AND
  • a short position in one option
• The value of the portfolio at maturity is $u{S}_0\mathrm{\Delta }-f_u$ or $d{S}_0\mathrm{\Delta }-f_d$.
• The portfolio is riskless when:
  $$u{S}_0\mathrm{\Delta }-f_u=d{S}_0\mathrm{\Delta }-f_d$$
  from which it follows that:
  $$\mathrm{\Delta }=\frac{f_u-f_d}{u{S}_0-d{S}_0}=\frac{f_u-f_d}{S_u-S_d}$$

• The terminal value of the portfolio is $u{S}_0\mathrm{\Delta }-f_u=d{S}_0\mathrm{\Delta }-f_d$.
• The initial value of the portfolio:
  • Since the final portfolio value is independent of the market direction, the portfolio is risk-free, thus: $$(u{S}_0\mathrm{\Delta }-f_u)e^{-rT}=(d{S}_0\mathrm{\Delta }-f_d)e^{-rT}$$
  • The cost to set up the portfolio today is: $\mathrm{\Delta }{S}_0-f_0$
• Thus by equating these two representations:
  $$f_0=\mathrm{\Delta }{S}_0-(u{S}_0\mathrm{\Delta }-f_u)e^{-rT}=\mathrm{\Delta }{S}_0-(d{S}_0\mathrm{\Delta }-f_d)e^{-rT}$$
• By substituting $\mathrm{\Delta }$ and simplifying algebraically:
  $$f_0=e^{-rT}[p{f}_u+(1-p)f_d]\text{with}p=\frac{e^{rT}-d}{u-d}$$
  Here $p$ is the risk-neutral probability of an up move (in Lecture 5 $p$ was the put price; this is a new meaning).

• The “risk-neutral” probability $p$ of an up movement satisfies ($E[\cdot ]$ is the expected value under the risk-neutral probabilities):
  $$E[S_T]=S_0{e}^{rT}⟹S_0=e^{-rT}E[S_T]=e^{-rT}[p{S}_u+(1-p)S_d]$$
• The real-world (actual) probability $p_a$ of stock going up is different to the risk-neutral probability $p$.
  • When $\mu$ is the expected return on the asset, then ($E_a[\cdot ]$ is the expected value under the actual real-world probabilities $p_a$): $$S_0=e^{-\mu T}{E}_a[S_T]=e^{-\mu T}[p_a{S}_u+(1-p_a)S_d]$$

• Substituting $S_0=100,S_u=120,S_d=80,r=0,T=1$ into equation (1) (note $e^0=1$):
  $$\mathrm{\$}120p+\mathrm{\$}80(1-p)=\mathrm{\$}100$$ $$p=\frac{\mathrm{\$}100-\mathrm{\$}80}{\mathrm{\$}120-\mathrm{\$}80}=\frac{20}{40}=0.5$$
• Equation (2) gives the option value ($f_u=\mathrm{\$}20,f_d=\mathrm{\$}0$):
  $$f_0=e^{-rT}[p{f}_u+(1-p)f_d]$$ $$=e^0[(0.5\times \mathrm{\$}20)+(0.5\times \mathrm{\$}0)]=\mathrm{\$}10+\mathrm{\$}0=\mathrm{\$}10$$

• In a “risk-neutral” world, the initial value of an asset is equal to its discounted expected payoff at maturity, discounted at $r$:
  $$E[S_T]=S_0{e}^{rT}⟹p{S}_u+(1-p)S_d=S_0{e}^{rT}(3)$$ $$E[f_T]=f_0{e}^{rT}⟹p{f}_u+(1-p)f_d=f_0{e}^{rT}(4)$$
• From (3) we solve for $p$:
  $$pu{S}_0+(1-p)d{S}_0=S_0{e}^{rT}⟹p=\frac{e^{rT}-d}{u-d}$$
• From (4) the general pricing formula is:
  $$f_0=e^{-rT}[p{f}_u+(1-p)f_d]$$

Setup: $S_0=\mathrm{\$}100$; in one year ($T=1$), $S_u=\mathrm{\$}120$, $S_d=\mathrm{\$}80$. Interest rate $r=0$. Value the $K=\mathrm{\$}100$ put option, whose price today is $p_0$ (an option price, not the probability $p$).

• In practice, $u$ and $d$ are usually calibrated from the asset price volatility $\sigma$ using the Cox, Ross and Rubinstein (1979) approach. For a small time step $\mathrm{\Delta }t$:
  $$u=e^{\sigma \sqrt{\mathrm{\Delta }t}}$$ $$d=\frac{1}{u}=e^{-\sigma \sqrt{\mathrm{\Delta }t}}$$
• Example: take $\sigma =20\mathrm{\%}$ and split the year into two steps of $\mathrm{\Delta }t=0.5$ year:
  $$u=e^{\sigma \sqrt{\mathrm{\Delta }t}}=e^{0.2\sqrt{0.5}}=e^{0.1414}=1.1519$$ $$d=\frac{1}{u}=\frac{1}{1.1519}=0.8681$$
• The two-step tree on the next pages uses this $u,d$. The $120 / $80 in the one-step examples were given moves (±20%, i.e. $\sigma =20\mathrm{\%}$), not computed from the CRR formula.

Per-step discount factor $e^{-0.05\times 0.5}=0.9753$. The textbook original calibrates $u,d$ from $\sigma =22.9\mathrm{\%}$. Original Hull problem with full solution: see Hull Exercises & Solutions, Lecture 7.

Price a call option with strike $K=\mathrm{\$}100$ and maturity $T=1$ year (below, $f_u$ and $f_d$ are the option values at the up and down nodes after six months, no longer terminal payoffs):

The price of the one-year European call option is $9.54.

Price a put option with strike $K=\mathrm{\$}100$ and maturity $T=1$ year:

The price of the one-year European put option is $4.66.

• American put, $K=\mathrm{\$}100$, $T=1$ year. At each node compare the continuation value with the intrinsic value $max[K-S_t,0]$ ($S_t$ is the stock price at that node):

• At node $d$ (stock price $S=\mathrm{\$}86.81$ there) early exercise is optimal! Exercising now gives $\mathrm{\$}13.19>\mathrm{\$}10.72$ (continuation value).
• The American put is worth $5.74, more than the corresponding European put at $4.66; the early-exercise premium is $5.74-4.66=\mathrm{\$}1.08$.

Setup: exactly the same two-step tree as the previous pages: $S_0=\mathrm{\$}100$, $u=1.1519$, $d=0.8681$, $r=5\mathrm{\%}$, $\mathrm{\Delta }t=0.5$.
$p=0.5539$, $1-p=0.4461$, per-step discount factor $e^{-0.025}=0.9753$.
Now change only one thing — a higher strike: a put option with $K=\mathrm{\$}110$.

• In a “risk-neutral” world, the expected return on a stock paying a continuously compounded dividend yield $q$ is $r-q$. Thus:
  $$E[S_T]=S_0{e}^{(r-q)T}⟹p{S}_u+(1-p)S_d=S_0{e}^{(r-q)T}(5)$$ $$E[f_T]=f_0{e}^{rT}⟹p{f}_u+(1-p)f_d=f_0{e}^{rT}(6)$$
• From (5) we have that:
  $$pu{S}_0+(1-p)d{S}_0=S_0{e}^{(r-q)T}⟹p=\frac{e^{(r-q)T}-d}{u-d}$$
• From (6) we discount the expected payoff at $r$:
  $$f_0=e^{-rT}[p{f}_u+(1-p)f_d]$$