Lesson 3 · Derivatives PricingCFA L2

Option Markets and Contracts

An option is the right — but not the obligation — to buy or sell. That asymmetry makes options the richest of all derivatives to value. This lesson builds option pricing from payoffs and no-arbitrage boundaries, through put-call parity and synthetics, to the two great pricing engines: the discrete-time binomial model and the continuous-time Black-Scholes-Merton model — plus the Greeks and the Black model for options on futures.

On this page

1–2 · Definitions & characteristics 3–4 · Markets & types 5.1 · Payoff values 5.2–5.4 · Boundary conditions 5.5 · Put–call parity & synthetics 5.6–5.9 · American, cash flows & sensitivities 6 · The binomial model 7 · Black–Scholes–Merton 8 · Options on forwards & futures

An option is a contract that gives one party the right to buy or sell an underlying asset at a fixed price on (or before) a fixed date. A call grants the right to buy; a put grants the right to sell. The buyer pays the seller (writer) an option premium up front for that right. Because the holder will only exercise when it is profitable, an option's payoff is one-sided — and pricing it means valuing that optionality.

Learning outcomes

Calculate and interpret the prices of a synthetic call, synthetic put, synthetic bond, and synthetic underlying, and explain why an investor would create them.
Calculate and interpret option prices using one- and two-period binomial models, including interest-rate options.
Explain the assumptions of the Black-Scholes-Merton model and how each input affects the option price (the Greeks).
Explain delta and dynamic hedging, the gamma effect, and the effect of the underlying's cash flows.
Estimate volatility (historical and implied), establish put-call parity for options on forwards/futures, and identify the appropriate European pricing model (the Black model).

1–2 · Basic definitions and characteristics

Every option is described by a handful of features:

Underlying and premium (price) — what the option is on, and what the buyer pays.
Exercise (strike) price $X$ — the fixed price at which the holder may buy (call) or sell (put).
Expiration date — European options exercise only at expiration; American options exercise at any time up to expiration.
Payoff — the option's value at expiration.
Intrinsic (exercise) value vs time value — intrinsic value is what exercising is worth now; the rest of the premium is time (speculative) value. Before expiration an option normally sells for more than intrinsic value.
Moneyness — whether the option has positive intrinsic value: in-the-money (ITM), at-the-money (ATM), or out-of-the-money (OTM).

Reading an option chain

For a stock at $16.25: a 15-strike call is ITM ($16.25>15$); a 17.50-strike call is OTM. A 17.50-strike put is ITM ($17.50>16.25$); a 15-strike put is OTM. Call premiums fall as the strike rises; put premiums rise as the strike rises. You might not exercise an ITM option, but you would never exercise an OTM one.

3–4 · The structure of options markets and types of options

An exchange-traded option is standardised, traded on an exchange, cleared, and guaranteed. Over-the-counter (OTC) options — often exotic — are privately negotiated off-exchange. Options exist on a wide range of underlyings:

Financial options: stock, index, bond, interest-rate, and currency options.
Options on futures, commodity options, and other exotic types.

5.1 · Payoff values

At expiration the option is worth its intrinsic value — zero, or the in-the-money amount:

Payoffs at expiration (3)(4)

$$ c_T = \max(0,\,S_T-X), \qquad p_T = \max(0,\,X-S_T) $$

Same form for European ($c_T,p_T$) and American ($C_T,P_T$) options at expiration. Prior to expiration, market price = intrinsic value + time value.

The four basic position diagrams are worth memorising: a long call has unlimited upside and limited downside (the premium); a short call is its mirror; a long put profits as the price falls toward zero; a short put is its mirror.

Worked example 1

Payoffs across underlyings, plus two strategies

A. Stock index at 5,601.19, multiplier 500.

Show solution

X = 5,500: call $\max(0,101.19)\times500=50{,}595$; put $=0$. X = 6,000: call $=0$; put $\max(0,398.81)\times500=199{,}405$.

5,500 → call 50,595 / put 0; 6,000 → call 0 / put 199,405

B. Bond at $1.035 per $1 par, $100,000 face.

Show solution

X = 1.00: call $0.035\times100{,}000=\$3{,}500$; put 0. X = 1.05: call 0; put $0.015\times100{,}000=\$1{,}500$.

1.00 → 3,500 / 0; 1.05 → 0 / 1,500

C. 90-day rate at 9%, notional $50m, 90/360.

Show solution

X = 8%: call $0.01\times\tfrac{90}{360}\times50\text{m}=\$125{,}000$. X = 10.5%: put $0.015\times\tfrac{90}{360}\times50\text{m}=\$187{,}500$.

8% → call 125,000; 10.5% → put 187,500

D. Swiss franc at $0.775, $500,000 per franc.

Show solution

X = 0.75: call $0.025\times500{,}000=\$12{,}500$. X = 0.81: put $0.035\times500{,}000=\$17{,}500$.

0.75 → call 12,500; 0.81 → put 17,500

E. Futures at 110.5, $1m, % of par.

Show solution

X = 110: call $0.5\times\tfrac{1}{100}\times1\text{m}=\$5{,}000$. X = 115: put $4.5\times\tfrac{1}{100}\times1\text{m}=\$45{,}000$.

110 → call 5,000; 115 → put 45,000

F. Covered call — buy stock at $40, sell the 40-strike call for $7.

Show solution

At $52: $52-\max(0,12)=40$. At $38: $38-\max(0,-2)=38$. Payoff is capped at $40 above the strike and declines with the stock below it — the same shape as a short put (Panel D).

Capped at 40 — a covered call

G. Protective put — buy stock at $60, buy the 60-strike put for $5.

Show solution

At $68: $68+\max(0,-8)=68$. At $50: $50+\max(0,10)=60$. Payoff is floored at $60 and rises with the stock — the same shape as a long call (Panel A).

Floored at 60 — a protective put

5.2–5.4 · Boundary conditions

Minimum and maximum values

No-arbitrage pins option prices between hard limits. Every option is worth at least zero. A call can never be worth more than the underlying; a European put can never exceed the present value of the strike, and an American put never exceeds the strike itself:

Min / max values (5)(6)(7)

$$ 0 \le c_0 \le S_0, \qquad 0 \le p_0 \le \frac{X}{(1+r)^{T}}, \qquad 0 \le P_0 \le X $$

Example ($S_0=52,\,X=50,\,r=5\%,\,T=0.5$): $0\le c_0\le52$; $0\le p_0\le 48.80$; $0\le P_0\le50$.

Lower bounds

The lower bound of an American option is its intrinsic value. For European calls and puts it is the underlying (or strike PV) net of the other leg — derived by comparing a call-plus-bond portfolio to the underlying:

Lower bounds (8)(9)(10)

$$ c_0 \ge \max\!\Big[0,\,S_0-\tfrac{X}{(1+r)^{T}}\Big], \qquad p_0 \ge \max\!\Big[0,\,\tfrac{X}{(1+r)^{T}}-S_0\Big] $$

$$ C_0 \ge \max\!\Big[0,\,S_0-\tfrac{X}{(1+r)^{T}}\Big], \qquad P_0 \ge \max\big(0,\,X-S_0\big) $$

The American call shares the European lower bound (so it is never optimal to exercise a non-dividend call early). The American put keeps the higher intrinsic-value bound.

Worked example 2

Lower bounds, European vs American

Options expire in 42 days ($T=0.1151$); underlying $72; risk-free 4.5%; no cash flows.

A. European bounds, strikes 70 and 75.

Show solution

70 call: $\max[0,\,72-70/1.045^{0.1151}]=\$2.35$. 75 call: $\max[0,-2.62]=\$0$.

70 put: $\max[0,-2.35]=\$0$. 75 put: $\max[0,\,75/1.045^{0.1151}-72]=\$2.62$.

Calls 2.35 / 0; Puts 0 / 2.62

B. American bounds.

Show solution

Calls identical (2.35 / 0). American puts use intrinsic value: 70 put $\max[0,70-72]=\$0$; 75 put $\max[0,75-72]=\$3$ — higher than the European 2.62.

Calls 2.35 / 0; Amer puts 0 / 3

Effect of exercise price and time to expiration

A call with a higher strike cannot be worth more than one with a lower strike; a put with a higher strike is worth at least as much as one with a lower strike.
A longer-term European/American call is worth at least as much as a shorter-term one; likewise the American put. The European put is the exception — a longer-term European put can be worth either more or less.

5.5 · Put-call parity and synthetics

The single most important result in options. A fiduciary call (a call plus a zero-coupon bond paying $X$) and a protective put (a put plus the underlying) have identical payoffs at expiration — $\max(S_T,X)$ — so they must cost the same today:

Put-call parity (13)

$$ c_0 + \frac{X}{(1+r)^{T}} = p_0 + S_0 $$

The no-arbitrage condition for options. Rearranging it manufactures any instrument synthetically.

Synthetic positions

$$ C_0 = P_0 + S_0 - \frac{X}{(1+r)^{T}}, \qquad P_0 = C_0 - S_0 + \frac{X}{(1+r)^{T}} $$

A synthetic call = long put + long underlying + short bond. A synthetic put = long call + short underlying + long bond. Synthetic underlying = long call + long bond − short put; synthetic bond = long put + long underlying − short call. Investors build synthetics to exploit mispricing or to trade an instrument that is cheaper/more liquid synthetically.

Why this matters

Whenever the actual price of an option diverges from its synthetic, an arbitrage exists: sell the expensive version, buy the cheap one, and the position is perfectly hedged to a riskless profit.

Worked example 3

Put-call parity arbitrage

European options, strike $45, expire in 115 days ($T=0.3151$); underlying $48, no cash flows; risk-free 4.5%. Put $3.75, call $8.00.

A. Identify the mispricing via the synthetic call.

Show solution

$$ c_0^{\text{synthetic}} = 3.75 + 48.00 - \frac{45.00}{1.045^{0.3151}} = \$7.37 < \$8.00 $$

The synthetic call is cheaper than the traded call — the traded call is overpriced.

Synthetic $7.37 vs traded $8.00

B. Execute the arbitrage.

Show solution

Sell the call for $8.00; buy the synthetic call (buy put $3.75, buy underlying $48, issue a bond paying $45 → brings in $44.38 today). The position nets $0.63 up front and pays nothing at expiration (perfectly hedged for all $S_T$).

Riskless +$0.63 up front

5.6–5.9 · American options, cash flows, rates, volatility

American options and early exercise

American prices are never below European prices ($C_0\ge c_0,\ P_0\ge p_0$). For a non-dividend call, $C_0 \ge S_0 - X/(1+r)^T > S_0 - X$ — selling beats exercising, so early exercise is never optimal and $C_0 = c_0$. The exception is a dividend-paying stock, where exercising captures the dividend. For an American put, early exercise is frequently optimal — e.g. a put on a bankrupt company should be exercised immediately to collect $X$ now.

Cash flows on the underlying

Income on the underlying (dividends, coupons) reduces its effective price. Replace $S_0$ with $S_0 - PV(CF,0,T)$ everywhere:

With cash flows

$$ c_0 \ge \max\!\Big[0,\,[S_0-PV(CF,0,T)]-\tfrac{X}{(1+r)^{T}}\Big] $$

$$ c_0 + \frac{X}{(1+r)^{T}} = p_0 + [S_0 - PV(CF,0,T)] $$

Interest rates and volatility

Higher interest rates raise call prices (the call is a leveraged claim on the underlying) and lower put prices (the opportunity cost of waiting to sell rises).
Higher volatility raises both call and put prices — more upside and downside both increase the value of one-sided payoffs.

The Greeks — option price sensitivities

Delta — sensitivity to the underlying price.
Gamma — how fast delta itself changes (how well delta approximates the move).
Rho — sensitivity to the risk-free rate.
Theta — the rate of time-value decay.
Vega — sensitivity to volatility.

6 · Discrete-time pricing: the binomial model

One-period model

The underlying moves to $S^+ = Su$ or $S^- = Sd$. Build a hedge of $n$ units of the underlying against one short call so the portfolio is riskless ($H^+ = H^-$); the hedge ratio is

Hedge ratio & option price (15)(16)(17)

$$ n^{*}=\frac{c^{+}-c^{-}}{S^{+}-S^{-}}, \qquad c=\frac{\pi c^{+}+(1-\pi)c^{-}}{1+r}, \qquad \pi=\frac{1+r-d}{u-d} $$

$\pi$ is the risk-neutral probability. The option price is the expected payoff under $\pi$, discounted at the risk-free rate — value as if investors were risk-neutral.

Worked illustration

$S=50,\ u=1.25,\ d=0.80,\ X=50,\ r=7\%$. Then $S^+=62.50\ (c^+=12.50)$, $S^-=40\ (c^-=0)$, $\pi=\tfrac{1.07-0.80}{1.25-0.80}=0.60$, and $c=\tfrac{0.6(12.5)+0.4(0)}{1.07}=\$7.01$. If the call traded at $8 (overpriced), sell 1,000 calls, buy $n=0.556$ units each (556 units), and lock in a 12.37% riskless return — above the 7–8% rate.

Worked example 4

One-period binomial call + arbitrage

$S=65$, up 30% / down 22%, $r=8\%$, call strike $70.

A. Price the call.

Show solution

$u=1.30,\ d=0.78$. $S^+=84.50\ (c^+=14.50)$, $S^-=50.70\ (c^-=0)$.

$$ \pi=\frac{1.08-0.78}{1.30-0.78}=0.5769 \;\Rightarrow\; c=\frac{0.5769(14.5)+0.4231(0)}{1.08}=\$7.75 $$ c = $7.75

B. The call trades at $9.00. Arbitrage with 10,000 calls.

Show solution

Overpriced. $n=\delta=\tfrac{14.50-0}{84.50-50.70}=0.4290$. Sell 10,000 calls (+$90,000), buy 4,290 units (−$278,850) → net invest $188,850. Either outcome grows to ≈$217,505, a return $217{,}505/188{,}850-1=15.17\%$ — well above 8%.

15.17% riskless

Two-period model

Extend the tree one more step. Value the option backward: compute terminal payoffs, then $c^+,c^-$ one period out, then $c$ today — each node using the same $\pi$. The hedge ratio $n$ differs at every node.

Two-period recursion (18)(19)(20)

$$ c^{+}=\frac{\pi c^{++}+(1-\pi)c^{+-}}{1+r}, \quad c^{-}=\frac{\pi c^{-+}+(1-\pi)c^{--}}{1+r}, \quad c=\frac{\pi c^{+}+(1-\pi)c^{-}}{1+r} $$

Worked example 5

Two-period binomial call & hedge ratios

$S=30$, up 14% / down 11% each period, $r=3\%$, strike 30, expires in two periods.

A. Value the call.

Show solution

$u=1.14,\ d=0.89$. Terminal: $S^{++}=38.99\ (c^{++}=8.99)$, $S^{+-}=30.44\ (c^{+-}=0.44)$, $S^{--}=23.76\ (c^{--}=0)$. $\pi=\tfrac{1.03-0.89}{1.14-0.89}=0.56$.

$$ c^{+}=\frac{0.56(8.99)+0.44(0.44)}{1.03}=5.08,\quad c^{-}=\frac{0.56(0.44)}{1.03}=0.24 $$ $$ c=\frac{0.56(5.08)+0.44(0.24)}{1.03}=2.86 $$ c = 2.86

B. Hedge ratios for 10,000 calls.

Show solution

$n=\tfrac{5.08-0.24}{34.20-26.70}=0.6453$; $n^{+}=\tfrac{8.99-0.44}{38.99-30.44}=1.0000$; $n^{-}=\tfrac{0.44-0}{30.44-23.76}=0.0659$. So 6,453 units today; 10,000 (up node) or 659 (down node) at time 1.

δ = 0.6453 → 6,453 units today

Worked example 6

Binomial put pricing

$S=65$, up 30% / down 22%, $r=8\%$, put strike 70.

Show solution

$p^+=\max(0,70-84.50)=0$, $p^-=\max(0,70-50.70)=19.30$. $\pi=0.5769$.

$$ p=\frac{0.5769(0)+0.4231(19.30)}{1.08}=7.56 $$ p = 7.56

Binomial interest-rate options & American options

For interest-rate options the risk-neutral probability is taken as $\pi=0.5$ and the discount rate changes at each node using a given binomial tree of interest rates. The model also handles American options naturally: at each node, compare the calculated value to the early-exercise value and keep whichever is larger. Adding more time periods makes the discrete binomial price converge to the continuous-time (Black-Scholes-Merton) value.

Worked example 7

Binomial interest-rate put on a coupon bond

Using a two-period interest-rate tree ($\pi=0.5$), price a European put (strike 1.01) on a three-period 6% coupon bond, $1 face.

Show solution

Bond prices at time 2 from the zero prices: $S^{++}=1.06(0.9210)=0.9763$, $S^{+-}=1.06(0.9443)=1.0010$, $S^{--}=1.06(0.9682)=1.0263$. Put payoffs: $p^{++}=0.0337,\ p^{+-}=0.0090,\ p^{--}=0$.

Discount back at each node's rate: $p^{+}=\tfrac{0.5(0.0337)+0.5(0.0090)}{1.064}=0.0201$, $p^{-}=\tfrac{0.5(0.0090)+0.5(0)}{1.0429}=0.0043$, then $p=\tfrac{0.5(0.0201)+0.5(0.0043)}{1.0513}=0.0116$.

p ≈ 0.0116

7 · Continuous-time pricing: Black-Scholes-Merton

The BSM model assumes the underlying follows a geometric lognormal diffusion; a known, constant risk-free rate and volatility; no taxes, transaction costs, or cash flows; and European exercise. The famous formula:

Black-Scholes-Merton (21)(22)

$$ c = S_0\,N(d_1) - X e^{-r^{c}T}N(d_2), \qquad p = X e^{-r^{c}T}\big[1-N(d_2)\big] - S_0\big[1-N(d_1)\big] $$

$$ d_1 = \frac{\ln(S_0/X) + \big(r^{c}+\sigma^{2}/2\big)T}{\sigma\sqrt{T}}, \qquad d_2 = d_1 - \sigma\sqrt{T} $$

$\sigma$ is the annualised standard deviation of the continuously-compounded return; $r^{c}$ the continuously-compounded risk-free rate; $N(\cdot)$ the standard normal CDF.

Worked example 8

BSM call and put

$T=0.75$, $S_0=52.75$, $\sigma=0.35$, $r^{c}=4.88\%$, $X=50$.

Show solution

$$ d_1=\frac{\ln(52.75/50)+(0.0488+0.35^2/2)0.75}{0.35\sqrt{0.75}}=0.4489,\quad d_2=0.4489-0.35\sqrt{0.75}=0.1458 $$

$N(0.45)=0.6736,\ N(0.15)=0.5596$. Then

$$ c=52.75(0.6736)-50e^{-0.0488(0.75)}(0.5596)=8.56 $$ $$ p=50e^{-0.0488(0.75)}(1-0.5596)-52.75(1-0.6736)=4.01 $$ c ≈ 8.56, p ≈ 4.01

Worked example 9

BSM with a second data set

$S_0=68.5$, $X=65$, $r^{c}=4\%$, $T=0.3014$, $\sigma=0.38$.

Show solution

$$ d_1=0.4135,\quad d_2=0.2049,\quad N(0.41)=0.6591,\ N(0.20)=0.5793 $$ $$ c=68.5(0.6591)-65e^{-0.04(0.3014)}(0.5793)=7.95,\qquad p=3.67 $$ c ≈ 7.95, p ≈ 3.67. (BSM is very sensitive to rounding in the normal table.)

Inputs and the Greeks

Delta = change in option price ÷ change in underlying — the first derivative w.r.t. the underlying. A delta-hedged position matches a short (long) call with a long (short) position of delta units of the underlying; because delta changes, the hedge must be rebalanced — dynamic hedging. Long-call delta > 0, short-call < 0, long-put < 0, short-put > 0.
Gamma = the second derivative — how sensitive delta is. Large gamma (at-the-money, near expiration) means delta changes fast and is a poor approximation, so a delta hedge needs frequent adjustment.
Rho (risk-free rate), Theta (time decay; usually negative), Vega (volatility; positive for calls and puts, largest at-the-money).
Exercise price: higher $X$ → lower call, higher put.

Cash flows on the underlying

BSM adjusted for cash flows

$$ c = \big[S_0-PV(CF,0,T)\big]N(d_1) - X e^{-r^{c}T}N(d_2) $$

$$ \text{dividend yield: } c = S_0 e^{-\delta^{c}T}N(d_1) - X e^{-r^{c}T}N(d_2); \quad \text{currency: } c = S_0 e^{-r^{fc}T}N(d_1) - X e^{-r^{c}T}N(d_2) $$

Worked example 10

BSM with cash flows on the underlying

$S_0=225$, $X=200$, $r^{c}=5.25\%$, $T=3$, $\sigma=0.15$.

A. PV of cash flows over the option life = 19.72.

Show solution

$S_0-PV=205.28$. $d_1=0.8364,\ d_2=0.5766$; $N(0.84)=0.7995,\ N(0.58)=0.7190$.

$$ c=205.28(0.7995)-200e^{-0.0525(3)}(0.7190)=41.28 $$ c ≈ 41.28

B. Continuously-compounded dividend yield = 2.7%.

Show solution

$S_0 e^{-0.027(3)}=207.49$. $d_1=0.8776,\ d_2=0.6178$; $N(0.88)=0.8106,\ N(0.62)=0.7324$.

$$ c=207.49(0.8106)-200e^{-0.0525(3)}(0.7324)=43.06 $$ c ≈ 43.06

The critical role of volatility

Volatility is the only BSM input not directly observable. Historical volatility is the sample standard deviation of recent continuously-compounded returns, $\sigma=\sqrt{\tfrac{1}{N-1}\sum(R_i^{c}-\bar R^{c})^2}$. Implied volatility is the volatility that, plugged into BSM, reproduces the option's market price — found by trial and error, it tells you the volatility the market is pricing in.

8 · Options on forwards and futures

When the underlying is a forward or futures price $f_0(T)$, the payoffs are $c_T=\max(0,f_T(T)-X)$ and $p_T=\max(0,X-f_T(T))$, with bounds analogous to asset options. Put-call parity adapts to use the forward price:

Put-call parity for options on forwards (28)

$$ c_0 + \frac{X - F(0,T)}{(1+r)^{T}} = p_0 $$

Worked example 11

Forward-option parity arbitrage

Strike 90, $r=5\%$, $T=2$, call 15.25, put 3.00, forward price 101.43.

Show solution

$$ p_0^{\text{parity}}=15.25+\frac{90-101.43}{1.05^{2}}=4.88 > 3.00 $$

The actual put (3.00) is underpriced. Buy put (−3.00), sell call (+15.25), buy bond worth $-(101.43-90)/1.05^2=-10.37$ → +1.88 up front, with zero cash at expiration for all $S_T$. A riskless gain.

Riskless +1.88 up front

Early exercise and the Black model

American options on futures may be worth exercising early (deep-in-the-money futures calls behave like the underlying), whereas options on forwards generate no cash before expiration, so early exercise is never justified. The standard model for pricing European options on futures is the Black model:

The Black model

$$ c = e^{-r^{c}T}\big[f_0(T)N(d_1) - X N(d_2)\big], \qquad p = e^{-r^{c}T}\big[X(1-N(d_2)) - f_0(T)(1-N(d_1))\big] $$

$$ d_1 = \frac{\ln(f_0/X) + (\sigma^{2}/2)T}{\sigma\sqrt{T}}, \qquad d_2 = d_1 - \sigma\sqrt{T} $$

Worked example 12

The Black model = BSM on the spot

Forward price 139.19, $T=0.5890$, $X=125$, $r^{c}=4.25\%$, $\sigma=0.15$.

A. Black-model call price.

Show solution

$$ d_1=\frac{\ln(139.19/125)+(0.15^2/2)0.5890}{0.15\sqrt{0.5890}}=0.9916,\quad d_2=0.8765 $$

$N(0.99)=0.8389,\ N(0.88)=0.8106$:

$$ c=e^{-0.0425(0.5890)}[139.19(0.8389)-125(0.8106)]=15.06 $$ c = 15.06

B. Show it equals BSM on the implied spot.

Show solution

With no cash flows and continuous compounding, $S_0=F(0,T)e^{-r^{c}T}=139.19e^{-0.0425(0.5890)}=135.75$. Plugging into BSM gives the same $d_1=0.9916,\ d_2=0.8765$ and $c=135.75(0.8389)-125e^{-0.0425(0.5890)}(0.8106)=15.06$ — identical.

Same price, 15.06

Worked example 13

Interest-rate put via the Black model

Expires in 280 days ($T=0.7671$); forward rate 6.8%; $r^{c}=6.25\%$; exercise rate 7%; $\sigma=0.02$; 180-day underlying rate; notional $10m.

Show solution

$$ d_1=\frac{\ln(0.068/0.07)+(0.02^2)0.7671}{0.02\sqrt{0.7671}}=-1.6461,\quad d_2=-1.6636 $$

$N(-1.65)=0.0495,\ N(-1.66)=0.0485$:

$$ p=e^{-0.0625(0.7671)}[0.07(1-0.0485)-0.068(1-0.0495)]=0.00187873 $$

The payoff is made 180 days after expiry, so discount at the forward rate: $e^{-0.068(180/365)}(0.00187873)=0.00181677$; the 180-day rate $=0.00181677\times\tfrac{180}{360}=0.00090839$; times notional $\$10\text{m}\times0.00090839=\$9{,}084$.

Put value ≈ $9,084

Key takeaway

Option valuation is one no-arbitrage idea applied at increasing resolution. Put-call parity ties calls, puts, the underlying, and a bond into one equation — break it and you have an arbitrage. The binomial model prices any option by working backward through a tree under the risk-neutral probability; refine the tree and it converges to Black-Scholes-Merton. The Greeks describe how that price moves, and the Black model is simply BSM written in terms of a forward or futures price.

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