There is an input missing from the first lesson, and it looks like the most important one. Nothing in the model asks how much the stock is expected to rise. Two stocks at $100 with the same volatility get the same option prices, even if one has analysts forecasting 30% a year and the other nothing. The only growth rate in the formula is the interest rate. This lesson is about why, and it is the single deepest idea in option pricing.
The hedge removes the direction
Start from what a market maker does. They sell you a call and immediately buy enough stock to cover its delta, say 50 shares per contract. If the stock rises $1, the call they sold costs them about $50 and the stock they own makes $50. If it falls $1, the reverse. Over the next small move, the combined position neither makes nor loses money on direction. Keep re-hedging as delta changes and the position stays that way.
A position with no direction has no risk from the stock's drift, whatever that drift is. So what should it earn? Exactly what any other riskless money earns: the interest rate. If it earned more, everyone would borrow at the rate and build the position; if less, everyone would do the reverse. That one argument pins the option's price down without anyone agreeing on where the stock is going. The expected return cancels out, because the hedge cancels the direction.
Pricing as if everyone were indifferent to risk
Since the stock's expected return drops out, the model is free to pick any value for it that makes the arithmetic easy. It picks the interest rate, less dividends. The result is called risk-neutral pricing: value the option as if the stock grows at the rate, then discount the expected payoff back to today at the same rate. It is not a claim that investors do not care about risk, or a forecast that the stock will earn the rate. It is a shortcut that gives the same price as the hedging argument, and it only works because the hedge exists.
In that risk-neutral world the center of the stock's future is the forward price from Module 2: today's price grown at the rate and trimmed by dividends. A $100 stock with a 4% rate and a 1% dividend yield has a one-year forward of about $103. The model builds its lognormal range around that forward, not around anyone's price target.
Long a $100 call, short a $100 put, on a $100 stock: a synthetic forward, worth the carry. It grows with time and with the rate, and a dividend yield pulls it down. Volatility would not change it at all, so there is no slider for it.
The chart shows the forward's footprint on the chain. A call minus a put at the same strike is a synthetic forward, so its value is the carry: the forward less the strike, in today's money. At the money with a month left it is cents. With two years left and a 5% rate it is dollars. Drag the rate and dividend and watch the line tilt. No expected-return slider exists, because the model has no place to put one.
What the market's expectations do instead
If the forecast is not in the formula, can it reach the price at all? Yes, through the one input that is not bookkeeping. A trader who fears a crash does not change the rate; they pay up for puts, which raises implied vol on the downside strikes. That is the skew from the last lesson. Views about direction and risk enter option prices through volatility, strike by strike, never through a drift term. It is also why the model's probabilities, like the chance an option finishes in the money, are risk-neutral probabilities: useful for pricing and hedging, but not the same as the real-world odds, which carry the stock's true drift and the market's appetite for risk.
One year at 25% volatility. The shaded bands are a fixed reference around today's price; drag the rate up and the curve itself drifts right, drag the dividend up and it drifts back. That drift is carry, not a forecast.
The maths: the replication argument
Hold one call, short Δ shares, where Δ = ∂C/∂S. Over a short time dt the stock moves by dS = μ S dt + v S dW, where μ is its true expected return and dW the random part. The call changes by dC = (∂C/∂t) dt + Δ dS + ½ Γ v² S² dt. Subtract Δ dS for the hedge and both μ and dW vanish: the portfolio changes by (∂C/∂t + ½ Γ v² S²) dt, a sure amount.
A sure return must equal the rate, so ∂C/∂t + ½ v² S² Γ + (r - q) S Δ - r C = 0. That is the Black-Scholes equation, and the Black-Scholes formula is its solution. μ appears nowhere in it. The same result written as an average: C = e^(-rT) × E[max(S_T - K, 0)], with the average taken as if the stock grew at r - q.
A check with numbers: a $100 stock, $100 strike, one year, 4% rate, 1% dividend yield. The forward is 100 × e^(0.03) = $103.05. Parity says call minus put is S e^(-qT) - K e^(-rT) = 99.00 - 96.08 = $2.93, and the model gives exactly that, $2.93, at any volatility. Change the stock's expected return from 0% to 30% and neither number moves.
Your turn
Open GOOGL and choose the expiration several months out that the questions name. Read the stock price, then build the forward from the at-the-money strike's call and put mids. The gap between the two is the carry the model uses in place of any forecast. The next lesson runs the same hedging argument one small step at a time, on a tree, and uses it to price the right to exercise early.