Case study: implied volatility by Newton's method

The price as a function of volatility

A call option gives the right to buy a stock at a fixed price $K$, the strike, at a fixed date, $T$ years away. The Black–Scholes formula prices it from the stock price $S$ and the interest rate $r$ (continuous, as a decimal). It also needs the volatility $\sigma$, the yearly standard deviation of the stock's log returns.

The Derivatives Modelling course derives the formula, and case study 2.6.5 rebuilds it as an integral. Here, take it as given:

$N$ is the standard normal cumulative distribution (Level 5; lesson 2.6.3 meets it as an area), norm.cdf in SciPy. Everything is known except $\sigma$, so this is a function of one variable, $c(\sigma)$.

For the option in the hook, $S = K = 100$, $T = 0.5$ and $r = 0.03$:

As $\sigma$ shrinks to 0 the price falls to $S - Ke^{-rT} = 1.4888$, the value of the option with no uncertainty. As $\sigma$ grows it rises towards $S = 100$. A market price outside that range has no implied volatility at all.

Vega: the derivative with respect to volatility

Newton's method needs the slope $c'(\sigma)$. Options traders call it vega. Differentiating the formula with the chain rule (lesson 2.3.3) leads, after a cancellation, to a short result:

Because vega is positive, $c(\sigma)$ rises everywhere. So each market price has at most one implied volatility, and the sign of $c(\sigma) - \text{price}$ says which side of it you are on.

Vega 27.77 is per unit of $\sigma$, a move from 0 to 100% volatility. Desks quote it per vol point, one percentage point: 0.2777, so each extra point adds about 28 cents.

Newton's method for implied volatility

Solve $c(\sigma) - \text{price} = 0$. The step is lesson 2.4.3's, with vega as the slope:

Two things make this easy. At the money the price curve is almost a straight line: vega barely changes, from 27.36 at 10% to 27.77 at 20% and 30%. And the start, 20%, is close to the answer.

Deep out of the money: where Newton breaks

Now take a strike of 150, 50% above the stock price, with the same $S$, $T$ and $r$. The call trades at $2.00. At 20% volatility the formula gives only 0.0149, and vega is 0.7564: the curve is almost flat there.

Newton's step divides the gap by that tiny slope. From 20% the first step lands on 282%, the second on $-9.87\%$, a negative volatility, and the third on 916,440,330%. The answer is 44.78%.

The shape of the curve

For this option $c(\sigma)$ is S-shaped. It is flat near 0, bends up, and later bends down: $c'' \approx 70.54$ at 30% and $-5.45$ at 200%. So it has an inflection point (lesson 2.4.2), and there vega, the slope, is at its largest.

Lesson 2.4.3 explains why, with concavity. If the answer is below $\sigma^*$, the start is to its right on a rising, concave-up curve, and each tangent lies below the curve. If the answer is above, the start is to its left on a concave-down stretch, where each tangent lies above the curve.

Either way, no step jumps past the answer. For strike 150, $\sigma^* = 1.2497$, or 125%. From there the guesses are 0.535213, 0.456067, 0.447934 and 0.447826: four steps, all from the right.

A safety net: keep a bracket

When $\ln(S/K) + rT = 0$, the formula gives $\sigma^* = 0$, where the price formula divides by zero. Production code therefore also keeps a bracket, as in lesson 2.4.3. Since $c(\sigma)$ rises, $[0.01, 3]$ brackets any volatility from 1% to 300%.

import math

from scipy.stats import norm


def _call(S, K, T, r, sigma):
    d1 = (math.log(S / K) + (r + sigma**2 / 2) * T) / (sigma * math.sqrt(T))
    return S * norm.cdf(d1) - K * math.exp(-r * T) * norm.cdf(d1 - sigma * math.sqrt(T))


def test_at_the_money():
    """A 6-month at-the-money call at $7.00 implies about 22.26%"""
    expect_close(implied_vol(7.0, 100, 100, 0.5, 0.03), 0.22264129444955746, what="implied_vol(7.0, ...)", rel=1e-8)


def test_out_of_the_money():
    """Strike 120 at $1.50 implies about 23.60%"""
    expect_close(implied_vol(1.5, 100, 120, 0.5, 0.03), 0.23596239715145917, what="implied_vol(1.5, K=120)", rel=1e-6)


def test_reprices():
    """The answer reprices the option: a 1-year call at $12.00"""
    sigma = implied_vol(12.0, 100, 95, 1.0, 0.02)
    expect_close(_call(100, 95, 1.0, 0.02, sigma), 12.0, what="call price at your volatility", rel=1e-9)


def test_round_trip():
    """Price an option at 35%, then recover 35%"""
    price = _call(50, 55, 0.25, 0.01, 0.35)
    expect_close(implied_vol(price, 50, 55, 0.25, 0.01), 0.35, what="recovered volatility", rel=1e-8)


def test_deep_out_of_the_money_good_start():
    """Strike 150 at $2.00, started at the inflection point 1.2497, implies about 44.78%"""
    expect_close(implied_vol(2.0, 100, 150, 0.5, 0.03, sigma0=1.2497), 0.4478258393239575, what="implied_vol(2.0, K=150, sigma0=1.2497)", rel=1e-8)


def test_deep_out_of_the_money_bad_start():
    """Strike 150 at $2.00 from 20%: a step goes negative, so ValueError"""
    try:
        implied_vol(2.0, 100, 150, 0.5, 0.03)
    except ValueError:
        return
    raise AssertionError("expected ValueError: from 20% the second Newton step gives a negative volatility")