Implied Volatility via Newton-Bisection

Worked Solution

How to Think About It: Implied vol is the market's way of quoting option prices in volatility units instead of dollars. Every trader's screen shows implied vol, not Black-Scholes prices, because vol is a more stable and comparable quantity. Finding implied vol means inverting the Black-Scholes formula -- you know the price, you need the $\sigma$ that produces it. The good news: Black-Scholes price is a strictly increasing function of $\sigma$ (vega is always positive), so the root is unique whenever it exists. The bad news: there is no closed-form inverse, so you need a numerical root-finder. Newton's method is fast (quadratic convergence) but can blow up if vega is tiny. Bisection always converges but is slow. The practical answer is to combine both.

Part (i): Definition and Uniqueness

The implied volatility $\hat{\sigma}$ is defined as the unique $\sigma > 0$ satisfying:

$C_{\text{BS}}(S_0, K, T, r, \hat{\sigma}) = C_{\text{mkt}}$

Existence: The call price $C_{\text{BS}}(\sigma)$ is continuous in $\sigma$ on $(0, \infty)$ with: - $\lim_{\sigma \to 0^+} C_{\text{BS}}(\sigma) = \max(S_0 - K e^{-rT}, 0)$ (the intrinsic value) - $\lim_{\sigma \to \infty} C_{\text{BS}}(\sigma) = S_0$ (the stock price itself)

So by the intermediate value theorem, a solution exists whenever:

$\max(S_0 - K e^{-rT}, 0) < C_{\text{mkt}} < S_0$

This is just the no-arbitrage bound on a European call. Any market price within these bounds has a well-defined implied vol.

Uniqueness: The vega of a European call is:

$\mathcal{V} = \frac{\partial C_{\text{BS}}}{\partial \sigma} = S_0 \sqrt{T} \, \phi(d_1) > 0$

where $\phi$ is the standard normal PDF and $d_1 = \frac{\ln(S_0 / K) + (r + \sigma^2/2)T}{\sigma \sqrt{T}}$. Since $\phi(d_1) > 0$ for all finite $d_1$, vega is strictly positive for all $\sigma > 0$. This means $C_{\text{BS}}(\sigma)$ is strictly increasing in $\sigma$, so the root is unique.

Part (ii): Newton-Raphson Iteration

We want to find the root of $f(\sigma) = C_{\text{BS}}(\sigma) - C_{\text{mkt}} = 0$. Newton's method gives:

$\sigma_{n+1} = \sigma_n - \frac{f(\sigma_n)}{f'(\sigma_n)} = \sigma_n - \frac{C_{\text{BS}}(\sigma_n) - C_{\text{mkt}}}{\mathcal{V}(\sigma_n)}$

Written out:

$\sigma_{n+1} = \sigma_n - \frac{C_{\text{BS}}(\sigma_n) - C_{\text{mkt}}}{S_0 \sqrt{T} \, \phi(d_1(\sigma_n))}$

This converges quadratically near the root. In practice, 3-5 iterations typically suffice to reach machine precision when starting from a reasonable initial guess.

Failure mode: If $\sigma_n$ is very large or very small, $d_1$ becomes extreme, $\phi(d_1) \to 0$, and vega nearly vanishes. The Newton step then becomes enormous, potentially producing a negative $\sigma_{n+1}$ or overshooting wildly. This is why pure Newton is not safe.

Part (iii): Robust Safeguarded Newton-Bisection

Practical initial bounds: - $\sigma_{\text{lo}} = 0.01$ (1% vol -- below any realistic equity or FX vol) - $\sigma_{\text{hi}} = 5.0$ (500% vol -- above any observed market vol)

Verify that $f(\sigma_{\text{lo}}) < 0$ and $f(\sigma_{\text{hi}}) > 0$ so the root is bracketed. If not, widen the bounds.

Algorithm (Brent-style safeguarded Newton):

1. Set $a = \sigma_{\text{lo}}$, $b = \sigma_{\text{hi}}$, and initial guess $\sigma_0 = 0.25$ (a common starting point for equity implied vol).

2. At each iteration: - Compute the Newton step: $\sigma_{\text{Newton}} = \sigma_n - f(\sigma_n) / \mathcal{V}(\sigma_n)$ - If $\sigma_{\text{Newton}} \in (a, b)$ and $|\sigma_{\text{Newton}} - \sigma_n|$ is less than half the previous step (ensuring we are making progress), accept the Newton step. - Otherwise, fall back to a bisection step: $\sigma_{\text{bisect}} = (a + b) / 2$. - Update the bracket: if $f(\sigma_{n+1}) < 0$, set $a = \sigma_{n+1}$; otherwise set $b = \sigma_{n+1}$.

1. Converge when $|f(\sigma_n)| < \epsilon$ (e.g., $\epsilon = 10^{-10}$) or $|b - a| < \delta$ (e.g., $\delta = 10^{-12}$).

Why this works: Bisection alone converges linearly -- it halves the bracket each step, so after 50 iterations the bracket has width $5 / 2^{50} \approx 4 \times 10^{-15}$, which is machine epsilon. But in the "good" region near the root where vega is bounded away from zero, Newton converges quadratically, reaching full precision in 3-5 steps. The safeguard ensures we never leave the bracket, so convergence is guaranteed regardless of the initial guess.

Practical refinements: - A better initial guess uses the Brenner-Subrahmanyam approximation: $\sigma_0 \approx \sqrt{2\pi / T} \cdot C_{\text{mkt}} / S_0$, which is often within 5-10% of the true implied vol. - For deep OTM options, work with the put instead (via put-call parity) to avoid numerical issues with very small call prices. - Cap the maximum Newton step size at, say, $0.5$ per iteration to prevent overshooting.

Answer: Implied volatility $\hat{\sigma}$ is the unique $\sigma > 0$ satisfying $C_{\text{BS}}(\sigma) = C_{\text{mkt}}$, and it exists and is unique whenever the market price lies within no-arbitrage bounds. Newton-Raphson with the update $\sigma_{n+1} = \sigma_n - (C_{\text{BS}}(\sigma_n) - C_{\text{mkt}}) / \mathcal{V}(\sigma_n)$ converges quadratically near the root. For robustness, bracket the root with $\sigma \in [0.01, 5.0]$ and use safeguarded Newton-bisection: accept Newton steps when they stay inside the bracket and make sufficient progress, otherwise bisect. This guarantees convergence for any valid market price.