Implied Volatility from Piecewise Constant Variance

Summary

For a diffusion whose volatility changes over time, the document explains how to calculate the Black–Scholes implied volatility for an option expiring at time T. It distinguishes implied variance from implied volatility: average the instantaneous variance over the option’s lifetime, then take the square root to obtain implied volatility. For piecewise constant volatility, if T falls between two volatility change dates, the accumulated variance consists of the variance through the last change date plus the variance accrued at the current level from that date to expiry. Dividing this total by T gives the implied variance.

The derivation clarifies that the next interval endpoint can lie after the option expiry; only volatility up to T contributes. The result assumes the stated diffusion model and deterministic volatility function. It does not address stochastic volatility, jumps, or other settings where this simple variance averaging may not apply.

Key ideas

• Implied variance is the time average of instantaneous variance over the option’s life.
• Implied volatility is the square root of implied variance.
• When expiry falls inside a constant-volatility interval, include only the portion of that interval ending at expiry.
• A later interval endpoint may exceed expiry because it defines the volatility schedule, not the option’s maturity.

Tags

# Explanation of formula for implied volatility given $\sigma(t)$


# Explanation of formula for implied volatility given $\sigma(t)$












If $dS = S\mu dt + S \sigma(t) dW$, then we know that the implied volatility is $\int_0^T \sigma^2(s)/T \ ds$.

However, if $\sigma(t)$ is a piecewise constant function, i.e. constant between $T_1, T_2$ and between $T_2, T_3$, and so on.

Then, according to some lecture notes, the implied vols are

That, I don't quite understand. Where does this formula come from? If $T$ is the expiry, then how can there be a $T_{i+1} > T$? I thought the expiry was the final such $T$ value?

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/33256

Note that the implied volatility is given by \begin{align*} \hat{\sigma}(T)=\sqrt{\frac{1}{T}\int_0^T \sigma^2(t) dt}, \end{align*} while $\frac{1}{T}\int_0^T \sigma^2(t) dt= \hat{\sigma}^2(T)$ is the implied variance.

For a piecewise constant volatility function $\sigma$, the implied variance for option with maturity $T$, where $T_i \le T \le T_{i+1}$, is given by \begin{align*} \hat{\sigma}^2(T)&=\frac{1}{T}\int_0^T \sigma^2(t) dt \\ &= \frac{1}{T}\left(\int_0^{T_i}\sigma^2(t) dt + \int_{T_i}^T\sigma^2(t) dt \right)\\ &=\frac{1}{T}\left(T_i\, \hat{\sigma}^2(T_i) + (T-T_i)\, \sigma_{i+1}^2\right), \end{align*} where \begin{align*} \hat{\sigma}^2(T_i) = \frac{1}{T_i}\int_0^{T_i}\sigma^2(t) dt \end{align*} is the implied variance for option with maturity $T_i$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.