Skip to content
All library documents

How Option Vega Changes with Expiry and Volatility

Article Quant Q&A · Author: Yilie Ma

Summary

The document examines two Black–Scholes sensitivities for an at-the-money European option: how Vega changes as time to expiry grows, and how option price changes as implied volatility rises. For the expiry question, it considers the derivative of Vega with respect to maturity, known as Veta. Under the stated assumptions of zero rates and dividends and a strike equal to spot, the sign of Veta changes at a maturity determined by volatility. At typical volatility levels, that change occurs far beyond the expiries of most traded instruments, so Vega generally increases over ordinary maturities.

For the volatility question, the answer explains why an option-price curve can look nearly linear over a short expiry even though its slope, Vega, varies. The exact curvature is governed by the second derivative of price with respect to volatility, called Volga or Vomma. The near-linearity is an approximation that works well for small maturities, not a general property. The derivation’s conclusions are anchored to the stated at-the-money setup and assumptions.

Key ideas

  • Veta measures how Vega changes as time to expiry changes.
  • For the stated at-the-money setup, Veta changes sign at a maturity determined by volatility.
  • At typical volatility levels, Vega tends to rise across maturities common in traded instruments.
  • A near-linear option-price response to volatility can be a useful approximation for short expiries.
  • Volga or Vomma describes the curvature of option price with respect to volatility.

Tags

Full text
# European option Vega with respect to expiry and implied volatility


# European option Vega with respect to expiry and implied volatility












I was told that the Vega of an European option always increases when its time to expiry increases (all else equal). I found this confusing and potentially wrong, but there doesn't seem to be relevant sources online about this. Let's take an ATM option for simplicity, its Vega is: $S\sqrt(\tau)N'(d1)$, which is just $1 \over \sqrt(2\pi)$$S\sqrt(\tau)e^{-(r+{\sigma^2\over2})^2\tau\over2}$. Now as $\tau$ increases to a large range, I found that Vega certainly decreases as we have a $-\tau$ term in the exponent. However in small ranges of $\tau$ for example between 0 and 1, Vega does increases as $\tau$ increases. Am I mistaken here?

I also would like to see the relationship of European option price with respect to volatility and plotted a graph where the $y$ axis is the option price computed from Black-Scholes, and the $x$ axis is $\sigma$ (also holding everything else equal and use ATM options for simplicity). The graph surprisingly looks like a straight line. But from the formula above, the local slope of this line should just be Vega at different values of $\sigma$, and thus should be decreasing, so theoretically the line should be concave. Am I mistaken here?

## Answer by Quantuple (score 3, accepted)

https://quant.stackexchange.com/a/37899

To keep notations uncluttered, consider that $r=q=0$ in what follows, while focusing on the particular case of an ATM option i.e. $K=S$ (otherwise use the same reasoning with $K=F(0,T)=Se^{(r-q)T}$ i.e. an ATMF option, the conclusion won't change that much).

In your first question, you're looking for the sign of the derivative of Vega with respect to the residual maturity $\tau$. The closed-form expression of this Greek is given here under the name Veta. In our particular configuration we get $$ \text{Veta} = - S \phi(d_1) \sqrt{\tau} \left[ \frac{1 + d_1d_2}{2\tau} \right] $$ with $d_1 = \frac{1}{2}\sigma\sqrt{\tau}$ and $d_2 = -d_1$ such that $$ \text{Veta} = - S \phi(d_1) \frac{1}{2\sqrt{\tau}} \left[ 1 - \frac{1}{4}\sigma^2\tau \right] $$ since the prefactor has always the same sign, we see that a sign change occurs at $\tau = \frac{4}{\sigma^2}$. For typical levels of volatility -- say 20% -- this gives $\tau \approx 100Y$. This is easily confirmed by a quick graph (I've used $\sigma=0.2$, $r=q=0$, $S=K=1$). Since most traded instruments generally expire long before that time, this justifies the common belief of considering that $\text{Vega}$ is 'always' increasing.

In the second question, you're looking at $C = f(\sigma)$ all the other parameters being fixed where $C$ is the price of an ATM call. You're observation that $C$ is a linear function of $\sigma$ is due to the fact that this well known approximation holds quite well in practice for small $\tau$. Of course if you want the exact result you should again work out the closed-form expression of the second derivative of $C$ with respect to $\sigma$ (Volga or Vomma) and look at when it is significant vs. negligible.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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