Why At-the-Money Straddle Vega Scales with Square Root of Time

Summary

The answer explains the square-root-of-time rule of thumb for the vega of at-the-money options. Under Black–Scholes, vega is proportional to the underlying price, the normal density evaluated at d1, and the square root of time to expiry. Near the money, for sufficiently small volatility over the option’s life, d1 is approximately zero, so the density term is close to a constant. This leaves vega approximately proportional to the square root of maturity.

For a straddle, the call and put vegas are added, doubling the single-option exposure under the stated approximation. The answer also notes that longer-dated implied volatility tends to have lower volatility itself, which can offset some of the larger cash vega when considering risk. The result is an approximation rather than a universal scaling law: it relies on near-the-money conditions and the Black–Scholes framework, and the density term can vary when those assumptions do not hold.

Key ideas

• Black–Scholes vega contains a square-root-of-time factor multiplied by the normal density at d1.
• For near-the-money options with modest volatility over the life of the option, d1 is approximately zero.
• When the density term is nearly constant, vega scales approximately with the square root of maturity.
• An at-the-money straddle combines the vega of its call and put legs.
• The approximation does not account for all changes in volatility exposure across maturities.

Tags

# Vega, square root of time, and ATM straddles


# Vega, square root of time, and ATM straddles












Could someone intuitively explain why for say a 1y EURUSD option - If you buy 100 (50/leg of straddle) of 1y at the money EUR vol, that = sq root of 12 x 100 = roughly 350k of EUR vol. If you buy 100 of 6m EUR vol, that = sq root of 6 x 100 = roughly 245k of EUR vol. If you buy 100 of 1m EUR vol, that = sq root of 1 x 100 = 100k of EUR vol. Why does that work?

Or starting from scratch, whats the right formula for calculating vega from the atm straddles?

Many thanks

## Answer by Newquant (score 2)

https://quant.stackexchange.com/a/77132

Black Scholes vega is given as $$ S \cdot \phi(d1)\cdot\sqrt{\tau} $$

With $\phi$ as $\frac{e^{-x^2}}{\sqrt{2\pi}}$ and d1 as $\frac{ln(\frac{S}{K}) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}$.

For ATM straddles, d1 is approximately 0 (or, at least for small $\sigma\sqrt{\tau}$, close to 0), which means that $\phi(d1)$ becomes ~$\frac{1}{\sqrt{2\pi}}$, and this approximation is locally insensitive to changes in time to maturity. This simplifies the vega formula to $$ S \cdot \frac{1}{\sqrt{2\pi}}\cdot\sqrt{\tau} $$

Which varies with the square root of time, explaining the rule of thumb that you've shown above in the EUR example.

It's worth noting though that even though cash vega exposure increases, the volatility of IV decreases as time to maturity (~sqrt(t)) increases, so the increased dollar risk is offset by decreased underlying risk.

Edit: For the straddle you multiply vega * 2, so it becomes $$ S \cdot \sqrt{\frac{2}{\pi}}\cdot\sqrt{\tau} $$

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