An at-the-money option and the square root of time
An at-the-money option is worth roughly its expected move, and the expected move grows with the square root of time. So its value decays slowly at first and very fast at the end.
- Price an at-the-money call with 0.4 × S × σ × √T, and read implied vol back out of a price
- Scale a price forward in time with √(days left ÷ days at the start)
- Compute theta per day and show why it accelerates into expiry
- Recover the days left from a price ratio
The intuition
A $100 stock with 30% volatility: its three-month at-the-money call is worth about 0.4 × 100 × 0.30 × √(90 ÷ 365) = $5.96. Halve the time to 45 days and the call keeps √0.5 = 71% of its value, $4.21, even though half the time has gone. Quarter the time and half the value is gone. Value tracks √T, so the early days shave little off and the decay is back-loaded into the final weeks.
Theta is the slope of that curve. Differentiate 0.4 S σ √T and you get θ ≈ C ÷ (2 × days left): at 90 days the call loses about 3.3 cents a day, at 45 days 4.7 cents, and the daily loss keeps rising as the days run out. Option sellers love the final weeks. Option buyers holding an at-the-money option through them pay dearly for it.
ATM call ≈ 0.4 × S × σ × √T with T = days ÷ 365. Value later: C₂ = C₁ × √(days₂ ÷ days₁). Theta per day ≈ C ÷ (2 × days left). Days left from a price ratio: days₂ = days₁ × (C₂ ÷ C₁)². Implied vol from a price: σ ≈ C ÷ (0.4 × S × √T).
Why it works
- The conventions here: an at-the-money call on a non-dividend stock, zero rates, the interview approximation C ≈ 0.4 S σ √T. 'All else equal' means the stock and the implied volatility stay put and only time passes. Prices are quoted to cents and later answers use the quoted price.
- Where 0.4 comes from. An at-the-money call pays half the absolute move on average, and the expected absolute move of a normal variable is about 0.8 standard deviations. The one-standard-deviation move by expiry is S σ √T.
- √T is the whole mechanism. Independent daily moves add in variance, so the expected size of the total move grows with the root of time. Value follows the expected move.
- Theta from the derivative. d(√T)/dT = 1 ÷ (2√T), so θ = C ÷ (2T), or C ÷ (2 × days) per day. It rises as days fall: the same value has fewer days left to lose it over.
- The ratio test. Theta at the later date over theta now equals √(days now ÷ days later), always above one for an at-the-money option.
- Not every option decays this way. Deep in- or out-of-the-money options carry little time value weeks before expiry; their theta peaks earlier and then fades. The acceleration is an at-the-money story.
| √T: √(90 ÷ 365) | 0.4966 |
| Call: 0.4 × $100 × 30% × 0.4966 | $5.96 |
| With 45 days left: $5.96 × √0.5 | $4.21, 71% of the value with half the time gone |
| Theta now: $5.96 ÷ (2 × 90) | $0.0331 a day |
| Theta at 45 days: $4.21 ÷ (2 × 45) | $0.0468 a day |
| Days left when the call is worth $4.21: 90 × (4.21 ÷ 5.96)² | 44.9 |
Implied vol from the $5.96 price: 5.96 ÷ (0.4 × 100 × 0.4966) = 30.0%. On 20 contracts the decay from 90 to 45 days is ($5.96 − $4.21) × 2,000 = $3,500.
The formulas
About 0.4 of the one-standard-deviation move by expiry.
Value scales with the root of the time left.
The slope of the square root: it steepens into expiry.
The two inverses: days from a price ratio, vol from a price.
Worked example
Take the ratio of the days, square-root it, multiply the price. The follow-up asks why the option lost so much less than the share of time that passed.
See it move
Same option. Change how much of the time is still left, how long the option had to start with, the volatility and the stock price.
- Lower the share of the days still left. The value falls, the decay rises, and theta per day rises: the last weeks cost the most.
- Raise the volatility. The value, the decay and both thetas rise in proportion.
- Start with more days. The value at the start rises with the root of the days, and theta per day at the start falls, because there are more days to spread the decay over.
Run it backwards
The call was worth one amount with a known number of days left and is now worth another, with nothing but time changed. How many days are left?
Value is proportional to √days, so days are proportional to value squared: days₂ = days₁ × (C₂ ÷ C₁)².
The other inverse is implied volatility from a price, σ ≈ C ÷ (0.4 S √T), which is the check question's version of the same approximation.
Traps
Say it in the interview
“Roughly what is a three-month at-the-money call worth, and how fast does it decay?”
Check yourself
5 fresh questions, with new numbers. Answer each one correctly to finish the chapter. Get one wrong and you will see the full working, then you can try it again with new numbers.
Answers within 1% are marked right. Type the number; $, %, x and M are fine. First tries count toward Learned: the topic is Learned once every chapter is done and 75% of first tries were right.
- ATM call ≈ 0.4 S σ √T; implied vol ≈ C ÷ (0.4 S √T).
- C₂ = C₁ √(days₂ ÷ days₁): 71% of the value with half the time gone.
- Theta ≈ C ÷ (2 × days): it accelerates into expiry for at-the-money options.
- Calendar days ÷ 365 for options; 252 is for returns.