At-the-Money-Forward Call Approximation Near Expiry

Summary

The document derives the near-expiry approximation for the price of an at-the-money-forward European call under Black–Scholes. It rewrites the call price using the forward price and discounts the resulting payoff expression at the risk-free rate. At the forward strike, the forward equals the strike, simplifying the two normal-distribution terms to values at equal and opposite arguments.

Expanding those terms for a short time to expiry yields a leading price proportional to volatility times the square root of remaining time. The scale factor is the discounted forward, which equals the spot price discounted by the dividend yield; this explains the replacement that prompted the question. The approximation relies on the Black–Scholes assumptions and is intended for short maturities and an exactly forward-at-the-money strike. A separate response notes that, for European options, this strike also corresponds to equal call and put prices.

Key ideas

• Express the Black–Scholes call price using the forward price before applying the at-the-money-forward condition.
• At the forward strike, the normal-distribution arguments become equal in magnitude and opposite in sign.
• A short-time expansion gives a leading price proportional to volatility and the square root of time to expiry.
• The discounted forward factor is equal to spot discounted by the dividend yield.
• The approximation is local to short maturities and the at-the-money-forward condition.

Tags

# At-The-Money-Forward option approximation


# At-The-Money-Forward option approximation












Given that the Black-Scholes formula for a European Call is given by:

$$C(S,t)=Se^{-D(T-t)}N(d_1)-Ke^{-r(T-t)}N(d_2)$$

$S$ is stock price, $K$ is strike price

An At-The-Money-Forward option is struck when $K=Se^{(r-D)(T-t)}$.

When $t\rightarrow T$, show that the approximation can be obtained to give $$C(S,t)\approx 0.4 Se^{-D(T-t)}\sigma\sqrt{T-t}$$

I noticed from this post HERE, I understood everything except the last step:

I have found that: $$C(S,t)\approx S(0.4\sigma\sqrt{T-t})$$ by using the Taylor's Series expansion of $N(x)$ around $0$, but do not know how can we use the following replacement? $$S=Se^{-D(T-t)}$$

## Answer by Quantuple (score 4)

https://quant.stackexchange.com/a/35435

Rewrite the call price as $$ C(S,t) = e^{-r(T-t)} \left( F N(d_1) - K N(d_2) \right) $$ where $F = Se^{(r-D)(T-t)}$ (forward price). Using $F$ you can also write that $$d_{1,2} = \frac{ \ln(F/K)\pm\frac{1}{2} \sigma^2(T-t) }{\sigma\sqrt{T-t}}$$

Now, by definition when the option is struct AMTF, $K=F$ and you have that $$ C(S,t) = e^{-r(T-t)} F \left[ N\left(\frac{1}{2}\sigma\sqrt{T-t}\right) - N\left(-\frac{1}{2}\sigma\sqrt{T-t}\right) \right] $$

You are then exactly in the same situation as in this question (see equation (1)), but with $Fe^{-r(T-t)}=Se^{-D(T-t)}$ instead of $S$. The rest of the development stays the same.

## Answer by REgg1 (score 1)

https://quant.stackexchange.com/a/69336

this implies that the ATMF implied from options prices, for European options, would be the strike where the price of call = price of put.

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