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


$S$ 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? $$S=Se^{-D(T-t)}$$


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.


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