# Derivation in Jaeckel's "By Implication" paper

In this paper by Jaeckel (2006), he derives the asymptotics for the option price $$b$$ as:

\begin{align*} \lim_{\sigma \to \infty}b= e^{\theta x/2} - \frac{4}{\sigma}\cdot \phi(\sigma/2) \tag{2.7}\\ \lim_{\sigma \to 0} b = \iota +x \cdot \phi(\frac{x}{\sigma}) \cdot \left(\frac{\sigma}{x}\right)^3 \tag{2.8} \end{align*}

For: $$\iota = h(\theta x)\cdot\theta\cdot\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right),$$ where $$h(\cdot)$$ is the heaviside function.

I get that the Abramovich-Stegun approximation for the gaussian cumulative distribution function(CDF) $$\Phi(z)$$ is perhaps used to derive expressions (2.7) and (2.8). But, I couldn't follow through and derive these myself.

I would like to ask for some help, to understand how Jaeckel derives this expressions.

As you mentioned, we know the Abramovich-Stegun approximation for the CDF $$\Phi(z) = h(z) − \dfrac{\varphi(z)}{z} \left[ 1 - \dfrac{1}{z^2} + \mathcal{O}\left(z^{-4} \right)\right], \quad \text{for} \; |z| \to \infty.$$

Moreover, note that we can develop the term $$e^{\pm x/2} \varphi \left(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right] \right) = \dfrac{1}{\sqrt{2\pi}}\exp \left\lbrace \pm \dfrac{x}{2} - \dfrac{1}{2}\left(\dfrac{x^2}{\sigma^2} + \dfrac{\sigma^2}{4}\right) \mp \dfrac{x}{2}\right\rbrace = \sqrt{2\pi} \varphi \left( \dfrac{x}{\sigma}\right) \varphi \left(\dfrac{\sigma}{2} \right),$$ where $$\sqrt{2\pi} \varphi \left( \dfrac{x}{\sigma}\right) \to 1$$ for $$\sigma \to \infty$$ and $$\sqrt{2\pi} \varphi \left( \dfrac{\sigma}{2}\right) \to 1$$ for $$\sigma \to 0$$.

I will split the derivation in the two cases, $$\sigma \to 0$$ and $$\sigma \to \infty$$.

First case: $$\sigma \to 0$$

In the $$\sigma\to0$$ case this means $$e^{\pm x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]\right) \simeq e^{\pm x/2} \left\lbrace h(\theta \cdot x) - \dfrac{\varphi(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right])}{\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]} + \mathcal{O}\right\rbrace.$$

Then, in the expression for $$b$$, we have \begin{aligned} b & = \theta \left\lbrace e^{x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} + \dfrac{\sigma}{2} \right]\right) − e^{−x/2} \Phi \left(\theta \left[\dfrac{x}{\sigma} - \dfrac{\sigma}{2} \right]\right)\right\rbrace \\ & \simeq \theta \cdot h(\theta \cdot x) \left( e^{x/2} − e^{−x/2} \right) + \theta \cdot \varphi(x/\sigma) \left[ \dfrac{-\theta}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{\theta}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right] \end{aligned} and doing simple algebra you can see that the last term in brackets equals (in the $$\sigma \to 0$$ limit) $$\left[ \dfrac{-1}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{1}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right] = \dfrac{x}{\left( x / \sigma \right)^3}.$$

Therefore, we have $$b \simeq \theta \cdot h(\theta \cdot x) \left( e^{x/2} − e^{−x/2} \right) + \varphi(x/\sigma) \cdot x \cdot \left(\sigma / x\right)^3.$$

Second case: $$\sigma \to \infty$$

In the $$\sigma\to\infty$$ case this means $$e^{\pm x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]\right) \simeq e^{\pm x/2} \left\lbrace h(\pm \theta) - \dfrac{\varphi(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right])}{\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]}+ \mathcal{O} \right\rbrace.$$

Then, in the expression for $$b$$, we have \begin{aligned} b & = \theta \left\lbrace e^{x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} + \dfrac{\sigma}{2} \right]\right) − e^{−x/2} \Phi \left(\theta \left[\dfrac{x}{\sigma} - \dfrac{\sigma}{2} \right]\right)\right\rbrace \\ & \simeq \theta \left( e^{x/2} h(\theta \cdot x) − h(-\theta \cdot x) e^{−x/2} \right) + \theta \cdot \varphi(\sigma / 2) \left[ \dfrac{-\theta}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{\theta}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right] \end{aligned}

similarly to the previous case, now in the $$\sigma \to \infty$$ limit, we have $$\left[ \dfrac{-1}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{1}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right] = - 4 / \sigma.$$

Therefore, we have $$b \simeq e^{\theta \cdot x/2} - 4 / \sigma \cdot \varphi(\sigma/2),$$ where I have used the definition of the Heaviside to cancel one of the two terms with $$h(\cdot)$$.

Hope this helps!