The cumulative distribution function of the standard normal distribution function is denoted with $\Phi$, and it is defined as:
$$\Phi(n)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x \exp(-\frac{x^2}{2})dn.$$
While several numerical approximation of $\Phi(n)$ are given, they all run into the problem that for n < -7, the function quickly evaluates to 0. As was the case when computing this function from the previous post:
$$E(P, t, m) = e^{\frac{\nu^2 t}{2}}\frac{\Phi(n(P,m)-\nu\sqrt{t})}{\Phi(n(P,m))}.$$
So for values of $n<-7$, we want to find an approximation that can be taken under the $\log$ for a reduction. We shall start with the guess of: $F(x) = -\frac{e^{-\frac{x^2}{2}}}{x}$, under derivatives, the function becomes $$F'(x) = e^{-\frac{x^2}{2}} + \frac{1}{x^2} e^{-\frac{x^2}{2}}.$$
If we integrated this from $-\infty$ to $x$ then, for $n<0$, the error between $\int_{-\infty}^n F'(x) dx$ and $ \int_{-\infty}^n e^{-\frac{x^2}{2}}dx$ is $$F(x)- \sqrt{2\pi} \Phi(n)=\int_{-\infty}^n \frac{1}{x^2} e^{-\frac{x^2}{2}} dx<\frac{1}{n^2} \int_{-\infty}^n e^{-\frac{x^2}{2}}dx=\frac{\sqrt{2\pi}}{n^2} \Phi(n).$$
Then we can see that this approximation is at most a fraction $\frac{1}{n^2}$ away from $\Phi(n)$, which promises to give better results as $n<0$ decreases and absoulte value of $n$ increases. Thus we shall take $\Phi$ to be, for $n<-7$:
$$\Phi(n) = -\frac{e^{-\frac{n^2}{2}}}{\sqrt{2\pi}n}.$$
So to obtain the log plot of the expression
$$E(P, t, m) = e^{\frac{\nu^2 t}{2}}\frac{\Phi(n(P,m)-\nu\sqrt{t})}{\Phi(n(P,m))},$$
we have
\begin{align*} &\log\left(e^{\frac{\nu^2 t}{2}}\frac{\Phi(n-\nu\sqrt{t})}{\Phi(n)}\right)\\
=&\frac{t\nu^2}{2}-\log(\nu \sqrt{t}-n)-\frac{(\nu\sqrt{t}-n)^2}{2}-\log(\Phi(n))-\log(\sqrt{2\pi}).\end{align*}