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So I wanted to find the probability (or, at least the magnitude of the probability), of something more than 70.5 standard deviations below the mean in a normal distribution. (It's nothing practical, it's just for a joke with my friends). However, there's no calculators I can find that have enough precision to calculate

$\frac{1}{2}\left[1 + \text{erf}\left(\frac{-70.5}{\sqrt{2}}\right)\right] $ or $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-70.5} e^{-\frac{t^2}{2}} dt\ $

The PDF can be found with some high precision calculators, and that's around $10^{-1080} $, but are there any ways to approximate the CDF? Like somehow taking log or something to find the approximate value? Or is $10^{-1080} $ also a close estimate to the CDF?

Mewtwo 2387
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  • Substitute $t=70.5+u/70.5$. Let $y=1/70.5$ and expand $\exp -(uy)^2/2$ as a taylor series in $y$. Finally integrate $\int e^{-u}u^{2k}du$ etc by parts – Empy2 Dec 20 '23 at 10:16

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If you want to compute $$A=\frac{1}{2} (1+\text{erf}(x))=1-\frac{1}{2}\text{erfc}(x)$$ for large values of $x$, use $$\text{erfc}(x)=1-\frac{e^{-x^2}}{\pi }\sum_{n=0}^\infty (-1)^n\frac{\Gamma \left(n+\frac{1}{2}\right)}{ x^{2 n+1}} $$ For your value of $x$ $$\frac{e^{-x^2}}{\pi }\sum_{n=0}^{10}(-1)^n\frac{\Gamma \left(n+\frac{1}{2}\right)}{ x^{2 n+1}}=-5.99224\times 10^{-1082}$$

Claude Leibovici
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