为什么我们渐近地将值0比值1渐近？

对于这里的人们来说，这可能是基本的。我只是计算机用户。

我愚弄了标准正态累积分布函数（CDF）的极值（0和1），并且注意到对于较大的负值变量，我们可以获得非常小的概率值，但没有达到相同的范围另一方面，对于较大的正值，对于较小（绝对值）的变量值，已经出现了值“ 1”。

From a theoretical point of view, the tail probabilities of the Standard Normal distribution are symmetric around zero, so the probability mass to the left of, say, X=-10, is the same as the probability mass to the right of X=10. So at X=-10 the distance of the CDF from zero is the same as is its distance from unity at X=10. But the computer/software complex doesn't give me this.

Is there something in the way our computers and software (usually) compute, that creates this asymmetric phenomenon, while the actual relation is symmetric?

Computations where done in "r", with an ordinary laptop.

This post is related, Getting high precision values from qnorm in the tail

Floating-point formats represent numbers as a sign s (+1 or ?1), a significand f, and an exponent e. Each format has some fixed base b, so the number represented is s•f•b^e, and f is restricted to be in [1, b) and to be expressible as a base-b numeral of some fixed number p of digits. These formats can represent numbers very close to zero by making e very small. But the closest they can get to 1 (aside from 1 itself) is where either f is as near 1 as it can get (aside from 1 itself) and e is 0 or f is as near b as it can get and e is ?1.

For example, in the IEEE-754 binary64 format, commonly used for double in many languages and implementations, b is two, and p is 53, and e can be as low as ?1022 for normal numbers (there are subnormal numbers that can be smaller). This means the smallest representable normal number is 2^?1022. But near 1, either e is 0 and f is 1+2^?52 or e is ?1 and f is 2?2^?52. The latter number is closer to 1; it is s•f•b^e = +1•(2?2^?52)•2^?1 = 1?2^?53.

因此，以这种格式，我们可以从零到2 ^？1022的距离（更接近于次正规数），但是从1 到2 ^？53的距离。