CDF of standard normal random variable never actually is 0 or 1, right?

The CDF of a standard normal random variable is never actually 0 or 1, they only approach 0 and 1, correct?

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The CDF for the normal distribution is defined as $F(x) = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^x e^{-(t-\mu)^2/(2\sigma^2)} \, dt$.

The integrand is the probability density function (PDF), which by definition must be non-negative everywhere. The integrand here is an exponential function, which clearly cannot be zero anywhere, and so is strictly positive. We already know (by definition of a CDF) $x \to \infty$, $F(x) \to 1$ monotonically from below. Based on the previous statements, there can be no finite value of $x$ such that $F(x) = 1$.

Likewise, $F(-x) = 1-F(x) \to 0$ as $x \to \infty$. By a similar rationale, there is also no finite $x$ such that $F(x) = 0$.

This is to say that the normal distribution function is defined on infinite support. The CDF is defined such that the integral of the PDF over the support is 1.

Since a normal RV $x$ can be any real number, you can always pick a number larger or smaller where the PDF -- the integrand in the CDF integral definition -- is non-zero. So you are correct, there are no finite values where the CDF of the normal is exactly $0$ or $1$.

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This is very strangely phrased. If there is a finite value of $x$ such that $F(x)=1$, then certainly $F(x)\to 1$ as $x\to \infty$, so the latter can't imply the former is not the case. –  Ben Millwood Sep 19 '12 at 1:02
Indeed, the "Thus" in the second phrase is plainly wrong. –  leonbloy Sep 19 '12 at 1:09
@leonbloy Not so if the integrand is strictly positive everywhere -- which is the case here, although I did not state that explicitly. I have edited the answer. –  Arkamis Sep 19 '12 at 3:02
@EdGorcenski it's ok for me now –  leonbloy Sep 19 '12 at 3:09