Is the error function and $\mathcal{N}(0,1)$ the same thing?

Question

Are $\mathcal{N}(0,1)$ and $\Phi(x)$ the same thing? It seems that the derivative $\Phi(x)$ and the density of $\mathcal{N}(0,1)$ are the same, but I'm not sure.

Answer 1

It's a matter of terminology. They are not the same, but they apply to the same concept.

If a random variable $X$ has the standard normal probability distribution $\mathcal N(0,1)$, then the cumulative probability function that $X$ is less than $x$ is $$\Phi(x)=P(X<x).$$

Since you are asking about the "error function", we also have the following.

If a random variable $Y$ has the normal probability distribution $\mathcal N(0,1/\sqrt 2)$, then the cumulative probability that $Y$ is between $-x$ and $x$ is the so called error function $$\operatorname{erf}(x)=P(-x<Y<x).$$

Answer 2

The following functions are essentially the same thing

• Error function $$ \operatorname{erf}(z)=\frac{2}{\sqrt{\pi}}\int_{\color{red}{0}}^ze^{-t^2}\,dt\,. $$

• Cumulative distribution function of the standard normal distribution ${\cal N}(0,1)$: $$ \Phi(z)=\frac{1}{\sqrt{2\pi}}\int_{\color{red}{-\infty}}^ze^{-\frac{t^2}{2}}\,dt\,. $$

• Heat kernel $$ p_t(z)=\frac{1}{\sqrt{4t\pi}}\int_{\color{red}{-\infty}}^ze^{-\frac{x^2}{4t}}\,dx\,. $$ Obviously, $$ \boxed{\quad\frac{1}{2}+\frac{1}{2}\operatorname{erf}\Big(\frac{z}{\sqrt{2}}\Big)=\Phi(z)=p_{\frac{1}{2}}(z)\,.\quad} $$