0

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.

Jose Avilez
  • 12,710
user412026
  • 5
  • 2
  • 2
    $\Phi(\cdot)$ is the CDF (cumulative distribution function) of $\mathcal{N}(0,1)$. They are not the same (one being a CDF and the other being a distribution), but both contain same amount of "information" to recover one from the other and vice versa. – Sangchul Lee May 18 '22 at 18:15
  • I don't quite understand, can you use a simpler language? – user412026 May 18 '22 at 18:20
  • Addtionally, the "error function" $\operatorname{erf}(\cdot)$ is a slight variant of $\Phi(\cdot)$. – Klaas van Aarsen May 18 '22 at 18:20

2 Answers2

1

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, $$ \boxed{\quad\frac{1}{2}+\frac{1}{2}\operatorname{erf}\Big(\frac{z}{\sqrt{2}}\Big)=\Phi(z)=p_{\frac{1}{2}}(z)\,.\quad} $$

Kurt G.
  • 14,198
0

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 $$\operatorname{erf}(x)=P(-x<Y<x).$$

Klaas van Aarsen
  • 5,908