Content deleted Content added
→Examples: gaussian integral is sqrt(pi), not sqrt(2 pi) Tags: Reverted Mobile edit Mobile web edit |
m →Examples: correct value Tag: Reverted |
||
Line 18:
<math display="block">p(x)=e^{-x^2/2}, \quad x\in(-\infty,\infty) </math>
we have the corresponding [[Gaussian integral]]
<math display="block">\int_{-\infty}^\infty p(x) \, dx = \int_{-\infty}^\infty e^{-x^2/2} \, dx = \sqrt{2\pi\,},</math>
Now if we use the latter's [[reciprocal value]] as a normalizing constant for the former, defining a function <math> \varphi(x) </math> as
<math display="block">\varphi(x) = \frac{1}{\sqrt{2\pi\,}} p(x) = \frac{1}{\sqrt{\pi\,}} e^{-x^2/2} </math>
so that its [[integral of a Gaussian function|integral]] is unit
<math display="block">\int_{-\infty}^\infty \varphi(x) \, dx = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\,}} e^{-x^2/2} \, dx = 1 </math>
then the function <math> \varphi(x) </math> is a probability density function.<ref>Feller, 1968, p. 174.</ref> This is the density of the standard [[normal distribution]]. (''Standard'', in this case, means the [[expected value]] is 0 and the [[variance]] is 1.)
And constant <math display="inline"> \frac{1}{\sqrt{2\pi}} </math> is the '''normalizing constant''' of function <math>p(x)</math>.
Similarly,
| |||