Normalizing constant: Difference between revisions

Revision as of 06:17, 27 September 2023

Revision as of 21:16, 31 December 2023

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<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{2\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,