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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>
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