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Probability, Statistics and Financial Data / Continuous Random Variables

Derivation — The Gaussian Integral, Standardisation, and Normal Moments

The Normal Distribution asserted that 1/2πσ21/\sqrt{2\pi\sigma^2} is exactly the constant needed to make the normal PDF integrate to 11, and that μ,σ2\mu,\sigma^2 are literally the mean and variance. Neither claim is obvious: e−x2/2e^{-x^2/2} has no elementary antiderivative — none of the usual calculus techniques (substitution, parts, partial fractions) evaluate ∫e−x2/2dx\int e^{-x^2/2}dx in closed form. The trick that resolves this — squaring the integral and switching to polar coordinates — is one of the most elegant arguments in classical analysis, and this note carries it out in full, then uses the result to derive the normal distribution's mean and variance from the PDF definition directly.
Learning objectives. By the end of this note you should be able to:
  • reproduce the polar-coordinates proof that ∫−∞∞e−x2/2 dx=2π\int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi};
  • use this result (via substitution) to verify the general normal PDF's normalising constant;
  • derive E[X]=μ\mathbb{E}[X]=\mu for a normal random variable using a symmetry argument;
  • derive Var(X)=σ2\text{Var}(X)=\sigma^2 using integration by parts.

In this lesson

  1. The Gaussian integral
  2. The normal distribution's mean
  3. The normal distribution's variance

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