How do you derive the variance of a Gaussian distribution?

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How do you derive the variance of a Gaussian distribution?

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Answer 1 · 2015-11-26T07:51:57

Suppose $x$ $x$ has a probability density function $f (x)$ $f(x)$ . The variance of $x$ $x$ is calculated by $\int_{- \infty}^{\infty} {(x - μ)}^{2} f (x) d x$ $int_-oo^oo (x-mu)^2f(x) dx$ , where $μ$ $mu$ is the expected value of $x$ $x$ and is calculated by $μ = \int_{- \infty}^{\infty} x f (x) d x$ $mu = int_-oo^oo xf(x) dx$ .

Explanation:

The graph of a Gaussian is a characteristic symmetric "bell curve" shape.

The simplest case of a normal distribution is known as the standard normal distribution , described by the probability density function: $g (z) = \frac{1}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}}$ $g(z) = frac{1}{sqrt{2pi}}e^{-frac{z^2}{2}}$ .

Since the area under the curve is given by $\int_{- \infty}^{\infty} g (z) d z$ $int_-oo^oo g(z) dz$ , The factor $\frac{1}{\sqrt{2 π}}$ $1/sqrt{2pi}$ in this expression ensures that the total area under $g (z)$ $g(z)$ is equal to $1$ $1$ , as should all PDF.

The mean of $z$ $z$ is given by:

$E (z) = \int_{- \infty}^{\infty} z g (z) d z$ $"E"(z) = int_-oo^oo zg(z) dz$

$= \int_{- \infty}^{0} z g (z) d z + \int_{0}^{\infty} z g (z) d z$ $= int_-oo^0 zg(z) dz+int_0^oo zg(z) dz$ .

$= 0$ $= 0$

( $g$ $g$ is an even function, so the first half of the integral is equal to the negative of the second half of the integral. Since $g (- z) = g (z)$ $g(-z)=g(z)$ , substituting $u = - z$ $u=-z$ for the second integral should do the trick.)

The variance of $z$ $z$ is given by:

$Var (z) = \int_{- \infty}^{\infty} {(z - E (z))}^{2} g (z) d z$ $"Var"(z) = int_-oo^oo (z-"E"(z))^2g(z) dz$

$= \int_{- \infty}^{\infty} z^{2} g (z) d z$ $= int_-oo^oo z^2g(z) dz$

$= \int_{- \infty}^{\infty} \frac{z^{2}}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}} d z$ $= int_-oo^oo frac{z^2}{sqrt{2pi}}e^{-frac{z^2}{2}} dz$

$= 1$ $= 1$

(There are special techniques involved in computing integrals of this kind. The details are not shown but the result can be easily verified with a calculator.)

Substituting $x = σ z + μ$ $x=sigmaz+mu$ , we get the probability density of the Gaussian distribution:

$f (x ∣ σ, μ) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{{(x - μ)}^{2}}{2 σ^{2}}}$ $f(x|sigma,mu) = frac{1}{sigmasqrt{2pi}}e^{-frac{(x-mu)^2}{2sigma^2}}$ .

$μ$ $mu$ determines the location of the maximum and $σ$ $sigma$ determines how narrow/tall the maximum should be.

The variance is given by

$Var (x) = Var (σ z + μ)$ $"Var"(x) = "Var"(sigmaz+mu)$

$= σ^{2} Var (z)$ $= sigma^2 "Var"(z)$

$= σ^{2} (1)$ $= sigma^2 (1)$

$= σ^{2}$ $= sigma^2$

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How do you derive the variance of a Gaussian distribution?

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