Gaussian Function

Definition

1.A Gaussian function of a real variable is a real multiple of the exponential function with a negative quadratic exponent. Its general form depends on three parameters:

$$f(x) = ae^{-\frac{(x-b)^{2}}{2c^{2}}}$$

2.Here $a$ and $b$ are real and $c \neq 0.$ Replacing $c$ by $-c$ leaves the function unchanged, so we take $c > 0.$ We assume $a > 0.$ Under these assumptions, $a$ is the maximum value, $b$ is the center, and $c$ controls the width.

4.The exponent $-(x-b)^{2}/(2c^{2})$ has maximum $0$ at $x = b.$ Since the exponential is strictly increasing, $f$ has maximum value $a$ at the same point. Its tails satisfy $\lim_{|x| \to +\infty}f(x)e^{k|x|} = 0$ for every $k > 0.$

5.Dividing by $a$ and setting $t = (x-b)/c$ gives the reduced Gaussian:

$$g(t) = e^{-\frac{t^{2}}{2}}$$

6.In the identity $f(x) = ag((x-b)/c),$ the parameter $b$ translates the graph horizontally, $c$ scales it horizontally, and $a$ scales it vertically. The properties of $g$ transfer to $f$ under this change of variable.

Derivative and monotonicity

7.Differentiating the exponent and applying the chain rule gives a derivative proportional to $f:$

$$f'(x) = -\frac{a(x-b)}{c^{2}}e^{-\frac{(x-b)^{2}}{2c^{2}}} = -\frac{x-b}{c^{2}}f(x)$$

8.Since $f(x) > 0$ for every $x,$ the sign of $f'$ is opposite to the sign of $x - b.$ The function is therefore strictly increasing on $(-\infty, b)$ and strictly decreasing on $(b, +\infty),$ with a single stationary point at $x = b.$ That point is an absolute maximum with value $a,$ and the function has no local minimum.

9.

The identity $f'(x) = -\dfrac{x-b}{c^{2}}f(x)$ shows that $f$ solves the separable differential equation $y' = -\dfrac{x-b}{c^{2}}y.$ Separating the variables and imposing $y(b) = a$ recovers $f$ uniquely. Its relative rate of change is $f'(x)/f(x) = -(x-b)/c^{2},$ a decreasing linear function of $x.$

Inflection points and concavity

10.Differentiating $f'(x) = -\dfrac{x-b}{c^{2}}f(x)$ with the product rule and substituting the expression for $f'$ gives the second derivative:

$$ \begin{align} f''(x) &= -\frac{1}{c^{2}}f(x) - \frac{x-b}{c^{2}}f'(x) \\[6pt] &= -\frac{1}{c^{2}}f(x) + \frac{(x-b)^{2}}{c^{4}}f(x) \\[6pt] &= \frac{f(x)}{c^{2}}\left(\frac{(x-b)^{2}}{c^{2}} - 1\right) \end{align} $$

11.Since $f(x)/c^{2} > 0,$ $f''$ has the sign of $(x-b)^{2} - c^{2}.$ This expression is negative on $(b-c, b+c),$ positive outside this interval, and zero at $x = b-c$ and $x = b+c.$ Hence the function is concave downward on $(b-c, b+c)$ and concave upward outside this interval, with inflection points at:

$$x = b - c \qquad x = b + c$$

12.The function has the same value at both inflection points:

$$f(b \pm c) = ae^{-\frac{1}{2}} \approx 0.6065a$$

13.The horizontal distance from the axis of symmetry to either inflection point is $c.$

Width at half maximum

14.A second measure of the spread is the width of the curve at half its height. Setting $f(x) = a/2$ and dividing by $a$ gives:

$$e^{-\frac{(x-b)^{2}}{2c^{2}}} = \frac{1}{2}$$

15.Taking logarithms of both sides and using $\ln(1/2) = -\ln 2$ gives $(x-b)^{2} = 2c^{2}\ln 2,$ so the two solutions are:

$$x = b \pm c\sqrt{2\ln 2}$$

16.Their distance is the full width at half maximum:

$$\mathrm{FWHM} = 2c\sqrt{2\ln 2} \approx 2.3548c$$

17.The full width at half maximum is independent of $a$ and $b.$ Multiplying $c$ by a positive factor $k$ multiplies the width by $k$ and leaves the maximum value $a$ unchanged.

The Gaussian integral

18.The area under the curve is finite despite its unbounded domain. We first evaluate the improper integral:

$$I = \int_{-\infty}^{+\infty} e^{-x^{2}} \ dx = \sqrt{\pi}$$

19.Since the integrand has no elementary antiderivative, we compute $I^{2}.$ Writing the second factor with a different variable of integration, the square is a double integral over the plane:

$$I^{2} = \left(\int_{-\infty}^{+\infty} e^{-x^{2}} \ dx\right)\left(\int_{-\infty}^{+\infty} e^{-y^{2}} \ dy\right) = \iint_{\mathbb{R}^{2}} e^{-(x^{2}+y^{2})} \ dx \ dy$$

20.Because the integrand is nonnegative, Tonelli's theorem justifies the double-integral identity.

21.The exponent depends on $x$ and $y$ only through $x^{2} + y^{2},$ so we use polar coordinates. With $x = r\cos\theta$ and $y = r\sin\theta,$ we have $x^{2} + y^{2} = r^{2},$ the area element is $r \ dr \ d\theta,$ and $\mathbb{R}^{2}$ corresponds to $r \in [0, +\infty)$ and $\theta \in [0, 2\pi]:$

$$I^{2} = \int_{0}^{2\pi} \int_{0}^{+\infty} e^{-r^{2}} r \ dr \ d\theta$$

22.The substitution $u = r^{2},$ with $du = 2r \ dr,$ gives:

$$\int_{0}^{+\infty} e^{-r^{2}} r \ dr = \frac{1}{2}\int_{0}^{+\infty} e^{-u} \ du = \frac{1}{2}$$

23.Since the inner integral equals $1/2,$ the remaining integral is:

$$I^{2} = \int_{0}^{2\pi} \frac{1}{2} \ d\theta = \pi$$

24.Since the integrand is positive, $I > 0,$ so $I = \sqrt{\pi}.$

25.

The change to polar coordinates works because the level curves of $e^{-(x^{2}+y^{2})}$ are circles centered at the origin.

Area under a general Gaussian

26.The value of $I$ determines the area under any curve of the form $ae^{-(x-b)^{2}/(2c^{2})}.$ The substitution $t = (x-b)/(c\sqrt{2}),$ with $dx = c\sqrt{2} \ dt,$ turns the exponent into $-t^{2}$ and leaves the limits at $\pm\infty$ unchanged:

$$ \begin{align} \int_{-\infty}^{+\infty} ae^{-\frac{(x-b)^{2}}{2c^{2}}} \ dx &= ac\sqrt{2}\int_{-\infty}^{+\infty} e^{-t^{2}} \ dt \\[6pt] &= ac\sqrt{2}\sqrt{\pi} \\[6pt] &= ac\sqrt{2\pi} \end{align} $$

27.The area is proportional to the height $a$ and to the width parameter $c,$ and it is independent of the center $b,$ since a translation leaves the area unchanged.

28.Unit area requires $ac\sqrt{2\pi} = 1,$ so $a = 1/(c\sqrt{2\pi}).$ The normalized function is:

$$f(x) = \frac{1}{c\sqrt{2\pi}}e^{-\frac{(x-b)^{2}}{2c^{2}}}$$

29.This is the density of the normal distribution with mean $b$ and standard deviation $c.$ It is nonnegative and has total integral $1,$ the two requirements for a probability density. Its axis of symmetry is $x = b,$ and the horizontal distance from this line to either inflection point is the standard deviation $c.$

The error function

30.The function $e^{-x^{2}}$ has no elementary antiderivative. The error function is its definite integral from $0$ to $x,$ scaled so that its limit at $+\infty$ is $1:$

$$\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^{2}} \ dt$$

31.Since $I = \sqrt{\pi}$ and the integrand is even, $\int_{0}^{+\infty} e^{-t^{2}} \ dt = \sqrt{\pi}/2.$ Thus the factor $2/\sqrt{\pi}$ makes $\mathrm{erf}(x)$ tend to $1$ as $x \to +\infty.$ The integral of an even function from $0$ to $x$ is odd, and the positive integrand makes $\mathrm{erf}$ strictly increasing. By oddness, $\mathrm{erf}(x)$ tends to $-1$ as $x \to -\infty,$ so its range is $(-1, 1),$ and $\mathrm{erf}(0) = 0.$

32.The cumulative distribution function $\Phi$ of the standard normal variable is a rescaling of $\mathrm{erf}.$ Splitting the integral defining $\Phi$ at the origin and substituting $t = u\sqrt{2}$ gives:

$$\Phi(z) = \frac{1}{2}\left(1 + \mathrm{erf}\left(\frac{z}{\sqrt{2}}\right)\right)$$

33.The standard normal Z table lists values of $\Phi$ obtained from this identity or from numerical integration of the density.

Moments of the Gaussian

34.For $\lambda > 0,$ the Gaussian integral determines $\int_{-\infty}^{+\infty} x^{n}e^{-\lambda x^{2}} \ dx$ for every nonnegative integer $n.$ When $n = 0,$ the substitution $t = x\sqrt{\lambda}$ gives:

$$J(\lambda) = \int_{-\infty}^{+\infty} e^{-\lambda x^{2}} \ dx = \sqrt{\frac{\pi}{\lambda}}$$

35.For $\lambda > 0,$ exponential decay justifies differentiation under the integral sign. Differentiating with respect to $\lambda$ brings down a factor $-x^{2}:$

$$\int_{-\infty}^{+\infty} x^{2}e^{-\lambda x^{2}} \ dx = -J'(\lambda) = \frac{1}{2}\sqrt{\frac{\pi}{\lambda^{3}}}$$

36.Each differentiation with respect to $\lambda$ raises the exponent of $x$ by two, so repeated differentiation gives every even case. When $n$ is odd, the integrand is odd and the integral vanishes.

37.Applying the case $n = 2$ with $\lambda = 1/2$ and dividing by the normalizing constant $\sqrt{2\pi}$ gives the variance of the standard normal variable:

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} x^{2}e^{-\frac{x^{2}}{2}} \ dx = \frac{1}{\sqrt{2\pi}}\sqrt{2\pi} = 1$$

38.For the general density, the substitution $x = b + ct$ reduces the integrals for the expected value and variance to standard normal integrals. Hence the expected value is $b$ and the variance is $c^{2}.$ For $n \geq 1,$ the even moments of the standard normal variable are the products of the odd integers up to $2n-1:$

$$E(X^{2n}) = (2n-1)!! = 1 \cdot 3 \cdot 5 \cdots (2n-1)$$

39.The odd moments are $0$ by symmetry. The substitution $t = u^{2}$ in $\Gamma(1/2) = \int_{0}^{+\infty} t^{-1/2}e^{-t} \ dt$ gives:

$$\Gamma\left(\frac{1}{2}\right) = 2\int_{0}^{+\infty} e^{-u^{2}} \ du = \sqrt{\pi}$$

40.Together with $\Gamma(s+1) = s\Gamma(s),$ this identity determines $\Gamma(n+1/2)$ for every nonnegative integer $n.$ These values occur in the normalizing constants of gamma distributions with half-integer shape parameters and chi-square distributions with odd degrees of freedom. A chi-square variable with $k$ degrees of freedom is the sum of the squares of $k$ independent standard normal variables.

Products and convolutions

41.The product of two Gaussian functions is again a Gaussian function. Multiplying $e^{-(x-b_1)^{2}/(2c_1^{2})}$ and $e^{-(x-b_2)^{2}/(2c_2^{2})}$ adds the exponents, and the sum is a quadratic polynomial in $x$ with negative leading coefficient. After completing the square, that polynomial has the form $-(x-b_3)^{2}/(2c_3^{2}) + k,$ where $k$ is independent of $x.$ The parameters satisfy:

$$\frac{1}{c_3^{2}} = \frac{1}{c_1^{2}} + \frac{1}{c_2^{2}} \qquad b_3 = c_3^{2}\left(\frac{b_1}{c_1^{2}} + \frac{b_2}{c_2^{2}}\right)$$

42.The factor $e^{k}$ multiplies the two height factors. The product has center $b_3$ and width parameter $c_3.$ The reciprocals of the squared widths add, so the product is narrower than either factor.

43.The convolution of two normalized Gaussians is a normalized Gaussian. If $f_1$ and $f_2$ have parameters $(b_1, c_1)$ and $(b_2, c_2),$ completing the square in the convolution integral and using the Gaussian integral shows that the convolution has parameters:

$$b = b_1 + b_2 \qquad c = \sqrt{c_1^{2} + c_2^{2}}$$

44.The centers add, and the squared widths add. The convolution of two densities is the density of the sum of two independent variables. Hence a sum of independent normal variables is normal. Its mean and variance are the sums of the component means and variances. Among nondegenerate distributions stable under sums up to a change of location and scale, the normal distribution is the only one with finite variance. For independent, identically distributed random variables with finite nonzero variance, the central limit theorem states that their centered and scaled sums converge in distribution to a normal law.

Example

45.Consider the function:

$$f(x) = 5e^{-2(x-3)^{2}}$$

46.Matching the exponent with the general form gives $b = 3$ and $\dfrac{1}{2c^{2}} = 2.$ Since $c > 0,$ the equation $c^{2} = \dfrac{1}{4}$ gives $c = \dfrac{1}{2}.$ The coefficient of the exponential gives $a = 5.$

47.The maximum is $f(3) = 5,$ attained at $x = 3,$ and the curve is symmetric about the line $x = 3.$ The inflection points are at distance $c$ from the center:

$$x = 3 - \frac{1}{2} = 2.5 \qquad x = 3 + \frac{1}{2} = 3.5$$

48.At both points the function has value $5e^{-1/2} \approx 3.033.$ The full width at half maximum is:

$$\mathrm{FWHM} = 2c\sqrt{2\ln 2} = \sqrt{2\ln 2} \approx 1.177$$

49.The area under the curve is $ac\sqrt{2\pi}:$

$$\int_{-\infty}^{+\infty} 5e^{-2(x-3)^{2}} \ dx = 5 \cdot \frac{1}{2} \cdot \sqrt{2\pi} = \frac{5\sqrt{2\pi}}{2} \approx 6.267$$

50.Dividing $f$ by this area gives a function with unit integral:

$$\frac{2}{5\sqrt{2\pi}}f(x) = \frac{2}{\sqrt{2\pi}}e^{-2(x-3)^{2}}$$

51.This function is the density of a normal distribution with mean $3,$ standard deviation $\dfrac{1}{2},$ and variance $\dfrac{1}{4}.$