Why Sums of Squared Normals Are Chi-Square

Why the Sum Has a Chi-Square Distribution

A sum of squared normal values is always nonnegative, but that fact alone does not determine its probability distribution. To identify the distribution, we need to know how squaring changes a normal density and what happens when independent squared values are added. Those steps explain why the chi-square family arises and why its degrees of freedom count the independent terms.

The result is useful whenever a statistic can be expressed as a sum of squared, standardized normal quantities. The conditions matter: the quantities must be standard normal and mutually independent. A sum that merely looks like a sum of squares does not automatically have a chi-square distribution.

Here, \(N(0,1)\) means a normal distribution with mean 0 and variance 1. The square of a standard normal has a gamma density with shape \(1/2\) and scale 2; adding independent variables with that density adds their shape parameters. The derivation below establishes both facts, rather than relying on the appearance of the sum.

From One Standard Normal to Its Square

Let \(Z\sim N(0,1)\), and define \(Y=Z^2\). Since every square is nonnegative, \(Y\) can only take values \(y\geq 0\). For any \(y>0\), there are two possible values of \(Z\) that produce \(Y=y\): \(z=\sqrt{y}\) and \(z=-\sqrt{y}\). The change-of-variables formula adds the density contributions from both roots, accounting for the derivative of each inverse transformation.

The two contributions are equal because the standard normal density is symmetric. For \(y<0\), the density is zero. The factor \(1/\sqrt{y}\) is important: it appears because the inverse transformation has derivative magnitude \(1/(2\sqrt{y})\), which changes with \(y\).

To recognize this density, define the gamma function for \(a>0\) by \(\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt\). In particular, \(\Gamma(1/2)=\sqrt{\pi}\). A gamma density with shape \(a\) and scale 2 is \[ f(y)=\frac{y^{a-1}e^{-y/2}}{2^a\Gamma(a)},\qquad y>0. \] Setting \(a=1/2\) gives exactly the density of \(Z^2\). Thus each squared standard normal is gamma with shape \(1/2\) and scale 2.

Why Adding the Squares Adds Degrees of Freedom

The density of one square is a gamma density with shape \(1/2\). To see what happens when two such variables are added, more generally consider independent gamma variables \(U\) and \(V\), both with scale 2 and with shapes \(a>0\) and \(b>0\). Their sum \(S=U+V\) has density given by convolution:

Substituting the two gamma densities and collecting the factors gives

Use the substitution \(t=x/s\), so \(x=st\) and \(dx=s\,dt\). The integral becomes \(s^{a+b-1}\int_0^1 t^{a-1}(1-t)^{b-1}\,dt\). The integral over \(t\) is the beta integral \(B(a,b)\), and \(B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)\). This identity follows by writing \(\Gamma(a)\Gamma(b)\) as a double integral and changing variables from its two positive integration variables \(u,v\) to \(r=u+v\) and \(t=u/(u+v)\). The Jacobian is \(r\), so the double integral factors into \(\Gamma(a+b)\) times the beta integral.

Substituting that identity into the convolution yields

This is a gamma density with shape \(a+b\) and scale 2. Therefore, when independent squared standard normals are added, their gamma shapes add. Each contributes \(1/2\), so \(k\) terms produce shape \(k/2\). The resulting density is

This is the chi-square density with \(k\) degrees of freedom. The argument starts with one squared normal and applies the addition result repeatedly. It is the distributional derivation: the chi-square form follows from the normal density, the squaring transformation, and independence in the convolution.

What the Degrees of Freedom Count

In this construction, the degrees of freedom have a direct interpretation: there are \(k\) independently varying standard normal components, and each contributes one squared term. The count is not a correction or a label chosen after calculating the sum. It determines the gamma shape, and therefore the density of the sum.

The expected value and variance also show the contribution of each term. For a standard normal variable, \(E(Z^2)=1\) and \(E(Z^4)=3\); these moments follow from integration by parts in the normal density. Thus \(\operatorname{Var}(Z^2)=E(Z^4)-[E(Z^2)]^2=3-1=2\). For independent terms, expectations and variances add, giving

Every added independent squared standard normal increases the mean of the sum by 1 and its variance by 2. These formulas agree with the gamma density above. They also help explain why the number of independent terms matters, although matching a mean and variance alone would not prove that a variable is chi-square.

Conditions and a Common Mistake

The derivation depends on two essential conditions: each term must be standard normal, and the terms must be mutually independent. If independent normal measurements have different means and standard deviations, standardizing each by its own mean and standard deviation produces standard normal terms. If the terms are dependent, the convolution argument does not apply. If they are not centered at zero after standardization, the central chi-square result given here does not apply.

Common mistake: “There are \(k\) squared values, so their sum must have a chi-square distribution with \(k\) degrees of freedom.” The correction is to check the model for each term and the relationship among terms. The result requires \(k\) independent standard normal variables. Merely squaring \(k\) numbers, or even \(k\) individually normal but dependent variables, does not establish a chi-square distribution.