The Chi-Square Distribution as a Sum of Squares

From Standard Normal Variables to a Chi-Square Variable

A standard normal variable can be positive or negative. Squaring it removes the sign and produces a nonnegative value. Adding several such squared values gives a random quantity with a distribution that appears throughout statistical theory: the chi-square distribution. Its defining ingredients are precise: the variables being squared must each be standard normal, and they must be independent.

The number of independent standard normal components determines the degrees of freedom. This gives a direct way to recognize the distribution from a formula, while the sum-of-squares construction also explains why a chi-square variable cannot be negative. The definition alone does not tell us how to calculate probabilities or describe how the curve changes with the degrees of freedom; those require separate results.

Definition: Let $Z_1,\ldots,Z_k$ be mutually independent standard normal random variables, each with distribution $N(0,1)$, where $k$ is a positive integer. The sum $$ Q=Z_1^2+Z_2^2+\cdots+Z_k^2 $$ has a chi-square distribution with $k$ degrees of freedom. We write $Q\sim\chi^2_k$, and call $k$ its degrees of freedom, or $df$.

The definition has three essential parts. First, there are $k$ random variables. Second, each has the standard normal distribution, with mean 0 and variance 1. Third, they are mutually independent: knowing the values of some components does not change the distribution of any of the others. When all three conditions hold, the sum of their squares is chi-square with $df=k$.

The degrees of freedom here count the independent standard normal components in the defining sum. They are not determined just by the number of values one happens to observe, and they are not automatically one less than a sample size. The same value of $k$ can arise in different settings, but the chi-square definition applies only when the stated distribution and independence conditions are met.

Reading the Definition in Formulas

To recognize a chi-square variable, look for a sum in which each term is the square of a distinct, independent standard normal variable. A minus sign inside a squared term does not affect the result: $(-Z_i)^2=Z_i^2$. The number of squared components—not their realized numerical values—is what sets the degrees of freedom.

For example, if $Z_1$ and $Z_2$ are independent and each has distribution $N(0,1)$, then $Z_1^2+Z_2^2\sim\chi^2_2$. If a third independent standard normal variable is added as another squared term, the resulting sum has $df=3$. Merely writing down three squares is not sufficient if one component is not standard normal or if the components are dependent.

Worked Example: Identifying the Degrees of Freedom

Suppose $A$, $B$, and $C$ are mutually independent random variables, each with distribution $N(0,1)$. Define $Q=A^2+B^2+C^2$. Identify the distribution and its degrees of freedom.

There are three components, and each is standard normal. The statement also gives mutual independence, so all requirements in the definition are satisfied. Thus $$ Q=A^2+B^2+C^2\sim\chi^2_3,\qquad df=3. $$

The answer is $df=3$ because three independent standard normal variables are squared and added. It does not depend on the particular values that $A$, $B$, and $C$ might take in any one outcome.

Worked Example: Calculating One Realized Sum of Squares

In a hypothetical calculation, three independent standard normal random variables $Z_1,Z_2,Z_3$ happen to take the values $0.5$, $-1.2$, and $2.0$. Find the realized value of $Q=Z_1^2+Z_2^2+Z_3^2$, and state the distribution of the random variable $Q$.

Substitute the given values into the sum and square each value: $$ Q=(0.5)^2+(-1.2)^2+(2.0)^2 =0.25+1.44+4.00 =5.69. $$

The realized sum is $5.69$. Separately, the random variable $Q$ has a chi-square distribution with three degrees of freedom, $Q\sim\chi^2_3$, because it is defined using three independent standard normal components. The number $5.69$ is one possible value of $Q$; it is not the distribution itself.

Why the Support Is Nonnegative

The support of a random variable is the set of values that it can take, or, more precisely, the values for which every neighborhood has positive probability. The sum-of-squares definition immediately rules out negative values. It also puts zero in the support, even though the probability that the sum equals exactly zero is zero.

Here is the justification. Each square satisfies $Z_i^2\geq 0$, so their sum satisfies $Q\geq 0$. Thus no negative number can be a value of $Q$. For $Q$ to equal zero, every squared term must equal zero, which requires $Z_1=\cdots=Z_k=0$. A standard normal variable has probability zero of equaling any one exact value, so $P(Q=0)=0$. But values arbitrarily close to zero do have positive probability: all $k$ variables can fall in a sufficiently small interval around zero, and that event has positive probability.

The support extends to every positive value as well. For any fixed $q>0$, the vector $(\sqrt q,0,\ldots,0)$ has sum of squares equal to $q$. The joint density of the independent standard normal variables is positive around that vector. Therefore, for any neighborhood of $q$, there is a neighborhood of this vector whose sums of squares fall in the given neighborhood; that event has positive probability. Taken together, these facts show that the support is $[0,\infty)$.

Result: If $Q\sim\chi^2_k$ for a positive integer $k$, then $Q\geq 0$, and its support is $[0,\infty)$. In particular, $P(Q<0)=0$ and $P(Q=0)=0$, even though zero is in the support.

Worked Example: Checking the Conditions Before Naming a Distribution

Consider two proposed sums. For the first, $U$ and $V$ are independent and each has distribution $N(0,1)$, and $R=U^2+V^2$. For the second, $X$ and $Y$ are independent, with $X\sim N(0,1)$ and $Y\sim N(0,4)$, and $S=X^2+Y^2$. Decide which sum has a chi-square distribution under the definition.

For $R$, both variables are standard normal and they are independent. There are two squared terms, so $$ R=U^2+V^2\sim\chi^2_2. $$

For $S$, independence holds, but $Y$ is not standard normal: its variance is 4 rather than 1. Consequently, $X^2+Y^2$ does not meet the defining conditions for a chi-square variable. It is not correct to label $S$ as $\chi^2_2$ just because it is a sum of two squares.

To see the role of standardization, set $W=Y/2$. Then $W\sim N(0,1)$, and $W$ is independent of $X$. The sum $X^2+W^2$ is $\chi^2_2$. But the original sum is $$ S=X^2+Y^2=X^2+4W^2, $$ which is not that sum of two standard normal squares. This check identifies why the original expression does not satisfy the definition; it does not require calculating the distribution of $S$.

What Degrees of Freedom Do—and Do Not—Tell You

For a variable constructed directly from independent standard normal components, $df$ is the count of terms in the defining sum. Once the number of terms is known, the notation $\chi^2_k$ names the distribution. It does not identify an observed value of the random variable: for instance, $Q\sim\chi^2_4$ says what distribution $Q$ follows, while a value such as $Q=6.23$ is one realization.

The degrees-of-freedom label also does not by itself establish that a formula is chi-square. Verify the components first. A sum involving nonstandard normal variables, dependent variables, or terms that are not squares of standard normal variables cannot be declared chi-square simply by counting terms. In some applications, transformations produce independent standard normal components; in that case, it is the transformed components that must be counted.

This construction matters because it provides a common probability model for nonnegative sums of standardized squared quantities. It is therefore useful to learn to recognize the defining structure before using any later calculation involving the distribution. At this stage, the key skill is identification: check the component distributions, check independence, count the terms, and record the nonnegative support.

Common Mistakes

Counting squares without checking their distributions. A formula with two squared variables is not automatically $\chi^2_2$. Each component must be standard normal.
Ignoring dependence. Even if each component separately has distribution $N(0,1)$, dependence means the definition's independence requirement has not been verified.
Confusing a value with a distribution. A computed sum such as $5.69$ is one realized value; it does not replace the statement $Q\sim\chi^2_k$.
Assuming the sum can equal zero with positive probability. Zero is in the support, but for a positive number of continuous standard normal components, the probability of all of them being exactly zero is zero.
Assuming degrees of freedom always equal sample size minus one. In this definition, $df$ is the number of independent standard normal terms. Other settings may have different reasoning for degrees of freedom.

Key takeaway: A chi-square variable with $k$ degrees of freedom is the sum of the squares of $k$ mutually independent standard normal variables. Check standard normality and independence, count the components, and remember that the support is $[0,\infty)$.

Check Your Understanding

Answer each question by using the defining sum-of-squares construction.

Suppose $Z_1,\ldots,Z_5$ are mutually independent standard normal variables. Write their sum of squares and state its distribution and degrees of freedom.
Two independent standard normal variables take the values $-0.8$ and $1.5$. Calculate their realized sum of squares.
Why is a sum of three squared variables not necessarily $\chi^2_3$? State both conditions that must be checked.
Can a chi-square variable be negative? Is zero in its support, and does it have positive probability of equaling zero?
If $X\sim N(0,1)$ and $Y\sim N(0,9)$ are independent, does $X^2+Y^2$ meet the definition of a chi-square variable? Explain.