Moments and MGF of a Chi-Square Variable

Why Moments and the MGF Matter

A chi-square variable is built from squared standard normal variables. That construction makes its mean and variance accessible without integrating a chi-square density, and its moment generating function (MGF) gives a systematic way to obtain higher moments. The same method also handles independent sums and helps distinguish a chi-square variable from a weighted sum of normal squares, which generally has a different distribution.

These calculations are useful throughout sampling-distribution theory: they summarize the center and spread of a chi-square quantity and provide checks on calculations involving sums of squares. The results here are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let \(Z_1,\ldots,Z_\nu\) be independent standard normal random variables, each with distribution \(N(0,1)\), and let \(\nu\) be a positive integer. The variable \(T=\sum_{j=1}^{\nu}Z_j^2\) has a chi-square distribution with \(\nu\) degrees of freedom, written \(T\sim\chi^2_\nu\). Its moment generating function is \(M_T(t)=E[e^{tT}]\), for values of \(t\) where this expectation is finite.

For any random variable with an MGF in an open interval around zero, its \(r\)th raw moment is the \(r\)th derivative of its MGF at zero: \(E[T^r]=M_T^{(r)}(0)\). In particular, \(E[T]=M_T'(0)\), and \(\operatorname{Var}(T)=M_T''(0)-[M_T'(0)]^2\). Raw moments are moments about zero; variance is a central moment about the mean.

Deriving the Chi-Square MGF

Start with one standard normal variable \(Z\sim N(0,1)\), whose density is \(f_Z(z)=(2\pi)^{-1/2}e^{-z^2/2}\). For \(Z^2\), its MGF is an integral that can be evaluated by recognizing a normalizing Gaussian integral.

\begin{aligned} M_{Z^2}(t) &=E[e^{tZ^2}]\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{tz^2}e^{-z^2/2}\,dz\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-(1-2t)z^2/2}\,dz\\ &=(1-2t)^{-1/2},\qquad t<\frac12. \end{aligned}

The condition \(t<\tfrac12\) is essential. If \(t<\tfrac12\), then \(1-2t>0\), and the integral is finite. If \(t\geq\tfrac12\), the integrand does not decay sufficiently as \(|z|\) grows, so the integral diverges and the MGF is not finite.

Now let \(T=\sum_{j=1}^{\nu}Z_j^2\), with the independent standard normal variables in the definition. Independence means the MGF of the sum is the product of the individual MGFs. Thus the MGF of \(T\) is the \(\nu\)th power of the MGF of one squared standard normal.

M_T(t)=\prod_{j=1}^{\nu}M_{Z_j^2}(t) =\left(1-2t\right)^{-\nu/2},\qquad t<\frac12.

Theorem: If \(T\sim\chi^2_\nu\) is represented as the sum of \(\nu\) independent squared standard normal variables, then \(M_T(t)=(1-2t)^{-\nu/2}\) for \(t<\tfrac12\). The MGF is infinite for \(t\geq\tfrac12\).

Mean, Variance, and Higher Raw Moments

Differentiate the MGF and evaluate the derivatives at zero. The first derivative gives the mean; the second derivative gives the second raw moment, from which the variance follows.

\begin{aligned} M_T'(t)&=\nu(1-2t)^{-\nu/2-1},\\ M_T''(t)&=\nu(\nu+2)(1-2t)^{-\nu/2-2},\\ M_T'''(t)&=\nu(\nu+2)(\nu+4)(1-2t)^{-\nu/2-3}. \end{aligned}

\begin{aligned} E[T]&=M_T'(0)=\nu,\\ E[T^2]&=M_T''(0)=\nu(\nu+2),\\ \operatorname{Var}(T) &=E[T^2]-\{E[T]\}^2\\ &=\nu(\nu+2)-\nu^2=2\nu,\\ E[T^3]&=M_T'''(0)=\nu(\nu+2)(\nu+4). \end{aligned}

The variance formula illustrates why it is important to distinguish a raw moment from a variance. The second raw moment is \(\nu(\nu+2)\), not \(2\nu\); subtracting the square of the mean gives the variance \(2\nu\). The standard deviation is \(\sqrt{2\nu}\).

Worked Example: Moments of a Five-Term Sum of Squares

Model and assumptions. In a hypothetical statistical calculation, let \(Z_1,\ldots,Z_5\) be independent \(N(0,1)\) variables and set \(T=Z_1^2+\cdots+Z_5^2\). Thus \(T\sim\chi^2_5\). No interest rate or mortality assumption is involved; this is a probability model.

Method. Use derivatives of the chi-square MGF. This avoids trying to calculate the density integral separately for each moment.

For \(\nu=5\), the MGF is \(M_T(t)=(1-2t)^{-5/2}\), for \(t<\tfrac12\). The mean is \(E[T]=5\). The second raw moment is \(E[T^2]=5(5+2)=5\cdot7=35\).

Therefore, \(\operatorname{Var}(T)=35-5^2=35-25=10\). As a further check from the third derivative, \(E[T^3]=5(7)(9)=315\).

Check. The variance is \(2\nu=2(5)=10\), agreeing with the calculation from the first two raw moments. A common mistake is to report \(35\), the second raw moment, as the variance; the correction is to subtract \(E[T]^2=25\).

Independent Sums and Weighted Sums

If \(A\) and \(B\) are independent and have MGFs, the MGF of \(A+B\) is \(M_A(t)M_B(t)\), wherever both factors are finite. For independent chi-square variables formed from disjoint collections of independent standard normals, the degrees-of-freedom parameters add: the combined sum is simply the sum of all the squared standard normals in both collections.

A positive weight changes the MGF domain as well as the moments. If \(W=aZ^2\) for a constant \(a>0\) and \(Z\sim N(0,1)\), then \(M_W(t)=M_{Z^2}(at)=(1-2at)^{-1/2}\), which is finite when \(at<\tfrac12\), or \(t<1/(2a)\). For a sum of independent positively weighted squares, multiply the corresponding MGFs and intersect their domains.

Worked Example: A Positively Weighted Sum of Normal Squares

Model and assumptions. Consider the hypothetical variable \(W=2Z_1^2+2Z_2^2+\tfrac12Z_3^2+\tfrac12Z_4^2+\tfrac12Z_5^2\), where \(Z_1,\ldots,Z_5\) are independent \(N(0,1)\) variables. The coefficients are fixed and positive. This is a probability calculation, so no interest rate or mortality assumption applies.

Method. Find the MGF by multiplying the component MGFs, then differentiate or use the mean and variance of each scaled square. The distinction from an ordinary chi-square variable is that the squared terms do not all have coefficient one.

For each of the first two terms, \(M_{2Z_j^2}(t)=(1-4t)^{-1/2}\), with domain \(t<\tfrac14\). For each of the last three terms, \(M_{\frac12Z_j^2}(t)=(1-t)^{-1/2}\), with domain \(t<1\). The common domain is the intersection \(t<\tfrac14\) and \(t<1\), namely \(t<\tfrac14\). Hence \(M_W(t)=(1-4t)^{-1}(1-t)^{-3/2}\).

For a standard normal square, \(E[Z_j^2]=1\) and \(\operatorname{Var}(Z_j^2)=2\). Scaling a variable by \(a\) multiplies its mean by \(a\) and its variance by \(a^2\). Therefore, \(E[W]=2(1)+2(1)+3(\tfrac12)(1)=4+\tfrac32=5.5\). By independence, \(\operatorname{Var}(W)=2(2^2)+2(2^2)+3\{2(\tfrac12)^2\}=8+8+\tfrac32=17.5\).

It follows that \(E[W^2]=\operatorname{Var}(W)+\{E[W]\}^2=17.5+5.5^2=17.5+30.25=47.75\).

Check. The variance is \(34\) from the two coefficient-2 terms and \(1.5\) from the three coefficient-\(\tfrac12\) terms, for a total of \(35.5\)? This sum would be incorrect because each coefficient-2 term contributes \(2(2^2)=8\), so the two contribute \(16\), not \(34\). Rechecking directly: the first two terms contribute \(8+8=16\); the last three contribute \(0.5+0.5+0.5=1.5\). Thus the correct variance is \(17.5\), and \(E[W^2]=47.75\). Treating this weighted sum as an ordinary chi-square variable would also give the wrong moments.

Combining Independent Chi-Square Variables

The MGF also provides a direct verification for independent chi-square sums. Suppose \(A\sim\chi^2_r\) and \(B\sim\chi^2_s\) are independent. Multiplying their MGFs gives \((1-2t)^{-(r+s)/2}\), the MGF of a chi-square variable with \(r+s\) squared standard normal terms. This conclusion depends on independence; the product rule cannot be used for dependent variables without further information.

Worked Example: Combining Two Independent Chi-Square Quantities

Model and assumptions. In a hypothetical model, let \(A\sim\chi^2_3\) and \(B\sim\chi^2_4\) be independent, and define \(R=A+B\). Each variable is a sum of the indicated number of independent squared \(N(0,1)\) variables; all underlying normal variables are independent. No interest rate or mortality assumption applies.

Method. Multiply the MGFs to obtain the MGF of the sum, then use its derivatives for the requested moments.

The component MGFs are \(M_A(t)=(1-2t)^{-3/2}\) and \(M_B(t)=(1-2t)^{-4/2}\), each finite for \(t<\tfrac12\). Thus \(M_R(t)=M_A(t)M_B(t)=(1-2t)^{-7/2}\), for \(t<\tfrac12\). This is the MGF of a chi-square variable with seven squared standard normal terms.

The mean is \(E[R]=7\), and the variance is \(\operatorname{Var}(R)=2(7)=14\). Hence the second raw moment is \(E[R^2]=14+7^2=14+49=63\). The third raw moment is \(E[R^3]=7(7+2)(7+4)=7\cdot9\cdot11=693\).

Check. Independence also gives \(E[A+B]=3+4=7\) and \(\operatorname{Var}(A+B)=2(3)+2(4)=6+8=14\), confirming the MGF results. Adding the two means but failing to add the variances would miss the contribution from both independent components.

Common MGF and Moment Errors

Using the wrong MGF domain. For \(aZ^2\), the condition is \(at<\tfrac12\). In particular, the MGF of \(\tfrac12Z^2\) is \((1-t)^{-1/2}\), finite for \(t<1\), not merely for \(t<\tfrac12\). In a product, use the intersection of all component domains.
Confusing variance with the second raw moment. For \(T\sim\chi^2_\nu\), \(E[T^2]=\nu(\nu+2)\), while \(\operatorname{Var}(T)=2\nu\). The conversion is \(\operatorname{Var}(T)=E[T^2]-\{E[T]\}^2\).
Forgetting that weights are squared in the variance. If \(W=aX\), then \(\operatorname{Var}(W)=a^2\operatorname{Var}(X)\), not \(a\operatorname{Var}(X)\).
Applying independence without checking it. The variance of a sum includes covariance terms unless the variables are independent (or otherwise known to have zero covariance). The product of MGFs likewise requires independence.

Key takeaway: For \(T\sim\chi^2_\nu\), \(M_T(t)=(1-2t)^{-\nu/2}\) on \(t<\tfrac12\), \(E[T]=\nu\), and \(\operatorname{Var}(T)=2\nu\). Derivatives at zero give higher raw moments, and independent MGFs multiply on the intersection of their domains.

Check Your Understanding

For each problem, use the stated independence and distribution assumptions, and distinguish raw moments from variance.

Let \(T\sim\chi^2_8\). Find its mean, variance, and second raw moment. (Answer: mean \(8\), variance \(16\), second raw moment \(80\))
Let \(A\sim\chi^2_3\) and \(B\sim\chi^2_5\) be independent. Find the MGF, its domain, the mean, and the variance of \(A+B\). (Answer: \(M(t)=(1-2t)^{-4}\) for \(t<\tfrac12\); mean \(8\); variance \(16\))
Let \(W=3Z_1^2+\tfrac14Z_2^2+\tfrac14Z_3^2\), where the three variables are independent \(N(0,1)\). Find its MGF and domain, mean, variance, and second raw moment. (Answer: \(M_W(t)=(1-6t)^{-1/2}(1-\tfrac12t)^{-1}\) for \(t<\tfrac16\); mean \(3.5\); variance \(18.25\); second raw moment \(30.5\))
Let \(Q\sim\chi^2_4\). Use an MGF derivative formula to calculate \(E[Q^3]\), and also give \(E[Q]\) and \(\operatorname{Var}(Q)\). (Answer: \(E[Q^3]=4\cdot6\cdot8=192\); mean \(4\); variance \(8\))
Let \(V=\tfrac12\sum_{j=1}^{6}Z_j^2\), where the \(Z_j\) are independent \(N(0,1)\). Find its MGF and domain, then calculate its mean, variance, and second raw moment. (Answer: \(M_V(t)=(1-t)^{-3}\) for \(t<1\); mean \(3\); variance \(3\); second raw moment \(12\))

What You'll Learn