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Moments and Generating Functions · Tutorial 513 of 994

Multivariate MGFs and Mixed Moments

Learn to define a multivariate MGF, determine when its mixed derivatives yield joint moments, and use those derivatives in concrete models.

Intermediate 10 min read

What You'll Learn

  • Define the joint MGF and its domain for a random vector.
  • State why finiteness near the origin permits differentiation under the expectation.
  • Recover mixed moments using multi-index partial derivatives.
  • Calculate mixed moments for discrete and continuous joint models.
  • Relate marginal and linear-transformation MGFs to the joint MGF.
  • Distinguish a mixed moment from a product of marginal moments.

Why Mixed Moments Need a Joint MGF

For a random vector, moments describe not only the behavior of each coordinate but also how coordinates vary together. For example, \(E[XY]\) is a joint, or mixed, moment: it involves both random variables at once. The joint moment-generating function packages these quantities into a function of several arguments, with a separate argument for each coordinate.

The key condition is that the MGF be finite in a neighborhood of the origin. Under that condition, its partial derivatives can be brought inside the expectation, and each derivative contributes a factor from the corresponding coordinate. This makes the derivative order and the exponents in the mixed moment match exactly.

Definition and Mixed-Moment Formula

Definition: Let \(X=(X_1,\ldots,X_d)^{\mathsf T}\) be a random vector with values in \(\mathbb{R}^d\). Its joint moment-generating function is
$$ M_X(t)=E\!\left[\exp(t^{\mathsf T}X)\right] =E\!\left[\exp\!\left(\sum_{j=1}^d t_jX_j\right)\right], \qquad t=(t_1,\ldots,t_d)^{\mathsf T}\in\mathbb{R}^d, $$
where the expectation is allowed to be \(+\infty\). The domain of \(M_X\) is the set of \(t\) for which this expectation is finite. The joint MGF is finite in a neighborhood of the origin if this domain contains an open set around \(0\).

For a multi-index \(\alpha=(\alpha_1,\ldots,\alpha_d)\) of nonnegative integers, write \(|\alpha|=\sum_{j=1}^d\alpha_j\), and use \(\partial^\alpha=\partial_{t_1}^{\alpha_1}\cdots\partial_{t_d}^{\alpha_d}\). The matching mixed moment is \(E[\prod_{j=1}^d X_j^{\alpha_j}]\). An exponent of zero contributes a factor of one.

Theorem: If \(M_X(t)\) is finite in an open neighborhood of the origin, then every mixed absolute moment \(E[\prod_{j=1}^d |X_j|^{\alpha_j}]\) is finite. Moreover, \(M_X\) has continuous partial derivatives of every order in some neighborhood of the origin, and
$$ \left.\partial^\alpha M_X(t)\right|_{t=0} =E\!\left[\prod_{j=1}^d X_j^{\alpha_j}\right]. $$

Proof. Choose \(b>0\) so that the closed cube \([-b,b]^d\) is inside a neighborhood on which \(M_X\) is finite. For each possible sign vector \(\varepsilon\in\{-1,1\}^d\), the value \(M_X(b\varepsilon)\) is finite. Since \[ \exp\!\left(b\sum_{j=1}^d|X_j|\right) =\max_{\varepsilon\in\{-1,1\}^d}\exp(b\varepsilon^{\mathsf T}X) \leq \sum_{\varepsilon\in\{-1,1\}^d}\exp(b\varepsilon^{\mathsf T}X), \] taking expectations shows that \(E[\exp(b\sum_j|X_j|)]<\infty\).

Fix \(a\) with \(0<a<b\). For \(\|t\|_\infty\leq a\), where \(\|t\|_\infty=\max_j|t_j|\), the absolute value of the formal derivative integrand is bounded by \[ \left|\prod_{j=1}^d X_j^{\alpha_j}\exp(t^{\mathsf T}X)\right| \leq \left(\sum_{j=1}^d|X_j|\right)^{|\alpha|} \exp\!\left(a\sum_{j=1}^d|X_j|\right). \] The right side is at most \(C\exp(b\sum_j|X_j|)\) for a finite constant \(C\): the function \(u^{|\alpha|}e^{-(b-a)u}\) is bounded for \(u\geq0\). This bound is integrable by the preceding paragraph. In particular it proves finiteness of the mixed absolute moments.

To justify differentiation, take a smaller cube \(\|t\|_\infty\leq a\) and choose \(\delta>0\) with \(a+\delta<b\). A difference quotient for one more derivative in coordinate \(j\) introduces the factor \((e^{hX_j}-1)/h\). For \(0<|h|\leq\delta\), the mean value theorem gives \[ \left|\frac{e^{hX_j}-1}{h}\right| \leq |X_j|e^{|h||X_j|}. \] Thus the difference quotient is dominated by a polynomial in \(\sum_j|X_j|\) times \(\exp((a+\delta)\sum_j|X_j|)\), which is in turn bounded by a constant times \(\exp(b\sum_j|X_j|)\). Dominated convergence permits the derivative to pass inside the expectation. Repeating this argument for each coordinate and each finite derivative order gives \[ \partial^\alpha M_X(t) =E\!\left[\prod_{j=1}^d X_j^{\alpha_j}\exp(t^{\mathsf T}X)\right] \] on a neighborhood of the origin. Setting \(t=0\) proves the formula. The same integrable bounds and dominated convergence give continuity of these derivatives. \(\square\)

Calculating Mixed Moments

Worked Example: A Bivariate Discrete Distribution

Suppose \((X,Y)\) takes the values \((0,0),(1,0),(0,1),(1,1)\) with respective probabilities \(0.2,0.3,0.1,0.4\). Find \(E[XY]\) from the joint MGF, then calculate \(\operatorname{Cov}(X,Y)\).

The joint MGF, for real \(s,t\), is the finite sum \[ M_{X,Y}(s,t)=0.2+0.3e^s+0.1e^t+0.4e^{s+t}. \] Differentiating once in each argument gives \[ \frac{\partial^2 M_{X,Y}}{\partial s\,\partial t} =0.4e^{s+t}, \qquad E[XY]=\left.\frac{\partial^2 M_{X,Y}}{\partial s\,\partial t}\right|_{s=t=0}=0.4. \] The marginal means are \(E[X]=P(X=1)=0.3+0.4=0.7\) and \(E[Y]=P(Y=1)=0.1+0.4=0.5\). Therefore \[ \operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y] =0.4-(0.7)(0.5)=0.05. \] In this example the mixed moment is not the product of the two means.

Worked Example: A Mixed Moment of Independent Normals

Let \(X\sim N(1,4)\) and \(Y\sim N(-1,1)\) be independent. Find \(E[X^2Y]\) by differentiating their joint MGF.

For a normal random variable with mean \(\mu\) and variance \(\sigma^2\), the MGF is \(\exp(\mu t+\sigma^2t^2/2)\). Independence therefore gives \[ M_{X,Y}(s,t)=\exp(s+2s^2)\exp(-t+t^2/2). \] The required derivative is \(\partial_s^2\partial_t M_{X,Y}(0,0)\). First, \[ \left.\frac{\partial^2}{\partial s^2}\exp(s+2s^2)\right|_{s=0} = \left.\left((1+4s)^2+4\right)\exp(s+2s^2)\right|_{s=0}=5, \] and \[ \left.\frac{\partial}{\partial t}\exp(-t+t^2/2)\right|_{t=0} =\left.(-1+t)\exp(-t+t^2/2)\right|_{t=0}=-1. \] The factors separate, so \(\partial_s^2\partial_t M_{X,Y}(0,0)=5(-1)=-5\). Hence \(E[X^2Y]=-5\). This agrees with independence: \(E[X^2]E[Y]=((1)^2+4)(-1)=-5\).

Worked Example: A Dependent Pair Built From One Exponential Variable

Let \(U\) have an exponential distribution with rate \(1\), and define \(X=U\), \(Y=2U\). Find \(E[X^2Y]\) by using the joint MGF.

For \(s+2t<1\), the joint MGF is \[ M_{X,Y}(s,t)=E[e^{sU+2tU}] =\int_0^\infty e^{(s+2t)u}e^{-u}\,du =\frac{1}{1-s-2t}. \] The domain includes an open neighborhood of \((0,0)\). Differentiating first with respect to \(t\), and then twice with respect to \(s\), gives \[ \frac{\partial}{\partial t}M_{X,Y}(s,t) =2(1-s-2t)^{-2}, \] \[ \frac{\partial^2}{\partial s^2}\frac{\partial}{\partial t}M_{X,Y}(s,t) =12(1-s-2t)^{-4}. \] Consequently \(E[X^2Y]=\partial_s^2\partial_t M_{X,Y}(0,0)=12\). Directly, \(X^2Y=2U^3\), and the rate-\(1\) exponential has \(E[U^3]=3!=6\), also giving \(12\).

Marginals and Linear Transformations

Result: If \(X\) has a joint MGF finite near the origin and \(A\) is a fixed \(r\times d\) real matrix, then the random vector \(Y=AX\) has joint MGF
$$ M_Y(u)=M_X(A^{\mathsf T}u) $$
for \(u\) in a neighborhood of \(0\). In particular, setting all but one coordinate of \(t\) to zero in \(M_X(t)\) gives the MGF of the corresponding marginal coordinate.

Proof. By the definition of \(Y\), \[ M_Y(u)=E[\exp(u^{\mathsf T}AX)] =E[\exp((A^{\mathsf T}u)^{\mathsf T}X)] =M_X(A^{\mathsf T}u). \] Because the linear map \(u\mapsto A^{\mathsf T}u\) is continuous and sends \(0\) to \(0\), all sufficiently small \(u\) map into a neighborhood on which \(M_X\) is finite. Thus the displayed identity holds with finite values near the origin. Taking \(A\) to select a single coordinate gives the marginal statement. \(\square\)

This identity makes the joint MGF useful for studying sums and other linear combinations of coordinates. For example, if \(Y=X_1+\cdots+X_d\), then \(M_Y(u)=M_X(u,\ldots,u)\) wherever the right side is finite. This substitution gives the MGF of the sum; it does not imply that the coordinates are independent.

Interpretation and a Common Pitfall

Mixed derivatives encode joint products. For instance, if \(X=(X_1,X_2,X_3)\), then \(\partial_{t_1}^2\partial_{t_3}M_X(0)\) is \(E[X_1^2X_3]\), while \(\partial_{t_1}\partial_{t_2}M_X(0)\) is \(E[X_1X_2]\). Repeated differentiation with respect to one coordinate creates a repeated power of that coordinate.

  • Incorrect: Replace \(E[X_1X_2]\) by \(E[X_1]E[X_2]\) without establishing independence.
  • Correction: The mixed derivative gives \(E[X_1X_2]\). Independence is a separate condition under which the expectation factors into marginal expectations.
  • Incorrect: Differentiate a joint MGF at the origin merely because the formula is finite at the origin.
  • Correction: Finiteness at the origin alone is automatic, since \(M_X(0)=1\). The differentiation result requires finiteness in a neighborhood of the origin.

The neighborhood condition is substantive: it supplies exponential integrability that dominates the polynomial factors created by differentiation. When it holds, the joint MGF provides a systematic way to compute mixed moments and to check the dependence structure represented by them.

Check Your Understanding

Use the definitions and derivative rule to answer the following.

  1. For a random vector \((X,Y)\), which derivative of the joint MGF at the origin gives \(E[X^3Y^2]\)?
  2. Suppose \(M_{X,Y}(s,t)=\exp(s+t+s^2+t^2)\). Use derivatives at the origin to find \(E[XY]\).
  3. Why does finiteness of \(M_X(0)\) alone not justify recovering moments by differentiation?
  4. If \(Y=3X_1-X_2\), express \(M_Y(u)\) in terms of the joint MGF of \((X_1,X_2)\).
  5. What additional condition, beyond existence of the mixed moment, allows \(E[X_1X_2]\) to be replaced by \(E[X_1]E[X_2]\)?