Moment-generating function

In probability theory and statistics, the moment-generating function of a random variable is an alternative specification of its probability distribution. Thus, it provides the basis of an alternative route to analytical results compared with working directly with probability density functions or cumulative distribution functions. There are particularly simple results for the moment-generating functions of distributions defined by the weighted sums of random variables. Note, however, that not all random variables have moment-generating functions.

In addition to univariate distributions, moment-generating functions can be defined for vector- or matrix-valued random variables, and can even be extended to more general cases.

The moment-generating function does not always exist even for real-valued arguments, unlike the characteristic function. There are relations between the behavior of the moment-generating function of a distribution and properties of the distribution, such as the existence of moments.

Definition

In probability theory and statistics, the moment-generating function of a random variable X is

$M_X(t) := \mathbb{E}\!\left[e^{tX}\right], \quad t \in \mathbb{R},$

wherever this expectation exists. In other terms, the moment-generating function can be interpreted as the expectation of the random variable $e^{tX}$.

$M_X(0)$ always exists and is equal to 1.

A key problem with moment-generating functions is that moments and the moment-generating function may not exist, as the integrals need not converge absolutely. By contrast, the characteristic function always exists (because it is the integral of a bounded function on a space of finite measure), and thus may be used instead.

More generally, where $\mathbf X = ( X_1, \ldots, X_n)$^T, an n-dimensional random vector, one uses $\mathbf t \cdot \mathbf X = \mathbf t^\mathrm T\mathbf X$ instead of tX:

$M_{\mathbf X}(\mathbf t) := \mathbb{E}\!\left(e^{\mathbf t^\mathrm T\mathbf X}\right).$

The reason for defining this function is that it can be used to find all the moments of the distribution. The series expansion of e^tX is:

$e^{t\,X} = 1 + t\,X + \frac{t^2\,X^2}{2!} + \frac{t^3\,X^3}{3!} + \cdots +\frac{t^n\,X^n}{n!} + \cdots.$

Hence:

$\begin{align} M_X(t) = \mathbb{E}(e^{t\,X}) &= 1 + t \,\mathbb{E}(X) + \frac{t^2 \,\mathbb{E}(X^2)}{2!} + \frac{t^3\,\mathbb{E}(X^3)}{3!}+\cdots + \frac{t^n\,\mathbb{E}(X^n)}{n!}+\cdots \\ & = 1 + tm_1 + \frac{t^2m_2}{2!} + \frac{t^3m_3}{3!}+\cdots + \frac{t^nm_n}{n!}+\cdots, \end{align}$

where m_n is the nth moment.

Differentiating M_X(t) i times with respect to t and setting t = 0 we hence obtain the ith moment about the origin, m_i, see Calculations of moments below.

Examples

Here are some examples of the moment generating function and the characteristic function for comparison. It can be seen that the characteristic function is a Wick rotation of the moment generating function Mx(t) when the latter exists.

Distribution	Moment-generating function MX(t)	Characteristic function φ(t)
Bernoulli $\, P(X=1)=p$	$\, 1-p+pe^t$	$\, 1-p+pe^{it}$
Geometric $(1 - p)^{k-1}\,p\!$	$\frac{p e^t}{1-(1-p) e^t}\!$  $\forall t<-\ln(1-p)\!$	$\frac{p e^{it}}{1-(1-p)\,e^{it}}\!$
Binomial B(n, p)	$\, (1-p+pe^t)^n$	$\, (1-p+pe^{it})^n$
Poisson Pois(λ)	$\, e^{\lambda(e^t-1)}$	$\, e^{\lambda(e^{it}-1)}$
Uniform (continuous) U(a, b)	$\, \frac{e^{tb} - e^{ta}}{t(b-a)}$	$\, \frac{e^{itb} - e^{ita}}{it(b-a)}$
Uniform (discrete) U(a, b)	$\, \frac{e^{at} - e^{(b+1)t}}{(b-a+1)(1-e^{t})}$	$\, \frac{e^{ait} - e^{(b+1)it}}{(b-a+1)(1-e^{it})}$
Normal N(μ, σ2)	$\, e^{t\mu + \frac{1}{2}\sigma^2t^2}$	$\, e^{it\mu - \frac{1}{2}\sigma^2t^2}$
Chi-squared χ2k	$\, (1 - 2t)^{-k/2}$	$\, (1 - 2it)^{-k/2}$
Gamma Γ(k, θ)	$\, (1 - t\theta)^{-k}$	$\, (1 - it\theta)^{-k}$
Exponential Exp(λ)	$\, (1-t\lambda^{-1})^{-1}, \, (t<\lambda)$	$\, (1 - it\lambda^{-1})^{-1}$
Multivariate normal N(μ, Σ)	$\, e^{t^\mathrm{T} \mu + \frac{1}{2} t^\mathrm{T} \Sigma t}$	$\, e^{i t^\mathrm{T} \mu - \frac{1}{2} t^\mathrm{T} \Sigma t}$
Degenerate δa	$\, e^{ta}$	$\, e^{ita}$
Laplace L(μ, b)	$\, \frac{e^{t\mu}}{1 - b^2t^2}$	$\, \frac{e^{it\mu}}{1 + b^2t^2}$
Negative Binomial NB(r, p)	$\, \frac{(1-p)^r}{(1-pe^t)^r}$	$\, \frac{(1-p)^r}{(1-pe^{it})^r}$
Cauchy Cauchy(μ, θ)	does not exist	$\, e^{it\mu -\theta|t|}$

Calculation

The moment-generating function is the expectation of a function of the random variable, it can be written as:

• In the general case: $M_X(t) = \int_{-\infty}^\infty e^{tx}\,dF(x)$, using the Riemann–Stieltjes integral, and where F is the cumulative distribution function.

• For a discrete probability mass function, $M_X(t)=\sum_{i=1}^\infty e^{tx_i}\, p_i$

• For a continuous probability density function, $M_X(t) = \int_{-\infty}^\infty e^{tx} f(x)\,dx$

Note that for the case where X has a continuous probability density function ƒ(x), M_X(−t) is the two-sided Laplace transform of ƒ(x).

$\begin{align} M_X(t) & = \int_{-\infty}^\infty e^{tx} f(x)\,dx \\ & = \int_{-\infty}^\infty \left( 1+ tx + \frac{t^2x^2}{2!} + \cdots + \frac{t^nx^n}{n!} + \cdots\right) f(x)\,dx \\ & = 1 + tm_1 + \frac{t^2m_2}{2!} +\cdots + \frac{t^nm_n}{n!} +\cdots, \end{align}$

where m_n is the nth moment.

Sum of independent random variables

If $S_n = \sum_{i=1}^{n} a_i X_i$, where the X_i are independent random variables and the a_i are constants, then the probability density function for S_n is the convolution of the probability density functions of each of the X_i, and the moment-generating function for S_n is given by

$M_{S_n}(t)=M_{X_1}(a_1t)M_{X_2}(a_2t)\cdots M_{X_n}(a_nt) \, .$

Vector-valued random variables

For vector-valued random variables X with real components, the moment-generating function is given by

$M_X(t) = E\left( e^{\langle t, X \rangle}\right)$

where t is a vector and $\langle \cdot, \cdot \rangle$ is the dot product.

Important properties

Moment generating functions are positive and log-convex, with M(0) = 1.

An important property of the moment-generating function is that if two distributions have the same moment-generating function, then they are identical at almost all points. That is, if for all values of t,

$M_X(t) = M_Y(t),\,$

then

$F_X(x) = F_Y(x) \,$

for all values of x (or equivalently X and Y have the same distribution). This statement is not equivalent to the statement "if two distributions have the same moments, then they are identical at all points." This is because in some cases, the moments exist and yet the moment-generating function does not, because the limit

$\lim_{n \rightarrow \infty} \sum_{i=0}^n \frac{t^im_i}{i!}$

may not exist. The lognormal distribution is an example of when this occurs.

Calculations of moments

The moment-generating function is so called because if it exists on an open interval around t = 0, then it is the exponential generating function of the moments of the probability distribution:

$m_n = E \left( X^n \right) = M_X^{(n)}(0) = \frac{d^n M_X}{dt^n}(0).$

Here n must be a nonnegative integer.

Other properties

Hoeffding's lemma provides a bound on the moment-generating function in the case of a zero-mean, bounded random variable.

Relation to other functions

Related to the moment-generating function are a number of other transforms that are common in probability theory:

characteristic function

The characteristic function $\varphi_X(t)$ is related to the moment-generating function via $\varphi_X(t) = M_{iX}(t) = M_X(it):$ the characteristic function is the moment-generating function of iX or the moment generating function of X evaluated on the imaginary axis. This function can also be viewed as the Fourier transform of the probability density function, which can therefore be deduced from it by inverse Fourier transform.

cumulant-generating function

The cumulant-generating function is defined as the logarithm of the moment-generating function; some instead define the cumulant-generating function as the logarithm of the characteristic function, while others call this latter the second cumulant-generating function.

probability-generating function

The probability-generating function is defined as $G(z) = E[z^X].\,$ This immediately implies that $G(e^t) = E[e^{tX}] = M_X(t).\,$

See also

• Factorial moment generating function
• Rate function
• Hamburger moment problem