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Moment-generating function

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Concept in probability theory and statistics

Inprobability theory andstatistics, themoment-generating function of a real-valuedrandom variable is an alternative specification of itsprobability distribution. Thus, it provides the basis of an alternative route to analytical results compared with working directly withprobability density functions orcumulative distribution functions. There are particularly simple results for the moment-generating functions of distributions defined by the weighted sums of random variables. However, not all random variables have moment-generating functions.

As its name implies, the moment-generating function can be used to compute a distribution’smoments: then-th moment about 0 is then-th derivative of the moment-generating function, evaluated at 0.

In addition to real-valued distributions (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 of a real-valued distribution does not always exist, unlike thecharacteristic 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

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Let $X {\displaystyle X}$ be arandom variable withCDF $F_{X}$ . The moment generating function (mgf) of $X {\displaystyle X}$ (or $F_{X}$ ), denoted by $M_{X}(t)$ , is

$M_{X}(t)=\operatorname {E} \left[e^{tX}\right]$

provided thisexpectation exists for $t {\displaystyle t}$ in some openneighborhood of 0. That is, there is an $h>0$ such that for all $t {\displaystyle t}$ in $-h<t<h$ , $\operatorname {E} \left[e^{tX}\right]$ exists. If the expectation does not exist in an open neighborhood of 0, we say that the moment generating function does not exist.^[1]

In other words, the moment-generating function ofX is theexpectation of the random variable $e^{tX}$ . More generally, when $\mathbf {X} =(X_{1},\ldots ,X_{n})^{\mathrm {T} }$ , an $n {\displaystyle n}$ -dimensionalrandom vector, and $\mathbf {t}$ is a fixed vector, one uses $\mathbf {t} \cdot \mathbf {X} =\mathbf {t} ^{\mathrm {T} }\mathbf {X}$ instead of $t X {\displaystyle tX}$ : $M_{\mathbf {X} }(\mathbf {t} ):=\operatorname {E} \left[e^{\mathbf {t} ^{\mathrm {T} }\mathbf {X} }\right].$

$M_{X}(0)$ always exists and is equal to 1. However, 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, thecharacteristic function or Fourier transform always exists (because it is the integral of a bounded function on a space of finitemeasure), and for some purposes may be used instead.

The moment-generating function is so named because it can be used to find the moments of the distribution.^[2] The series expansion of $e^{tX}$ is

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

Hence, ${\begin{aligned}M_{X}(t)&=\operatorname {E} [e^{tX}]\\[1ex]&=1+t\operatorname {E} [X]+{\frac {t^{2}\operatorname {E} [X^{2}]}{2!}}+{\frac {t^{3}\operatorname {E} [X^{3}]}{3!}}+\cdots +{\frac {t^{n}\operatorname {E} [X^{n}]}{n!}}+\cdots \\[1ex]&=1+tm_{1}+{\frac {t^{2}m_{2}}{2!}}+{\frac {t^{3}m_{3}}{3!}}+\cdots +{\frac {t^{n}m_{n}}{n!}}+\cdots ,\end{aligned}}$

where $m_{n}$ is the $n {\displaystyle n}$ -thmoment. Differentiating $M_{X}(t)$ $i {\displaystyle i}$ times with respect to $t {\displaystyle t}$ and setting $t=0$ , we obtain the $i {\displaystyle i}$ -th moment about the origin, $m_{i}$ ; see§ Calculations of moments below.

If $X {\displaystyle X}$ is a continuous random variable, the following relation between its moment-generating function $M_{X}(t)$ and thetwo-sided Laplace transform of its probability density function $f_{X}(x)$ holds:

$M_{X}(t)={\mathcal {L}}\{f_{X}\}(-t),$

since the PDF's two-sided Laplace transform is given as

${\mathcal {L}}\{f_{X}\}(s)=\int _{-\infty }^{\infty }e^{-sx}f_{X}(x)\,dx,$

and the moment-generating function's definition expands (by thelaw of the unconscious statistician) to $M_{X}(t)=\operatorname {E} \left[e^{tX}\right]=\int _{-\infty }^{\infty }e^{tx}f_{X}(x)\,dx.$

This is consistent with the characteristic function of $X {\displaystyle X}$ being aWick rotation of $M_{X}(t)$ when the moment generating function exists, as the characteristic function of a continuous random variable $X {\displaystyle X}$ is theFourier transform of its probability density function $f_{X}(x)$ , and in general when a function $f(x)$ is ofexponential order, the Fourier transform of $f {\displaystyle f}$ is a Wick rotation of its two-sided Laplace transform in the region of convergence. Seethe relation of the Fourier and Laplace transforms for further information.

Examples

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Here are some examples of the moment-generating function and the characteristic function for comparison. It can be seen that the characteristic function is aWick rotation of the moment-generating function $M_{X}(t)$ when the latter exists.

Distribution	Moment-generating function $M_{X}(t)$	Characteristic function $\varphi (t)$
Degenerate $\delta _{a}$	$e^{ta}$	$e^{ita}$
Bernoulli $P(X=1)=p$	$1-p+pe^{t}$	$1-p+pe^{it}$
Binomial $B(n,p)$	$\left(1-p+pe^{t}\right)^{n}$	$\left(1-p+pe^{it}\right)^{n}$
Geometric $(1-p)^{k}\,p$	${\frac {p}{1-(1-p)e^{t}}},~t<-\ln(1-p)$	${\frac {p}{1-(1-p)\,e^{it}}}$
Negative binomial $\operatorname {NB} (r,p)$	$\left({\frac {p}{1-e^{t}+pe^{t}}}\right)^{r},~t<-\ln(1-p)$	$\left({\frac {p}{1-e^{it}+pe^{it}}}\right)^{r}$
Poisson $\operatorname {Pois} (\lambda )$	$e^{\lambda (e^{t}-1)}$	$e^{\lambda (e^{it}-1)}$
Uniform (continuous) $\operatorname {U} (a,b)$	${\frac {e^{tb}-e^{ta}}{t(b-a)}}$	${\frac {e^{itb}-e^{ita}}{it(b-a)}}$
Uniform (discrete) $\operatorname {DU} (a,b)$	${\frac {e^{at}-e^{(b+1)t}}{(b-a+1)(1-e^{t})}}$	${\frac {e^{ait}-e^{(b+1)it}}{(b-a+1)(1-e^{it})}}$
Laplace $L(\mu ,b)$	${\frac {e^{t\mu }}{1-b^{2}t^{2}}},~\|t\|<1/b$	${\frac {e^{it\mu }}{1+b^{2}t^{2}}}$
Normal $N(\mu ,\sigma ^{2})$	$e^{t\mu +\sigma ^{2}t^{2}/2}$	$e^{it\mu -\sigma ^{2}t^{2}/2}$
Chi-squared $\chi _{k}^{2}$	${\left(1-2t\right)}^{-k/2},~t<1/2$	${\left(1-2it\right)}^{-{k}/{2}}$
Noncentral chi-squared $\chi _{k}^{2}(\lambda )$	$e^{\lambda t/(1-2t)}{\left(1-2t\right)}^{-k/2}$	$e^{i\lambda t/(1-2it)}{\left(1-2it\right)}^{-k/2}$
Gamma $\Gamma (k,{\tfrac {1}{\theta }})$	${\left(1-t\theta \right)}^{-k},~t<{\tfrac {1}{\theta }}$	${\left(1-it\theta \right)}^{-k}$
Exponential $\operatorname {Exp} (\lambda )$	$\left(1-t\lambda ^{-1}\right)^{-1},~t<\lambda$	$\left(1-it\lambda ^{-1}\right)^{-1}$
Beta	$1+\sum _{k=1}^{\infty }\left(\prod _{r=0}^{k-1}{\frac {\alpha +r}{\alpha +\beta +r}}\right){\frac {t^{k}}{k!}}$	${}_{1}F_{1}(\alpha ;\alpha +\beta ;i\,t)\!$ (seeConfluent hypergeometric function)
Multivariate normal $N(\mathbf {\mu } ,\mathbf {\Sigma } )$	$\exp \left[\mathbf {t} ^{\mathrm {T} }\left({\boldsymbol {\mu }}+{\tfrac {1}{2}}{\boldsymbol {\Sigma }}\mathbf {t} \right)\right]$	$\exp \left[\mathbf {t} ^{\mathrm {T} }\left(i{\boldsymbol {\mu }}-{\tfrac {1}{2}}{\boldsymbol {\Sigma }}\mathbf {t} \right)\right]$
Cauchy $\operatorname {Cauchy} (\mu ,\theta )$	Does not exist	$e^{it\mu -\theta \|t\|}$
Multivariate Cauchy $\operatorname {MultiCauchy} (\mu ,\Sigma )$ ^[3]	Does not exist	$\exp \left(i\mathbf {t} ^{\mathrm {T} }{\boldsymbol {\mu }}-{\sqrt {\mathbf {t} ^{\mathrm {T} }{\boldsymbol {\Sigma }}\mathbf {t} }}\right)$

Calculation

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

For a discreteprobability mass function, $M_{X}(t)=\sum _{i=0}^{\infty }e^{tx_{i}}\,p_{i}$
For a continuousprobability density function, $M_{X}(t)=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx$
In the general case: $M_{X}(t)=\int _{-\infty }^{\infty }e^{tx}\,dF(x)$ , using theRiemann–Stieltjes integral, and where $F {\displaystyle F}$ is thecumulative distribution function. This is simply theLaplace-Stieltjes transform of $F {\displaystyle F}$ , but with the sign of the argument reversed.

Note that for the case where $X {\displaystyle X}$ has a continuousprobability density function $f(x)$ , $M_{X}(-t)$ is thetwo-sided Laplace transform of $f(x)$ .

${\begin{aligned}M_{X}(t)&=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx\\[1ex]&=\int _{-\infty }^{\infty }\left(1+tx+{\frac {t^{2}x^{2}}{2!}}+\cdots +{\frac {t^{n}x^{n}}{n!}}+\cdots \right)f(x)\,dx\\[1ex]&=1+tm_{1}+{\frac {t^{2}m_{2}}{2!}}+\cdots +{\frac {t^{n}m_{n}}{n!}}+\cdots ,\end{aligned}}$

where $m_{n}$ is the $n {\displaystyle n}$ thmoment.

Linear transformations of random variables

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If random variable $X {\displaystyle X}$ has moment generating function $M_{X}(t)$ , then $\alpha X+\beta$ has moment generating function $M_{\alpha X+\beta }(t)=e^{\beta t}M_{X}(\alpha t)$

$M_{\alpha X+\beta }(t)=\operatorname {E} \left[e^{(\alpha X+\beta )t}\right]=e^{\beta t}\operatorname {E} \left[e^{\alpha Xt}\right]=e^{\beta t}M_{X}(\alpha t)$

Linear combination of independent random variables

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If ${\textstyle S_{n}=\sum _{i=1}^{n}a_{i}X_{i}}$ , where theX_i are independent random variables and thea_i are constants, then the probability density function forS_n is theconvolution of the probability density functions of each of theX_i, and the moment-generating function forS_n is given by

$M_{S_{n}}(t)=M_{X_{1}}(a_{1}t)M_{X_{2}}(a_{2}t)\cdots M_{X_{n}}(a_{n}t)\,.$

Vector-valued random variables

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Forvector-valued random variables $\mathbf {X}$ withreal components, the moment-generating function is given by

$M_{X}(\mathbf {t} )=\operatorname {E} \left[e^{\langle \mathbf {t} ,\mathbf {X} \rangle }\right]$

where $\mathbf {t}$ is a vector and $\langle \cdot ,\cdot \rangle$ is thedot product.

Important properties

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Moment generating functions are positive andlog-convex,^{[citation needed]} withM(0) = 1.

An important property of the moment-generating function is that it uniquely determines the distribution. In other words, if $X {\displaystyle X}$ and $Y {\displaystyle Y}$ are two random variables and for all values of t,

$M_{X}(t)=M_{Y}(t),$ then $F_{X}(x)=F_{Y}(x)$

for all values ofx (or equivalentlyX andY 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

$\lim _{n\to \infty }\sum _{i=0}^{n}{\frac {t^{i}m_{i}}{i!}}$

may not exist. Thelog-normal distribution is an example of when this occurs.

Calculations of moments

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The moment-generating function is so called because if it exists on an open interval aroundt = 0, then it is theexponential generating function of themoments of theprobability distribution:

$m_{n}=\operatorname {E} \left[X^{n}\right]=M_{X}^{(n)}(0)=\left.{\frac {d^{n}M_{X}}{dt^{n}}}\right|_{t=0}.$

That is, withn being a nonnegative integer, then-th moment about 0 is then-th derivative of the moment generating function, evaluated att = 0.

Other properties

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Jensen's inequality provides a simple lower bound on the moment-generating function: $M_{X}(t)\geq e^{\mu t},$ where $\mu$ is the mean ofX.

The moment-generating function can be used in conjunction withMarkov's inequality to bound the upper tail of a real random variableX. This statement is also called theChernoff bound. Since $x\mapsto e^{xt}$ is monotonically increasing for $t>0$ , we have $\Pr(X\geq a)=\Pr(e^{tX}\geq e^{ta})\leq e^{-at}\operatorname {E} \left[e^{tX}\right]=e^{-at}M_{X}(t)$ for any $t>0$ and anya, provided $M_{X}(t)$ exists. For example, whenX is a standard normal distribution and $a>0$ , we can choose $t=a$ and recall that $M_{X}(t)=e^{t^{2}/2}$ . This gives $\Pr(X\geq a)\leq e^{-a^{2}/2}$ , which is within a factor of1+a of the exact value.

Various lemmas, such asHoeffding's lemma orBennett's inequality provide bounds on the moment-generating function in the case of a zero-mean, bounded random variable.

When $X {\displaystyle X}$ is non-negative, the moment generating function gives a simple, useful bound on the moments: $\operatorname {E} [X^{m}]\leq \left({\frac {m}{te}}\right)^{m}M_{X}(t),$ For any $X,m\geq 0$ and $t>0$ .

This follows from the inequality $1+x\leq e^{x}$ into which we can substitute $x'=tx/m-1$ implies $tx/m\leq e^{tx/m-1}$ for any $x,t,m\in \mathbb {R}$ .Now, if $t>0$ and $x,m\geq 0$ , this can be rearranged to $x^{m}\leq (m/(te))^{m}e^{tx}$ .Taking the expectation on both sides gives the bound on $\operatorname {E} [X^{m}]$ in terms of $\operatorname {E} [e^{tX}]$ .

As an example, consider $X\sim {\text{Chi-Squared}}$ with $k {\displaystyle k}$ degrees of freedom. Then from theexamples $M_{X}(t)=(1-2t)^{-k/2}$ .Picking $t=m/(2m+k)$ and substituting into the bound: $\operatorname {E} [X^{m}]\leq {\left(1+2m/k\right)}^{k/2}e^{-m}{\left(k+2m\right)}^{m}.$ We know thatin this case the correct bound is $\operatorname {E} [X^{m}]\leq 2^{m}\Gamma (m+k/2)/\Gamma (k/2)$ .To compare the bounds, we can consider the asymptotics for large $k {\displaystyle k}$ .Here the moment-generating function bound is $k^{m}(1+m^{2}/k+O(1/k^{2}))$ ,where the real bound is $k^{m}(1+(m^{2}-m)/k+O(1/k^{2}))$ .The moment-generating function bound is thus very strong in this case.

Relation to other functions

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

Characteristic function: Thecharacteristic 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 ofiX or the moment generating function ofX evaluated on the imaginary axis. This function can also be viewed as theFourier transform of theprobability density function, which can therefore be deduced from it by inverse Fourier transform.
Cumulant-generating function: Thecumulant-generating function is defined as the logarithm of the moment-generating function; some instead define the cumulant-generating function as the logarithm of thecharacteristic function, while others call this latter thesecond cumulant-generating function.
Probability-generating function: Theprobability-generating function is defined as $G(z)=\operatorname {E} \left[z^{X}\right].$ This immediately implies that $G(e^{t})=\operatorname {E} \left[e^{tX}\right]=M_{X}(t).$

References

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Citations

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^Casella, George; Berger, Roger L. (1990).Statistical Inference. Wadsworth & Brooks/Cole. p. 61.ISBN 0-534-11958-1.
^Bulmer, M. G. (1979).Principles of Statistics. Dover. pp. 75–79.ISBN 0-486-63760-3.
^Kotz et al.^{[full citation needed]} p. 37 using 1 as the number of degree of freedom to recover the Cauchy distribution

Sources

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Casella, George; Berger, Roger (2002).Statistical Inference (2nd ed.). Thomson Learning. pp. 59–68.ISBN 978-0-534-24312-8.

v t e Theory ofprobability distributions
probability mass function (pmf) probability density function (pdf) cumulative distribution function (cdf) quantile function
raw moment central moment mean variance standard deviation skewness kurtosis L-moment
moment-generating function (mgf) characteristic function probability-generating function (pgf) cumulant combinant