The MGF of a Continuous Uniform Variable

Why the Continuous Uniform MGF Has a Special Case at Zero

A continuous uniform model assigns equal density to every point in a finite interval. Its MGF can be derived directly from the density, producing a compact formula that is useful for evaluating exponential expectations and extracting moments. One detail requires care: the formula for nonzero $t$ has $t$ in its denominator, so its value at zero must be established separately.

This tutorial considers $X\sim\operatorname{Uniform}(a,b)$, where $a<b$. The interval endpoints may be positive, negative, or on opposite sides of zero. Because the interval is finite, the MGF exists for every real value of $t$.

Definition: If $X\sim\operatorname{Uniform}(a,b)$, its density is $f_X(x)=1/(b-a)$ for $a\leq x\leq b$, and $f_X(x)=0$ otherwise. Its MGF is $M_X(t)=E[e^{tX}]$. For this distribution, $$ M_X(t)= \begin{cases} \dfrac{e^{bt}-e^{at}}{(b-a)t}, & t\ne 0,\\[6pt] 1, & t=0. \end{cases} $$

Deriving the MGF from the Density

For a continuous random variable, the expectation of a function $g(X)$ is the integral of $g(x)f_X(x)$ over the support, provided the integral exists. Here $g(x)=e^{tx}$. Since $X$ is bounded, $e^{tX}$ is bounded for each fixed real $t$, so the expectation is finite.

\begin{aligned} M_X(t) &=E[e^{tX}]\\ &=\int_a^b e^{tx}\frac{1}{b-a}\,dx. \end{aligned}

When $t\ne 0$, an antiderivative of $e^{tx}$ is $e^{tx}/t$. Evaluating the integral at the endpoints gives

\begin{aligned} M_X(t) &=\frac{1}{b-a}\left[\frac{e^{tx}}{t}\right]_{a}^{b}\\ &=\frac{e^{bt}-e^{at}}{(b-a)t}, \qquad t\ne 0. \end{aligned}

At $t=0$, it is simplest to return to the expectation: $e^{0X}=1$, so $M_X(0)=E[1]=1$. The nonzero-$t$ expression also approaches 1 as $t$ approaches zero. Both the direct expectation and the limit are useful checks.

\lim_{t\to 0}\frac{e^{bt}-e^{at}}{(b-a)t} = \lim_{t\to 0} \frac{b e^{bt}-a e^{at}}{b-a} = \frac{b-a}{b-a} =1.

The second equality follows from l’Hôpital’s rule: both numerator and denominator in the original ratio approach zero. Thus the MGF is continuous at zero, even though its displayed nonzero-$t$ expression cannot be evaluated there by direct substitution.

Worked Example: Deriving the MGF and Moments on a Positive Interval

Suppose a hypothetical service duration $X$, measured in hours, is modeled as uniform on $[2,10]$. Find its MGF, mean, and variance. Use the limiting value at $t=0$ to check the MGF.

Method. Identify $a=2$, $b=10$, and $b-a=8$, then substitute into the uniform MGF. The parameter $t$ is measured per hour.

M_X(t)= \begin{cases} \dfrac{e^{10t}-e^{2t}}{8t}, & t\ne 0,\\[6pt] 1, & t=0. \end{cases}

To find the moments, expand the exponentials near zero, or use the general expansion derived below. The coefficient of $t$ is $E[X]$, and twice the coefficient of $t^2$ is $E[X^2]$:

M_X(t) =1+\frac{2+10}{2}t +\frac{2^2+2(10)+10^2}{6}t^2+\cdots =1+6t+22t^2+\cdots.

Therefore $E[X]=6$ hours and $E[X^2]=2(22)=44$ hours squared. The variance is

\operatorname{Var}(X)=E[X^2]-(E[X])^2=44-6^2=8\text{ hours}^2.

As a check, the direct uniform formulas give mean $(2+10)/2=6$ and variance $(10-2)^2/12=64/12=16/3$. These variance values do not agree—so the expansion calculation needs checking. In fact, the $t^2$ coefficient is $(a^2+ab+b^2)/6=(4+20+100)/6=124/6=62/3$, not $22$. Thus $E[X^2]=124/3$, and

\operatorname{Var}(X)=\frac{124}{3}-36=\frac{16}{3}\text{ hours}^2.

The check catches an arithmetic error before it becomes a final answer. The MGF at zero is also consistent: the limit of $(e^{10t}-e^{2t})/(8t)$ is $(10-2)/8=1$. A common mistake is to substitute $t=0$ into the quotient and treat the resulting $0/0$ as the MGF value; the correct value is the limit, which is 1.

Getting the Mean and Variance from the MGF

The MGF derivatives at zero give the raw moments: $M_X'(0)=E[X]$ and $M_X''(0)=E[X^2]$. These identities follow by differentiating $e^{tX}$ with respect to $t$, then evaluating at zero. For a bounded $X$, the relevant expectations are finite. Equivalently, expand the exponential inside the defining integral.

For the uniform variable, the expansion makes the moments especially transparent. Using $e^{tx}=1+tx+t^2x^2/2+\cdots$ in the integral gives

\begin{aligned} M_X(t) &=\frac{1}{b-a}\int_a^b \left(1+tx+\frac{t^2x^2}{2}+\cdots\right)\,dx\\ &=1+t\frac{1}{b-a}\int_a^b x\,dx +\frac{t^2}{2}\frac{1}{b-a}\int_a^b x^2\,dx+\cdots. \end{aligned}

Evaluating the two integrals yields $E[X]=(a+b)/2$ and $E[X^2]=(a^2+ab+b^2)/3$. The variance is the second raw moment minus the square of the mean:

\begin{aligned} \operatorname{Var}(X) &=\frac{a^2+ab+b^2}{3}-\left(\frac{a+b}{2}\right)^2\\ &=\frac{4a^2+4ab+4b^2-3a^2-6ab-3b^2}{12}\\ &=\frac{(b-a)^2}{12}. \end{aligned}

This derivation also explains why the interval’s location affects the mean but not the variance: shifting both endpoints by the same amount shifts the center, while the spread depends only on the interval length $b-a$.

Theorem: If $X\sim\operatorname{Uniform}(a,b)$ with $a<b$, then $$ E[X]=\frac{a+b}{2},\qquad E[X^2]=\frac{a^2+ab+b^2}{3},\qquad \operatorname{Var}(X)=\frac{(b-a)^2}{12}. $$ These results hold for any finite real endpoints $a<b$. The MGF formula for $t\ne 0$ must be supplemented by $M_X(0)=1$.

Worked Example: Evaluating the MGF at a Positive Argument

A hypothetical processing time $T$, in hours, is modeled as uniform on $[0,2]$. Find $M_T(1)$, $E[T]$, and $\operatorname{Var}(T)$. Here the MGF argument $t=1$ is per hour.

Method. Substitute $a=0$, $b=2$, and $t=1$ into the nonzero-$t$ formula:

M_T(1)=\frac{e^{2}-e^{0}}{(2-0)(1)} =\frac{e^2-1}{2} =\frac{7.389056-1}{2} =3.194528.

Thus $M_T(1)=3.1945$ to four decimal places. The moments are

E[T]=\frac{0+2}{2}=1\text{ hour},\qquad \operatorname{Var}(T)=\frac{(2-0)^2}{12}=\frac{1}{3}\text{ hours}^2.

As a basic check, $M_T(0)=1$, while $M_T(1)>1$, as expected because $e^T\geq 1$ on this interval. The common mistake here is to use the interval width alone in the exponent; the endpoint terms are $e^{bt}$ and $e^{at}$, so both endpoints must be substituted.

Negative Arguments and Interval Location

The MGF parameter $t$ need not be positive. A negative argument weights smaller values of $X$ more heavily through $e^{tX}$; it does not change the distribution of $X$. The same formula applies for every nonzero real $t$, and the quotient remains positive because numerator and denominator have matching signs.

Worked Example: A Negative MGF Argument on a Shifted Interval

A hypothetical completion time $T$, in hours, is modeled as uniform on $[4,12]$. Find $M_T(-0.1)$, $E[T]$, and $\operatorname{Var}(T)$. The argument $-0.1$ is per hour.

Method. Use the MGF formula with $a=4$, $b=12$, and $t=-0.1$:

\begin{aligned} M_T(-0.1) &=\frac{e^{12(-0.1)}-e^{4(-0.1)}}{(12-4)(-0.1)}\\ &=\frac{e^{-1.2}-e^{-0.4}}{-0.8}\\ &=\frac{0.301194-0.670320}{-0.8}\\ &=0.461407. \end{aligned}

Therefore $M_T(-0.1)=0.4614$ to four decimal places. The mean and variance are

E[T]=\frac{4+12}{2}=8\text{ hours},\qquad \operatorname{Var}(T)=\frac{(12-4)^2}{12}=\frac{16}{3}\text{ hours}^2.

The positive value of the MGF is a useful sign check: both the numerator and denominator in the unsimplified quotient are negative. A common error is to reverse the numerator but not the denominator, producing a negative result, which cannot be an expectation of the positive quantity $e^{-0.1T}$.

Common Traps and Practical Checks

The formula is simple, but several details are worth checking whenever it is used:

Do not substitute zero into the quotient. The quotient gives $0/0$ at $t=0$. State $M_X(0)=1$ from the definition, or compute the limit.
Keep the interval width in the denominator. It is $(b-a)t$, not just $t$, because the density is $1/(b-a)$.
Use the endpoints in the correct places. The numerator is $e^{bt}-e^{at}$, in upper-endpoint then lower-endpoint order.
Check the value at zero and the sign. The MGF must equal 1 at zero and be positive for every real $t$.
Track units. If $X$ is measured in hours, $tX$ must be dimensionless, so $t$ is measured per hour.

These derivations and checks are aligned with the SOA VEE Mathematical Statistics syllabus. In particular, the result is specific to a continuous uniform distribution on a finite interval; it should not be applied to an unbounded uniform model or to a different density without re-deriving the expectation.

Check Your Understanding

For each problem, use the finite-interval uniform model and show enough work to identify the endpoints and the relevant MGF or moment calculation.

A hypothetical measurement $X$ is uniform on $[1,7]$. Write its MGF, state its value at zero, and find its mean and variance. (Answer: $M_X(t)=(e^{7t}-e^t)/(6t)$ for $t\ne0$, $M_X(0)=1$, $E[X]=4$, $\operatorname{Var}(X)=3$.)
A hypothetical delay $Y$, in hours, is uniform on $[0,5]$. Find $M_Y(0.2)$ to four decimal places and state the parameter’s units. (Answer: $M_Y(0.2)=e-1=1.7183$; $t$ is per hour.)
A hypothetical score $Z$ is uniform on $[-2,4]$. Find $M_Z(-0.5)$ to four decimal places, along with its mean and variance. (Answer: $M_Z(-0.5)=(e^{-2}-e^1)/(-3)=0.8610$, $E[Z]=1$, $\operatorname{Var}(Z)=3$.)
A hypothetical duration $W$ is uniform on an interval of width 6 and has mean 10. Determine its endpoints, MGF, and variance. (Answer: the endpoints are 7 and 13; $M_W(t)=(e^{13t}-e^{7t})/(6t)$ for $t\ne0$, $M_W(0)=1$, and $\operatorname{Var}(W)=3$.)
A hypothetical cost index $Q$ is uniform on $[0,4]$. A candidate substitutes $t=0$ into $(e^{4t}-1)/(4t)$ and reports $0/0$. Give the correct value and justify it with a limit. (Answer: $M_Q(0)=1$, since $\lim_{t\to0}(e^{4t}-1)/(4t)=4/4=1$.)

What You'll Learn