What Is the MGF of a Uniform Distribution?

For a continuous uniform random variable, the moment-generating function depends on the endpoints of its interval. If $X$ is uniform on $[a,b]$, its MGF has a simple exponential form for nonzero $t$ and equals 1 at $t=0$. The formula also gives the mean and variance by differentiation.

Formula for a continuous uniform distribution

Let $X\sim\operatorname{Uniform}(a,b)$, with $a<b$. Its density is $f_X(x)=1/(b-a)$ for $a\le x\le b$ and zero elsewhere. By definition, the moment-generating function is

$$M_X(t)=\mathbb{E}[e^{tX}].$$

Therefore,

$$M_X(t)=\begin{cases} \dfrac{e^{bt}-e^{at}}{t(b-a)}, & t\ne 0,\\[6pt] 1, & t=0. \end{cases}$$

The MGF exists for every real value of $t$. This is true because $X$ is bounded, so the expectation of $e^{tX}$ is finite for any fixed real $t$.

Deriving the formula

Use the density to express the expectation as an integral:

$$\begin{aligned} M_X(t)&=\int_{-\infty}^{\infty}e^{tx}f_X(x)\,dx\\ &=\frac{1}{b-a}\int_a^b e^{tx}\,dx. \end{aligned}$$

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

$$M_X(t)=\frac{1}{b-a}\left[\frac{e^{tx}}{t}\right]_a^b =\frac{e^{bt}-e^{at}}{t(b-a)}.$$

At $t=0$, the integrand in the expectation is $e^{0X}=1$, so $M_X(0)=\mathbb{E}[1]=1$. The nonzero-$t$ expression also approaches 1 as $t$ approaches zero, but writing the value separately avoids an apparent division by zero.

Getting the mean and variance

When the MGF exists around zero, its derivatives at zero give the moments:

$$\mathbb{E}[X]=M_X'(0),\qquad \mathbb{E}[X^2]=M_X''(0).$$

For $X\sim\operatorname{Uniform}(a,b)$, these yield

$$\mathbb{E}[X]=\frac{a+b}{2},\qquad \operatorname{Var}(X)=\frac{(b-a)^2}{12}.$$

The variance follows from $\operatorname{Var}(X)=M_X''(0)-[M_X'(0)]^2$. The MGF is useful for more than moments: for independent random variables, the MGF of their sum is the product of their MGFs, whenever the MGFs exist.

Worked example: uniform on [2,5]

Suppose $X\sim\operatorname{Uniform}(2,5)$. Here $a=2$ and $b=5$, so

$$M_X(t)=\begin{cases} \dfrac{e^{5t}-e^{2t}}{3t}, & t\ne 0,\\[6pt] 1, & t=0. \end{cases}$$

Expanding the exponentials near zero makes the moments visible:

$$M_X(t)=1+\frac{7}{2}t+\frac{13}{2}t^2+\cdots.$$

An MGF expansion has the form $1+\mathbb{E}[X]t+\mathbb{E}[X^2]t^2/2+\cdots$. Thus $\mathbb{E}[X]=7/2=3.5$ and $\mathbb{E}[X^2]=13$. The variance is

$$\operatorname{Var}(X)=13-(3.5)^2=0.75.$$

This agrees with the uniform variance formula: $(5-2)^2/12=9/12=0.75$.

Common special cases and conventions

For the standard uniform distribution on $[0,1]$, set $a=0$ and $b=1$:

$$M_X(t)=\frac{e^t-1}{t}\quad(t\ne0),\qquad M_X(0)=1.$$

For a uniform distribution on $[-c,c]$, where $c>0$, the formula simplifies to $M_X(t)=\sinh(ct)/(ct)$ for $t\ne0$, with value 1 at zero.

The formula above is for a continuous uniform distribution. If instead $X$ is discrete and equally likely to be one of the integers $a,a+1,\ldots,b$, then

$$M_X(t)=\frac{1}{b-a+1}\sum_{k=a}^{b}e^{kt}.$$

The continuous formula does not apply to that integer-valued case.

Common mistakes

• Substituting $t=0$ into the quotient: The quotient appears to divide by zero there. Use $M_X(0)=1$, or take the limit.
• Using the endpoints as the mean: The mean is their midpoint, $(a+b)/2$, not either endpoint.
• Confusing width and variance: The interval width is $b-a$; the variance is its square divided by 12.
• Forgetting the density factor: The integral must be divided by $b-a$. Omitting it gives a result that is not an MGF, since it will not equal 1 at zero.