# Generate Moments of Continuous Uniform Distribution with Moment Generating Functions

I am having trouble generating moments from the moment generating function of the uniform.

By the definition of M.G.F, we can calculate: $$M(t) = \begin{cases} \frac{e^{tb} - e^{ta}}{tb-ta} : t \ne 0 \\ 1 : t=0 \end{cases}$$

However, generating moments involves taking the nth derivative and then setting t=0. If I take the derivative of the $$t \ne 0$$ case, the derivative is not defined when $$t=0$$; if I take the derivative of the $$t=0$$ case, the derivative is 0. The mgf is not generating moments for uniform distribution.

As you say, the derivatives of $$M(t)$$ are not defined at $$t=0$$.

For $$t\ne 0$$, the first derivative for example is $$M'(t)=\frac{1}{t^2(b-a)}\left[e^{tb}(tb-1)-e^{ta}(ta-1)\right]$$

But note that $$M'(t)\to \frac{a+b}{2}$$ as $$t\to 0$$, so $$M'(t)$$ has a removable discontinuity at $$t=0$$.

So just like $$M(t)$$ itself, we define

$$M'(t)=\begin{cases} \frac{1}{t^2(b-a)}\left[e^{tb}(tb-1)-e^{ta}(ta-1)\right]&,\text{ if }t\ne 0 \\\frac{a+b}{2}&,\text{ if }t=0 \end{cases}$$

Hence the first moment is given by $$E[X]=M'(0)=\lim_{t\to 0}M'(t)$$

If you do this from definition, you will end up with the same result:

$$M'(0)=\lim_{t\to 0}\frac1t \left[M(t)-M(0)\right]=\frac{a+b}{2}$$

We define the $$r$$th order derivative $$M^{(r)}(t)$$ similarly so that it is continuous at $$0$$.

And the $$r$$th order moment of $$X$$ for $$r\in \mathbb N$$ is given by

$$E[X^r]=M^{(r)}(0)=\lim_{t\to 0}M^{(r)}(t)$$

The numerator is an entire function, which means you can expand it as a Taylor series around any point you like, it will converge absolutely, and you can compute with this (infinite) sum term by term. Since for any $$z,$$

$$e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \cdots = \sum_{n=0}^\infty \frac{z^n}{n!},$$

compute the Taylor series of the numerator term by term. Since the first term is always $$1,$$ the first term of the difference is $$1-1=0,$$ allowing us to begin the sum at $$n=1$$ instead of $$n=0.$$ You will quickly notice that every term in the difference is a multiple of $$bt-at,$$ so we may factor this out:

\eqalign{ e^{bt} - e^{at} &= \sum_{n=0}^\infty \left(\frac{(bt)^n}{n!} - \frac{(at)^n}{n!}\right)\\ &= \sum_{n=1}^\infty \frac{(bt)^n-(at)^n}{n!} \\ &= \sum_{n=1}^\infty (bt-at) \frac{(b^{n-1}+b^{n-2}a + \cdots + b a^{n-2} + a^{n-1})t^{n-1}}{n!} \\ &= (bt-at)\sum_{n=1}^\infty \frac{(b^{n-1}+b^{n-2}a + \cdots + b a^{n-2} + a^{n-1})t^{n-1}}{n!}. }

Therefore $$M$$ is uniquely defined at $$0$$ as $$M(0)=\lim_{z\to 0}M(z)$$ and that limit takes no work at all to compute because

$$M(t) = \frac{e^{bt}-e^{at}}{bt - at} = \sum_{n=1}^\infty \frac{(b^{n-1}+b^{n-2}a + \cdots + b a^{n-2} + a^{n-1})t^{n-1}}{n!} = \sum_{n=0}^\infty \frac{(b^{n+1}-a^{n+1})}{(b-a)(n+1)}\frac{t^n}{n!}$$

also is an entire function (even in the case $$b=a,$$ by the way). You can read the $$n^\text{th}$$ moment directly off the last expression because it is the coefficient of $$t^n/n!,$$ given by

$$\mu_n = \frac{b^{n+1}-a^{n+1}}{(b-a)(n+1)}.$$

Although technically we did take a limit, we did not have to compute it, and neither did we need to compute any derivatives. (The expansion of $$e^z$$ is the definition of the exponential function: see Walter Rudin, Real and Complex Analysis, 1986.)

Let's check. The first few of these moments are

$$\mu_0 = 1;\ \mu_1 = \frac{b^2-a^2}{2(b-a)} = \frac{a+b}{2};\ \mu_2 = \frac{b^3-a^3}{3(b-a)} = \frac{b^2+ab+a^2}{3}.$$

We can easily compute these from the corresponding (raw) moments of a Uniform$$(0,1)$$ distribution, which are $$1,$$ $$1/2,$$ and $$1/3,$$ respectively, because when a variable with probability element $$f(x)\mathrm{d}x$$ is scaled by a factor $$\sigma,$$ $$\mu_n$$ is multiplied by $$\sigma^n$$ and when a variable is shifted by an amount $$a$$ the new moment is given by the Binomial Theorem as

$$\int_{\mathbb{R}} (x+a)^nf(x)\,\mathrm{d}x = \sum_{i=0}^n \binom{n}{i}a^{n-i}\,\int_{\mathbb{R}} x^if(x)\,\mathrm{d}x = \sum_{i=0}^n \binom{n}{i}a^{n-i}\mu_i.$$

Thus, scaling by $$\sigma=b-a$$ and shifting by $$a$$ gives

\eqalign{ \mu_0 &= 1\\ \mu_1 &= a(b-a)^0\mu_0 + (b-a)^1\mu_1 = a(1)+\frac{b-a}{2} = \frac{a+b}{2}\\ \mu_2 &= a^2(b-a)^0\mu_0 + 2a(b-a)^1\mu_1 + (b-a)^2\mu_2 = a^2(1) + \frac{2a(b-a)}{2} + \frac{(b-a)^2}{3} \\ &= \frac{b^2+ab+a^2}{3}, }

confirming the expressions given by $$M.$$ You can see how the calculations for higher moments are going to involve algebraic simplification of ever more complicated polynomials: the moment generating function approach spared us that work.

• (+1) I guess I should have emphasized that we need not compute derivatives and take their limits to find moments from the MGF. – StubbornAtom Feb 23 '20 at 19:08