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

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

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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.

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

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