Moment of order k in the uniform distribution

Question

To calculate the moment of order k of any random distribution knowing the characteristic function G is: E[x^k] = dⁿ/dt G (t = 0)/i^k. For the uniform distribution with density function f(x) = 1/(b-a) with a <= x <= b. I found that the characteristic function is G(t) = (e^itb - e^ita)/(itb - ita). Then E(x) = G'(t = 0)

G' = ((bt + i)e^ibt-(at+i)e^iat)/((b-a)t²)

if it evaluated at 0 this expression is indeterminate. I know how to calculate this moment through other methods (I know that E(x) = (a + b)/2) .

summarizing: The question is using the characteristic function how to arrive at the moment of order k in the uniform distribution. Thanks in advance.

Answer 1

First proof: Using Taylor expansion

You can shortcut the computation of $G'(t)$.

Indeed, let us work on the numerator of

$$G(t)=\frac{e^{it}-e^{ia}}{it(b-a)}$$

Replacing in this numerator the exponentials with their second order expansion, we get:

$$e^{it}-e^{ia}=\left(1+itb+\frac{i^2t^2b^2}{2}\right)-\left(1+ita+\frac{i^2t^2a^2}{2}\right)+o(t^2)$$

$$e^{it}-e^{ia}=it(b-a)+t^2\frac{a^2-b^2}{2}+o(t^2)$$

Dividing now by the denominator $it(b-a)$, we get:

$$G(t)=1+t\frac{(a-b)(a+b)}{2i(b-a)}+o(t)$$

$$G(t)=1+i\frac{a+b}{2}t+o(t)\tag{1}$$

Comparing (1) with the expansion :

$$G(t)=\underbrace{1}_{G(0)}+\underbrace{i \mu_1 }_{G'(0)} t + \underbrace{-\frac12 \mu_2}_{G''(0)} t^2 + ...$$

by unicity of the Taylor decomposition, we indeed retrieve the fact that :

$$\mu_1=\frac{a+b}{2}$$

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Second proof Applying l'Hospital rule to your expression:

$$G'(t) = \frac{(bt + i)e^{ibt}-(at+i)e^{iat}}{(b-a)t^2}$$

which is indeterminate when $t=0$ (form $\frac00$).

L'Hospital rule says that (under certain conditions that are generally verified), such a fraction $\frac{N(t)}{D(t)}$ has the same limit when $t$ tends to $0$ as the ratio of its derivatives

$$\frac{N'(t)}{D'(t)}=it \frac{b^2e^{ibt}-a^2e^{iat}}{2t(b-a)}=i \frac{b^2e^{ibt}-a^2e^{iat}}{2(b-a)}$$

which, when $t$ tends to $0$ tends to:

$$i \frac{b^2e^{ib0}-a^2e^{ia0}}{2(b-a)}=i \frac{b^2-a^2}{2(b-a)}=i\frac{1}{2}(b+a)$$

as desired.