Question

Let X Unif[0; 1]. Parts (a) and (b) below ask you to nd the nth moment of X in two di erent ways. (a) Use the formula E(Xn) = R 1 0 xn dx. (b) Compute the moment generating function MX(t) (or use the result of Exercise 5.3). Find the Taylor series expansion of MX(t) and identify the coecients. Hint. MX(t) will be written in terms of et. Substitute the Taylor series for et into the formula and simplify the expression.

Answer 1

$X$ follows the standard uniform distribution, so $P(X=x)$ is 1 if $x\in[0,1]$ and 0 otherwise. Then the $n$th moment of $X$ would be

$E[X^n]=\displaystyle\int_{-\infty}^\infty x^nP(X=x)\,\mathrm dx=\int_0^1x^n\,\mathrm dx=\dfrac{x^{n+1}}{n+1}\bigg|_0^1=\boxed{\frac1{n+1}}$

The moment generating function is

$M_X(t)=E[e^{tX}]=\displaystyle\int_{-\infty}^\infty e^{tx}P(X=x)\,\mathrm dx=\int_0^1e^{tx}\,\mathrm dx=\frac{e^{tx}}t\bigg|_0^1=\frac{e^t-1}t$

Recall that

$e^t=\displaystyle\sum_{n=0}^\infty\frac{t^n}{n!}$

Then

$\dfrac{e^t-1}t=\displaystyle\sum_{n=1}^\infty\frac{t^{n-1}}{n!}=\sum_{n=0}^\infty\frac{t^n}{(n+1)!}$

$\dfrac{e^t-1}t=1+\dfrac t2+\dfrac{t^2}6+\dfrac{t^3}{24}+\dfrac{t^4}{120}+\cdots$

The $n$th moment (denoted $\mu_n$) is then the $n$th derivative of the MGF about $t=0$, so that

$\mu_0=M_X(0)=1$

$\mu_1={M_X}'(0)=\dfrac12$

$\mu_2={M_X}''(0)=\dfrac13$

and so on; in general,

$\boxed{\mu_n={M_X}^{(n)}(0)=\dfrac1{n+1}}$