Uniform Random Variable Moment

Question

Let $X$ be a uniformly distributed random variable on the interval [0, 1].

What is $E[X^2]$?

How to apply the moment generating formula (if applicable?)

Answer 1

I give two solutions.

We can first avoid using generating functions by noting that the variance of $X$ is the mean of the square minus the square of the mean - that is,

$$ V[X] = E[X^2] - E^2[X] $$

For a $(0, 1)$ uniform random variable the mean is $1/2$ and the variance is $1/12$ (check this). Rearranging this in the above relation gives that $E[X^2] = 1/3$.

Alternatively we can use the generating function, which (by searching) is

$$ M_X(t) = \frac{e^{bt} - e^{at}}{t(b-a)}, \ t \neq 0. $$

To find the $n$-th moment we take $n$ derivatives and evaluate at zero. This will yield the same answer.

Answer 2

Given that $X\sim U(0,1)$. Then, \begin{equation*} E(X^2)=\int_{0}^{1}x^2 f(x) dx = \int_{0}^{1}x^2 dx = \left[\dfrac{x^3}{3}\right]_{0}^{1}=\dfrac{1}{3} \end{equation*} $E(X^2)$ Using MGF:

The moment generating function of the given $U(0,1)$ distribution is \begin{eqnarray*} M_{X}(t)&=& \int_{0}^{1}e^{tx} f(x) dx = \int_{0}^{1}e^{tx} dx = \left[\dfrac{e^{tx}}{t}\right]_{0}^{1}=\dfrac{e^t}{t}-\dfrac{1}{t}=\dfrac{1}{t}(e^t-1),\qquad t\neq 0.\\ &=&\dfrac{1}{t}\left(1+t + \dfrac{t^2}{2!} + \dfrac{t^3}{3!} + \cdots -1\right)=\dfrac{1}{t}\left(t + \dfrac{t^2}{2!} + \dfrac{t^3}{3!} + \cdots \right)\\ &=&\left(1+ \dfrac{t}{2!} + \dfrac{t^2}{3!} + \cdots \right)\\ E(X^2)&=&M_{X}^{(2)}(0)=\dfrac{2}{3!}=\dfrac{1}{3} \end{eqnarray*}

Answer 3

For some random variable $X \sim \mathrm{Uniform}(a,b)$. The moment generating function is: $$ M_X(t) = \begin{cases} \frac{\mathrm{e}^{bt}-\mathrm{e}^{at}}{t(b-a)} \quad &t \ne 0 \\ 1 \quad &t = 0 \end{cases} $$

To tackle the problem of calculating the moments from m.g.f., one may differentiate and then take the limit as $t\rightarrow0$. See, for example, details at Wolfram Math.