# Second Moment of Uniform Distribution – (With Proof)

-

Let X be a random variable following uniform distribution with values lying between ‘a’ and ‘b’. The pdf of X is given by the formula, f(x) = \begin{cases} \frac{1}{b-a} & a \leq x \leq b \\ 0 & \text{otherwise}\end{cases}

The second moment of the random variable is given by the formula, E(X^2)= \frac{b^2+ba+a^2}{3}

#### Proof (Without using MGF) :

\begin{align*} E(X^2) &= \int_{a}^{b}x^2f(x)dx \\ &= \frac{1}{b-a}\int_{a}^{b}x^2dx \\ &= \frac{1}{b-a}\left[\frac{x^3}{3}\right]_a^b \\ &= \frac{1}{b-a}\left[\frac{b^3-a^3}{3}\right] \\ &= \frac{1}{b-a}\left[\frac{(b-a)(b^2 + ba +a^2)}{3}\right] \\ &=\frac{b^2+ba+a^2}{3} \end{align*}

#### Proof (Using MGF) :

In order to find the second random moment of the uniform distribution using the MGF we should differentiate the moment generating function twice and then set t=0. The MGF (Moment Generating Function) of the uniform distribution is given as, M_X(t)= \frac {e^{tb}-e^{ta}}{t(b-a)} On differentiating it twice we get, \begin{align*} \frac{d^2}{dt^2}M_X(t) &= \frac{d^2}{dt^2}\left[ \frac {e^{tb}-e^{ta}}{t(b-a)}\right] \\ &= \frac{d^2}{dt^2}\left[ \frac {(1 + \frac{tb}{1!} + \frac{(tb)^2}{2!} + \frac{(tb)^3}{3!} + \frac{(tb)^4}{4!} + \ldots)-(1 + \frac{ta}{1!} + \frac{(ta)^2}{2!} + \frac{(ta)^3}{3!} + \frac{(ta)^4}{4!} + \ldots)}{t(b-a)}\right] \\ &= \frac{1}{b-a}\left[(b^3 + \frac{(tb)}{1} + \ldots)-(a^3 + \frac{(ta)}{1} + \ldots)\right] \end{align*}

Putting t=0 on both sides we get that, \begin{align*} E(X^2) = \frac{d^2}{dt^2}M_X(t) |_{t=0} &= \frac{1}{b-a}\left[(b^3 + \frac{(tb)}{1} + \ldots)-(a^3 + \frac{(ta)}{1} + \ldots)\right]_{t=0} \\ &= \frac{1}{b-a}\left[\frac{b^3-a^3}{3}\right] \\ &= \frac{1}{b-a}\left[\frac{(b-a)(b^2 + ba +a^2)}{3}\right] \\ &=\frac{b^2+ba+a^2}{3} \end{align*}

Hey 👋

I have always been passionate about statistics and mathematics education.

I created this website to explain mathematical and statistical concepts in the simplest possible manner.

If you've found value from reading my content, feel free to support me in even the smallest way you can.