Method of moments on uniform distributions

Question

I need help on how to find the estimates $a$ and $b$ in the uniform distribution $\mathcal U[a,b]$ using the method of moments. This is where I am at: I have found $U_1=\overline X$ and $m_1=\frac{a+b}2$ Also, $m_2=\frac1{n(E(X_i^2)}$ and $u_2=E(X_i^2)$ when I equate $m_1$ to $u_1$ and $m_2$ to $u_2$ I should find the estimates,right? I am unable to solve that.Kindly help.

Answer 1

The population mean is $\dfrac{a+b}{2}$. The population variance is $\dfrac{(b-a)^2}{12}$. If you have already found that the popuation variance of the $\mathcal U[0,1]$ distribution is $1/12$, just notice that the length of the interval has been multiplied by $b-a$, and that is a scale factor, so you multiply the variance by $(b-a)^2$. Hence the second moment is $$ \frac{(b-a)^2}{12}+\left(\frac{a+b}{2}\right)^2 = \frac{a^2+b^2+ab}{3}. $$

So the equations to be solved for $a$ and $b$ are: \begin{align} \frac{x_1+\cdots+x_n}{n} & = \frac{a+b}{2} \\[10pt] \frac{x_1^2+\cdots+x_n^2}{n} & = \frac{a^2+b^2+ab}{3} \end{align}

One way would be to solve the first equation for $b$ and then substitute that for $b$ in the second equation, getting a quadratic equation in $a$:

$$ b = 2\bar x - a, $$

$$ \frac{x_1^2+\cdots+x_n^2}{n} = \frac{a^2+(2\bar x - a)^2+a(2\bar x-a)}{3} $$

$$ \frac{x_1^2+\cdots+x_n^2}{n} = \frac{a^2 + 2\bar x a + 4\bar x^2}{3} $$

$$ a^2 + 2\bar x a + \left(4\bar x^2 - 3\frac{x_1^2+\cdots+x_n^2}{n}\right)=0. $$

Then proceed the way you usually do with quadratic equations.

PS: Completing the square gives $$ \begin{align} a^2 + 2\bar x a + \bar x^2 + 3\left(\bar x^2 - \frac{x_1^2+\cdots+x_n^2}{n}\right)& = 0\\[10pt] (a+\bar x)^2 & = 3\left(\frac{x_1^2+\cdots+x_n^2}{n} - \bar x^2\right) \\[10pt] (a+\bar x)^2 & = 3s^2 \end{align} $$ where this is taken to define the notation $s^2$. That means $s^2$ is the sample variance if that is taken to mean the version with $n$ rather than $n-1$ in the denominator.