Let $X_1, X_2, . . . , X_n$ be independent and identically distributed random
variables with probability density function uniform ($\theta_1, \theta_2)$. $-\infty < \theta_1 < \theta_2 < \infty$.
Find a method of moments estimator for ($\theta_1, \theta_2$).
My attempt: since we have two parameters. I compute EX, and $EX^2$. I find (where $m_1 $ denotes the population mean, that is $\frac{1}{n}\Sigma X_i$; and $m_2$ denotes the population second moment, that is $\frac{1}{n}\Sigma X_i^2$)
$EX = \frac{\theta_1 + \theta_2}{2} = m_1$
$EX^2 = \frac{1}{3} (\theta_1^2 + \theta_1 \theta_2 + \theta_2^2) = m_2$
But the problem I face is I cannot get unique solutions for $\theta_1 , \theta_2$.
Best Answer
Starting with isolating $\theta_1$ in the first equation we get $$\theta_1 = 2m_1-\theta_2.$$ Now substitute this expression for $\theta_1$ in the second equation. We then get $$\frac{1}{3}(\theta_2^2 + (2m_1 - \theta_2)\theta_2 + (2m_1 -\theta_2)^2) = m_2$$
This is a quadratic equation, which can be rewritten as $$\theta_2^2 - 2m_1\theta_2 + 4m_1^2 - 3m_2 = 0,$$ which by the quadratic formula has solutions $$\theta_2 = \frac{ 2m_1\pm\sqrt{4m_1^2-4(4m_1^2 - 3m_2) }}{2}=m_1 \pm \sqrt{3 }\sqrt{m_2-m_1^2}.$$ Now we can insert this expression in the equation for $\theta_1$ and get $$\theta_1 = m_1 \mp \sqrt{3 }\sqrt{m_2-m_1^2},$$ and finally we note, that in order for the estimator to make sense we must have $\theta_1 < \theta_2$, so we conclude $$\hat{\theta_1} = m_1 - \sqrt{3 }\sqrt{m_2-m_1^2} \quad \text{ and } \quad \hat{\theta_2} = m_1 + \sqrt{3 }\sqrt{m_2-m_1^2}.$$ Worth noting that $m_2 - m_1^2\geq 0$ due to Jensen's inequality so the estimator is well defined.