Let 5 numbers 2, 3, 5, 9 and 10 come from a uniform distribution on the interval $[\alpha,\beta]$.

Find the method of moments estimators of $\alpha$ and $\beta$.

Any help would be appreciated, thank you!

  • $\begingroup$ Do you know what the mean and second moment of the given distribution is (in terms of $\alpha$ and $\beta$)? $\endgroup$ – Minus One-Twelfth Feb 22 at 7:46
  • $\begingroup$ Yes for $X~Uniform(\alpha,\beta)$ the mean is $E[X] $ = $ 1/2 * (\alpha + \beta) $ and the second moment is $E[X^2] = 1/3 *( \alpha^2 + \alpha * \beta + \beta^2)$. So if I knew the value of $E[X]$ and $E[X^2]$, I could solve the system of equations for $\alpha$ and $\beta$... But I don't, do I? $\endgroup$ – CruZ Feb 22 at 7:56
  • 2
    $\begingroup$ $\newcommand{\E}{\mathbb{E}}$If you know the expressions for $\E[X]$ and $\E[X^2]$, then you need to do is equate these expressions to the sample average and second moment (these will just be two numbers for the given sample). This will give you two equations in two unknowns ($\alpha$ and $\beta$). Solving these will give you the method of moments estimates. $\endgroup$ – Minus One-Twelfth Feb 22 at 7:59
  • $\begingroup$ Thank you, that makes sense! $\endgroup$ – CruZ Feb 22 at 8:16

$\newcommand{\E}{\mathbb{E}}$Hints: The point of method of moments is that we calculate the sample moments and equate the theoretical moments to these. The sample moments are just numbers you can calculate from the sample (e.g. the average of the data is to be equated to the expression for $\E[X]$ and the average of the squares of the data to the expression for $\E[X^2]$). After doing this equating, solve simultaneously for $\alpha$ and $\beta$ to get the method of moments estimates.

For a sample $\{2,3,5,9,10\}$, the sample average is $\frac{1}{5}(2+3+5+9+10)$ and the sample second moment is $\frac{1}{5}\left(2^2+3^2+5^2+9^2+10^2\right)$.


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