Suppose that y1 = 8.3, y2 = 4.9, y3 = 2.6, y4 = 6.5 is a random sample of size 4 from the two parameter uniform pdf,

Question

Suppose that $y_1 = 8.3, y_2 = 4.9, y_3 = 2.6, y_4 = 6.5$ is a random sample of size 4 from the two-parameter uniform pdf,

$f_y(y;\theta_1,\theta_2) = \frac{1}{2\theta_2}, \theta_1 - \theta_2 \le y \le \theta_1 + \theta_2$

Use the method of moments to calculate $\theta_{1e}$ and $\theta_{2e}$

So I understand how to solve the question when it comes to one parameter. Where we essentially solve for $E(y)$. But I'm not sure how to go about solving for two parameters.

Answer 1

This is not the usual parameterization of a uniform distribution, so looking at standard references may not be immediately helpful,

Begin by showing That $E(X) = \theta_1$ and $Var(X) = \theta_2^2/3.$

Then, for the given sample, find $\bar X = 0.375$ and $S^2 = 6.3825.$ Computations in R:

x = c(8.3, 4.1, 2.6, 6.5)
mean(x); var(x)
[1] 5.375
[1] 6.3825

Finally, set $E(X) = \mu = \theta_1 = \bar X$ and $Var(X) = \sigma^2 = \theta_2^2/3 = S^2.$ Solve for $\theta_1, \theta_2$ in terms of data to get numerical MOM estimates $\hat\theta_1, \hat\theta_2,$ respectively. Obviously, $\hat\theta_1 - \bar X =5.375.$ What is $\hat\theta_2 ?$

Note: Later you may study maximum likelihood estimates. Sometimes MOM (method-of-moments) estimators are the same as MLEs (maximum likelihood estimators), and sometimes not. When sampling from the uniform distribution of this problem, there are differences between the two kinds of estimators.