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$(X_1,\dots, X_n)$ is a random sample extracted from a uniform distribution on the interval $$(\alpha-\beta, \alpha+\beta) \ \ \ \ \alpha \in \mathbb{R}, \beta \in \mathbb{R}^{+}$$

  1. Demonstrate that $E[X_1]$ depends only from $\alpha$ and $Var[X_1]$ depends only from $\beta$
  2. Calculate the estimators $T_1$ of $E[X_1](\alpha)$ and $T_2$ of $Var[X_1](\beta)$ using the method of moments

For 1. $X_1$ is a random variable of uniform distribution, so: $$E[X_1]=\frac{\alpha-\beta+\alpha+\beta}{2}=\alpha$$ and $$Var[X_1]=\frac{(\alpha+\beta-\alpha + \beta)^2}{12}=\frac{\beta^2}{3}$$

For 2. \begin{align}M_1'&=\overline{X}=T_1\\M_2'&=T_2+\overline{X}^2\end{align} How can I continue the study of estimators?

Thanks!

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First moment of the distribution is $μ_1=E[X]=α$ and the first moment of the sample is $m_1=\bar{X}$. So, set $$μ_1=m_1 \implies α=m_1$$ This is the first moment estimator for $α$. (Note: This method is confusing at the beginnning, because you think, ok so what? But think that $m_1$ is your sample mean and so it is known, it will be realized when you collect the sample and will give you a number.) Now, the second moment of the distribution is $$μ_2=E[X^2]=Var(X)+E[X]^2=\frac{β^2}{3}+α^2$$ and the second moment of the sample is $$m_2=\frac1n\sum_{i=1}^{n}X_i^2$$ Equating the second moments gives $$μ_2=m_2 \implies \frac{β^2}{3}+α^2=m_2$$ So, you obtained the system with two unknowns $α,β$ (I repeat that $m_1,m_2$ are known!) which you can solve \begin{cases}m_1=α\\m_2=\dfrac{β^2}{3}+α^2\end{cases}

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  • $\begingroup$ Thanks! $\frac{\beta^2}{3}=\frac{1}{n} \ \sum_{i=1}^n \ (X_i-\overline{X})^2$ ? How can I calculate the distortion of estimator $T_2$ ? $\endgroup$ – Elsa Feb 4 '16 at 12:30

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