Using the result from the previous problem, we know that [mathjaxinline]\mu = m_1[/mathjaxinline] and [mathjaxinline]\sigma^2 = m_2 - (m_1)^2[/mathjaxinline].
Using the sample moments, we have:
[mathjaxinline]\widehat{m_1} = \frac{1}{n}\sum_{i=1}^{n}X_i = \frac{0.5+1.8-2.3+0.9}{4} = 0.225[/mathjaxinline]
[mathjaxinline]\widehat{m_2} = \frac{1}{n}\sum_{i=1}^{n}(X_i - \widehat{m_1})^2 = \frac{(0.5-0.225)^2 + (1.8-0.225)^2 + (-2.3-0.225)^2 + (0.9-0.225)^2}{4} \approx 3.951[/mathjaxinline]
Therefore, the method of moments estimator for [mathjaxinline]\mu^*[/mathjaxinline] evaluated on this data-set is:
[mathjaxinline]\widehat{\mu}^{\text{MM}} = \widehat{m_1} = 0.225[/mathjaxinline]
We let [mathjaxinline]X_1, \ldots , X_ n \stackrel{iid}{\sim } N(\mu ^*, (\sigma ^*)^2)[/mathjaxinline] and consider the associated statistical model [mathjaxinline](\mathbb {R}, \{ N(\mu , \sigma ^2)\} _{\mu \in \mathbb {R}, \sigma > 0})[/mathjaxinline]. Let
[mathjaxinline]\displaystyle \psi : \mathbb {R}[/mathjaxinline] [mathjaxinline]\displaystyle \times (0, \infty ) \to \mathbb {R}^2[/mathjaxinline]
[mathjaxinline]\displaystyle (\mu , \sigma )[/mathjaxinline] [mathjaxinline]\displaystyle \mapsto (m_1(\mu , \sigma ), m_2(\mu , \sigma )).[/mathjaxinline]
denote the moments map considered in the previous problem, where [mathjaxinline]m_ k(\mu , \sigma )[/mathjaxinline] denotes the [mathjaxinline]k[/mathjaxinline]-th moment of the distribution [mathjaxinline]N(\mu , \sigma ^2 )[/mathjaxinline].
To answer the next question, you should recall:
your result on writing [mathjaxinline]\mu[/mathjaxinline] and [mathjaxinline]\sigma[/mathjaxinline] in terms of [mathjaxinline]m_1[/mathjaxinline] and [mathjaxinline]m_2[/mathjaxinline] obtained in the previous problem, and
the estimators [mathjaxinline]\widehat{m_1}[/mathjaxinline] and [mathjaxinline]\widehat{m_2}[/mathjaxinline] (the sample moments) for the true moments [mathjaxinline]m_1[/mathjaxinline] and [mathjaxinline]m_2[/mathjaxinline].
Suppose we observe the data-set [mathjaxinline]X_1 = 0.5 , X_2 = 1.8 , X_3 = -2.3, X_4 = 0.9[/mathjaxinline].
What is the method of moments estimator [mathjaxinline]\widehat{\mu }^{\text {MM}}[/mathjaxinline] for [mathjaxinline]\mu ^*[/mathjaxinline] evaluated on this data-set? (You are encouraged to use whatever computational tools may be helpful.)
[mathjaxinline]\widehat{\mu }^{\text {MM}}(0.5, 1.8, -2.3, 0.9) =[/mathjaxinline]
3 answers
What is the method of moments [mathjaxinline]\widehat{\sigma }^{\text {MM}}[/mathjaxinline] estimator for [mathjaxinline]\sigma ^*[/mathjaxinline] evaluated on this data-set?
Using the same result from the previous problem, we have [mathjaxinline]\mu = m_1[/mathjaxinline] and [mathjaxinline]\sigma^2 = m_2 - (m_1)^2[/mathjaxinline]. Substituting [mathjaxinline]\mu = \widehat{m_1}[/mathjaxinline] and [mathjaxinline]\sigma^2 = \widehat{m_2} - (\widehat{m_1})^2[/mathjaxinline] (using the sample moments), we get:
[mathjaxinline]\widehat{\sigma }^{\text {MM}} = \sqrt{\widehat{m_2} - (\widehat{m_1})^2} = \sqrt{3.951 - (0.225)^2} \approx 1.238[/mathjaxinline]
Therefore, the method of moments estimator for [mathjaxinline]\sigma^*[/mathjaxinline] evaluated on this data-set is approximately 1.238.
[mathjaxinline]\widehat{\sigma }^{\text {MM}} = \sqrt{\widehat{m_2} - (\widehat{m_1})^2} = \sqrt{3.951 - (0.225)^2} \approx 1.238[/mathjaxinline]
Therefore, the method of moments estimator for [mathjaxinline]\sigma^*[/mathjaxinline] evaluated on this data-set is approximately 1.238.