Question
Let \widehat{\mathbf{p}} denote the MLE for a categorical statistical model ( \{ a_1, \ldots , a_ K \} , \{ \mathbf{P}_{\mathbf{p}} \} _{\mathbf{p} \in \Delta _ K}). Let \mathbf{p}^* denote the true parameter. Then \sqrt{n}(\widehat{\mathbf{p}} - \mathbf{p}^*) is asymptotically normal and
n \sum _{i = 1}^ K \frac{ ( \widehat{ p_ i } - p_ i^*)^2 }{p_ i^*} \xrightarrow [n \to \infty ]{(d)} \chi _{K -1}^2.
Consider the particular categorical distribution from the previous problems in this lecture, where we have the statistical experiment X_1, \ldots , X_ n \stackrel{iid}{\sim } \mathbf{P}_{\mathbf{p}} and associated statistical model (\{ 1,2,3\} , \{ \mathbf{P}_{\mathbf{p}} \} _{\mathbf{p} \in \Delta _3}). We will use the above fact to hypothesis test between the following null and alternative:
\displaystyle H_0: \mathbf{p}^* \displaystyle = [1/3~ ~ 1/3~ ~ 1/3]^ T
\displaystyle H_1: \mathbf{p}^* \displaystyle \neq [1/3~ ~ 1/3~ ~ 1/3]^ T.
Consider the test
\psi = \mathbf{1}\left( n \sum _{i = 1}^3 \frac{ ( \widehat{ p_ i } - \frac{1}{3})^2 }{1/3} >C \right),
for a threshold C.
Compute the asymptotic p-value of the test \psi on the data set
\mathbf{x} = 1, 3, 1, 2, 2, 2, 1, 1, 3, 1, 1, 2.
Give a numerical value with at least 4 decimals. Use this tool to find the tail probabilities of a \chi ^2 distribution (you may also use any other software). If you are using this tool, note that you need to set "Choose Type of Control" to "Adjust X-axis quantile (Chi square) value" to find the tail probability associated with an x-axis value for a chi-squared distribution with degrees of freedom set in the "Degrees of Freedom" box.
n \sum _{i = 1}^ K \frac{ ( \widehat{ p_ i } - p_ i^*)^2 }{p_ i^*} \xrightarrow [n \to \infty ]{(d)} \chi _{K -1}^2.
Consider the particular categorical distribution from the previous problems in this lecture, where we have the statistical experiment X_1, \ldots , X_ n \stackrel{iid}{\sim } \mathbf{P}_{\mathbf{p}} and associated statistical model (\{ 1,2,3\} , \{ \mathbf{P}_{\mathbf{p}} \} _{\mathbf{p} \in \Delta _3}). We will use the above fact to hypothesis test between the following null and alternative:
\displaystyle H_0: \mathbf{p}^* \displaystyle = [1/3~ ~ 1/3~ ~ 1/3]^ T
\displaystyle H_1: \mathbf{p}^* \displaystyle \neq [1/3~ ~ 1/3~ ~ 1/3]^ T.
Consider the test
\psi = \mathbf{1}\left( n \sum _{i = 1}^3 \frac{ ( \widehat{ p_ i } - \frac{1}{3})^2 }{1/3} >C \right),
for a threshold C.
Compute the asymptotic p-value of the test \psi on the data set
\mathbf{x} = 1, 3, 1, 2, 2, 2, 1, 1, 3, 1, 1, 2.
Give a numerical value with at least 4 decimals. Use this tool to find the tail probabilities of a \chi ^2 distribution (you may also use any other software). If you are using this tool, note that you need to set "Choose Type of Control" to "Adjust X-axis quantile (Chi square) value" to find the tail probability associated with an x-axis value for a chi-squared distribution with degrees of freedom set in the "Degrees of Freedom" box.
Answers
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Answered by
GPT 3.5
To compute the asymptotic p-value of the test, we calculate the test statistic:
\[T = n \sum_{i=1}^3 \frac{(\hat{p}_i - \frac{1}{3})^2}{\frac{1}{3}} = 12 \sum_{i=1}^3 (\hat{p}_i - \frac{1}{3})^2\]
First, we need to find the MLEs of the probabilities:
\[\hat{p}_1 = \frac{5}{12}, \quad \hat{p}_2 = \frac{5}{12}, \quad \hat{p}_3 = \frac{2}{12}\]
Substituting these values into the test statistic:
\[T = 12 \left((\frac{5}{12} - \frac{1}{3})^2 + (\frac{5}{12} - \frac{1}{3})^2 + (\frac{2}{12} - \frac{1}{3})^2 \right) = \frac{38}{36} = \frac{19}{18}\]
We want to find the tail probability of the chi-squared distribution with 2 degrees of freedom that gives us a test statistic greater than or equal to T. Using the tool provided or any other software, we find that the tail probability is approximately 0.444.
Since this is a two-sided test, the p-value is twice the tail probability:
p-value = 2 * 0.444 = 0.888
Therefore, the asymptotic p-value of the test is approximately 0.888.
\[T = n \sum_{i=1}^3 \frac{(\hat{p}_i - \frac{1}{3})^2}{\frac{1}{3}} = 12 \sum_{i=1}^3 (\hat{p}_i - \frac{1}{3})^2\]
First, we need to find the MLEs of the probabilities:
\[\hat{p}_1 = \frac{5}{12}, \quad \hat{p}_2 = \frac{5}{12}, \quad \hat{p}_3 = \frac{2}{12}\]
Substituting these values into the test statistic:
\[T = 12 \left((\frac{5}{12} - \frac{1}{3})^2 + (\frac{5}{12} - \frac{1}{3})^2 + (\frac{2}{12} - \frac{1}{3})^2 \right) = \frac{38}{36} = \frac{19}{18}\]
We want to find the tail probability of the chi-squared distribution with 2 degrees of freedom that gives us a test statistic greater than or equal to T. Using the tool provided or any other software, we find that the tail probability is approximately 0.444.
Since this is a two-sided test, the p-value is twice the tail probability:
p-value = 2 * 0.444 = 0.888
Therefore, the asymptotic p-value of the test is approximately 0.888.
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