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helaidudu's Question
Math
Posted 2 months ago

1 In the simple regression model under MLR. 1 through MLR.4, we argued that the slope estimator, β^1\hat{\beta}_{1}, is consistent for β1\beta_{1}. Using β^0=yˉβ^1x11\hat{\beta}_{0}=\bar{y}-\hat{\beta}_{1} \mathrm{x}_{1}{ }_{1}, show that plim β^0=β0\hat{\beta}_{0}=\beta_{0}. [You need to use the consistency of β^1\hat{\beta}_{1} and the law of large numbers, along with the fact that β0=\beta_{0}= E(y)=β1E(x1).]\left.\mathrm{E}(\mathrm{y})=\beta_{1} \mathrm{E}\left(\mathrm{x}_{1}\right).\right]
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
We start with the given regression model and the consistency of the slope estimator β^1\hat{\beta}_{1}
step 2
The slope estimator β^1\hat{\beta}_{1} is consistent for β1\beta_{1}, which means plimβ^1=β1\text{plim} \, \hat{\beta}_{1} = \beta_{1}
step 3
Using the formula β^0=yˉβ^1xˉ1\hat{\beta}_{0} = \bar{y} - \hat{\beta}_{1} \bar{x}_{1}, we need to show that plimβ^0=β0\text{plim} \, \hat{\beta}_{0} = \beta_{0}
step 4
By the law of large numbers, yˉE(y)\bar{y} \rightarrow \mathrm{E}(y) and xˉ1E(x1)\bar{x}_{1} \rightarrow \mathrm{E}(x_{1}) as the sample size increases
step 5
Substituting the limits, we get plimβ^0=plim(yˉβ^1xˉ1)=E(y)β1E(x1)\text{plim} \, \hat{\beta}_{0} = \text{plim} \, (\bar{y} - \hat{\beta}_{1} \bar{x}_{1}) = \mathrm{E}(y) - \beta_{1} \mathrm{E}(x_{1})
step 6
Given that β0=E(y)β1E(x1)\beta_{0} = \mathrm{E}(y) - \beta_{1} \mathrm{E}(x_{1}), we conclude that plimβ^0=β0\text{plim} \, \hat{\beta}_{0} = \beta_{0}
Answer
plimβ^0=β0\text{plim} \, \hat{\beta}_{0} = \beta_{0}
Key Concept
Consistency of estimators
Explanation
The consistency of the slope estimator β^1\hat{\beta}_{1} and the law of large numbers ensure that the intercept estimator β^0\hat{\beta}_{0} converges in probability to the true intercept β0\beta_{0}.

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