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Probability: Difference between revisions
→Delta Method
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Suppose <math>\sqrt{n}(X_n - \theta) \xrightarrow{D} N(0, \sigma^2)</math>.<br> | Suppose <math>\sqrt{n}(X_n - \theta) \xrightarrow{D} N(0, \sigma^2)</math>.<br> | ||
Let <math>g</math> be a function such that <math>g'</math> exists and <math>g'(\theta) \neq 0</math><br> | Let <math>g</math> be a function such that <math>g'</math> exists and <math>g'(\theta) \neq 0</math><br> | ||
Then <math>\sqrt{n}(g(X_n) - g(\theta)) \xrightarrow{D} N(0, \sigma^2 g'(\theta)^2)</math> | Then <math>\sqrt{n}(g(X_n) - g(\theta)) \xrightarrow{D} N(0, \sigma^2 g'(\theta)^2)</math> | ||
Multivariate:<br> | Multivariate:<br> | ||
<math>\sqrt{n}(B - \beta) \xrightarrow{D} N(0, \Sigma) \implies \sqrt{n}(h(B)-h(\beta)) \xrightarrow{D} N(0, h'(\theta)^T \Sigma h'(\theta))</math> | <math>\sqrt{n}(B - \beta) \xrightarrow{D} N(0, \Sigma) \implies \sqrt{n}(h(B)-h(\beta)) \xrightarrow{D} N(0, h'(\theta)^T \Sigma h'(\theta))</math> | ||
;Notes | ;Notes | ||
* You can think of this like the Mean Value theorem for random variables. | * You can think of this like the Mean Value theorem for random variables. | ||