How can in prove the following statement with delta method: "If I divide a variable by its deviation, the deviation of the transformed variable is 1."
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This doesn't mean anything. I suppose you had standard deviation in mind, in that case delta method is pointless. Actually, it's always pointless for linear transformations – carlo Nov 02 '19 at 19:30
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@carlo I think it is also useful for linear transformations, as even for transformations of the form $f(x) = a + bx$ it is true that $\sigma^{2}{f(x)} \ne f(\sigma^{2}{x})$, right? (I mean, the Delta method might be trivial, but still technically useful?) – Alexis Nov 02 '19 at 20:43
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delta method is an approximation based on Tailor decomposition. You don't need it for linear transformation. But yes, sure that it is true. – carlo Nov 02 '19 at 23:17