logl::makegrads, creates a group of the gradients (first derivatives) of the log likelihood at the estimated parameter values. These gradients are often used in constructing Lagrange multiplier tests.
I am puzzled. Textbooks say, for LM test, the gradients are computed from the unrestricted model at the estimated value from restricted model. It seems that the gradients from makegrads is not the correct one in LM test on likelihood functions.
logl::makegrads, LM test
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EViews Glenn
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Re: logl::makegrads, LM test
Your statement about the unrestricted graidents is correct.
But unfortunately we are not psychic and don't know what unrestricted model you are thinking of so we can't compute the gradients for that model.
What does happen is that in a variety of cases, the unrestricted gradients are functions of the restricted gradients, or the restricted gradients are cleverly used in calculation of the LM statistic. Exclusion restrictions are a notable example...Various textbooks (Davidson and MacKinnon is a notable example) devote large sections to the use of the gradient matrix in forming test statistics.
But unfortunately we are not psychic and don't know what unrestricted model you are thinking of so we can't compute the gradients for that model.
What does happen is that in a variety of cases, the unrestricted gradients are functions of the restricted gradients, or the restricted gradients are cleverly used in calculation of the LM statistic. Exclusion restrictions are a notable example...Various textbooks (Davidson and MacKinnon is a notable example) devote large sections to the use of the gradient matrix in forming test statistics.
Re: logl::makegrads, LM test
Thanks!
If I have known the unrestricted (U) and restricted (R) model. Then makegrads from both U and R are improper for LM test. For they are grads from U at estimated value of U, and grads from R at estimated value of R, respectively. Not grads from U at estimated value of R. Right?
If I have known the unrestricted (U) and restricted (R) model. Then makegrads from both U and R are improper for LM test. For they are grads from U at estimated value of U, and grads from R at estimated value of R, respectively. Not grads from U at estimated value of R. Right?
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EViews Glenn
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Re: logl::makegrads, LM test
Correct. We provide the grads for the specified model. There is nothing in the gradients that refers to an alternative model (since you haven't specified one when you compute the gradients). As I note, what does happen is that you have in mind a more general U model which has gradients which are function so the gradients that we provide.
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