By Creel M.

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Given some regularity conditions on boundedness of Dθ f , we can switch the order of integration and differentiation, by the dominated convergence theorem. This gives Eθ [gt (θ)] = Dθ Z f (yt |xt , θ)dyt = Dθ 1 = 0 where we use the fact that the integral of the density is 1. • So Eθ (gt (θ) = 0 : the expectation of the score vector is zero. • This hold for all t, so it implies that Eθ gn (Y, θ) = 0. 4. 4. Asymptotic normality of MLE Recall that we assume that sn (θ) is twice continuously differentiable.

Note that the data are sorted by output level (the third column). 1) using OLS. err. 339 t-stat. 518 ********************************************************* • Do the theoretical restrictions hold? • Does the model fit well? • What do you think about RTS? While we will use Octave programs as examples in this document, since following the programming statements is a useful way of learning how theory is put into practice, 3If you are running the bootable CD, you have all of this installed and ready to run.

Let y = 1(heads) be a binary variable that indicates whether or not a heads is observed. 1. THE LIKELIHOOD FUNCTION 57 The derivative of this is ∂ ln fY (y, p) = ∂p = y (1 − y) − p (1 − p) y− p p (1 − p) Averaging this over a sample of size n gives ∂sn (p) 1 n yi − p = ∑ ∂p n i=1 p (1 − p) Setting to zero and solving gives pˆ = y¯ So it’s easy to calculate the MLE of p0 in this case. Now imagine that we had a bag full of bent coins, each bent around a sphere of a different radius (with the head pointing to the outside of the sphere).