This report demonstrates the consistency and asymptotic normality of minimum-disparity estimators based on conditional density estimates. In particular, it provides L1 consistency results for conditional density estimators and conditional density estimators under homoscedastic model restrictions. It defines a generic formulation of disparity estimates in conditionally-specified models and demon...

Problem 2 (Parallel Channels and Choice of channels) (a) Find the capacity C of the union of two channels (X1, p1(y1|x1),Y1) and (X2, p2(y2|x2),Y2), where at each time, one can send a symbol over channel 1 or channel 2 but not both. Assume that the output alphabets are distinct and do not intersect. Show that 2 = 21 + 22 . Thus 2 is the effective alphabet size of a channel with capacity C. (b) ...

and Applied Analysis 3 where yi t is the state of the ith neuron at timet, ci > 0 denotes the passive decay rate, wij1, wij2, aij , and bij are the interconnection matrices representing the weight coefficients of the neurons, fj · , gj · , and vj · are activation functions, and Ii is an external constant input. The delay kj is a real valued continuous nonnegative function defined on 0, ∞ , whic...

Nathaniel Dean asks the following: Is it possible to find four nonconcyclic points on the parabola y = x2 such that each of the six distances between pairs of points is rational? We demonstrate that there is a correspondence between all rational points satisfying this condition and orbits under a particular group action of rational points on a fiber product of (three copies of) an elliptic surf...

I’ll now sketch a proof. It basically follows by by recalling the definition of the probability of an even conditioned on X2 = a2. We first note that the set of choices for (X 1, X2, 3) given that X2 = a2 is def Sa2 = {(X1, a2, X3) : (X1, a2) ∈ C12 and (a2, X3) ∈ C23} . Conditioning on X2 = a2, we obtain a sample chosen uniformly from Sa2 . Thus, for (a 1, a2, a3) ∈ Sa2 , 1 P [(X1, X2, X3) = (a...

Solution: (a) We show: (∗) If 0 < x < y, then x2 < y2. Indeed, we may multiply the inequality x < y with the positive quantity x and get x2 < xy. We also may multiply x < y with the positive quantity y and get xy < y2. By transitivity, we obtain x2 < y2. From (∗), uniqueness follows: Assuming x2 = y2 = a and x, y > 0, we can rule out x < y (which would imply x2 < y2) and y < x (which would impl...

Solution. The Markov blanket of a node in a undirected graphical model equals the set of its neighbours: MB(x4) = ne(x4) = ne4 = {x1, x5}. This implies, for example, that x4 ⊥ x2, x3 | x1, x5. (e) On which minimal set of variables A do we need to condition to have x1 ⊥ x5 | A? Solution. We first identify all trails from x1 to x5. There are three such trails: (x1, x2, x5), (x1, x3, x2, x5), and ...