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centrally located scores and that uses less weight for more extreme values. This category includes trimmed statistics in which values past a certain point are weighted to 0 and thereby dropped from any estimation procedures. Less drastic approaches include the use of the biweight, in which values are weighted as an exponential function of their distance from the median. A third approach is to reduce the scope of the data one chooses to model on the basis of knowledge about extreme scores and the processes they represent. Dealing with Outliers The goal of EDA is to develop understandings and descriptions of data. This work is always set in some context and always presents itself with some assumptions about the scope of the work, even when these assumptions are unrecognized. Consider, for example, the task described in Behrens and Smith (1996) of developing a model of state-level economic aspects of education in the United States. In this analysis,

(8.95) [Hint: Construct a random variable X such that Pr(X = i) = pi . Let U be a uniform (0,1] random variable and let Y = X + U , where X and U are independent. Use the maximum entropy bound on Y to obtain the bounds in the problem. This bound is due to Massey (unpublished) and Willems (unpublished).] 8.8 Channel with uniformly distributed noise. Consider a additive channel whose input alphabet X = {0, 1, 2} and whose output Y = X+Z, where Z is distributed uniformly over the interval [ 1, 1]. Thus, the input of the channel is a discrete random variable, whereas the output is continuous. Calculate the capacity C = maxp(x) I (X; Y ) of this channel. Gaussian mutual information. Suppose that (X, Y, Z) are jointly Gaussian and that X Y Z forms a Markov chain. Let X and Y have correlation coef cient 1 and let Y and Z have correlation coef cient 2 . Find I (X; Z). Shape of the typical set. Let Xi be i.i.d. f (x), where f (x) = ce x .

0 0 0 0 0

= P(S2 = 0/S1 = 0)P(X 2 = 0/{S1 = 0, S2 = 0})P(Y2 / X 2 = 0) +P(S2 = 0/S1 = 0)P(X 2 = 1/{S1 = 0, S2 = 0})P(Y2 / X 2 = 1) = 0.5 1 0.5 + 0.5 0 0.05 = 0.25 2 (0, 1) = P(S2 = 1/S1 = 0)P(X 2 = 0/{S1 = 0, S2 = 1})P(Y2 / X 2 = 0) +P(S2 = 1/S1 = 0)P(X 2 = 1/{S1 = 0, S2 = 1})P(Y2 / X 2 = 1) = 0 0 0.5 + 0 0 0.05 =0 2 (0, 2) = P(S2 = 2/S1 = 0)P(X 2 = 0/{S1 = 0, S2 = 2})P(Y2 / X 2 = 0) +P(S2 = 2/S1 = 0)P(X 2 = 1/{S1 = 0, S2 = 2})P(Y2 / X 2 = 1) = 0.5 0 0.5 + 0.5 1 0.05 = 0.025 2 (0, 3) = P(S2 = 3/S1 = 0)P(X 2 = 0/{S1 = 0, S2 = 3})P(Y2 / X 2 = 0) +P(S2 = 3/S1 = 0)P(X 2 = 1/{S1 = 0, S2 = 3})P(Y2 / X 2 = 1) = 0 0 0.5 + 0 0 0.05 =0 In the same way, the following values are also determined: 2 (1, 0) = 0 0 0.5 + 0 0 0.05 = 0 2 (1, 1) = 0.5 1 0.5 + 1 0 0.05 = 0.25 2 (1, 2) = 0 0 0.5 + 0 0 0.05 = 0 2 (1, 3) = 0.5 0 0.5 + 0.5 1 0.05 = 0.025 3 (0, 0) = 0.5 1 0.15 + 0.5 0 0.3 = 0.075 3 (0, 1) = 0 0 0.15 + 0 0 0.3 = 0

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