Thus e y 2 x u u2 1 2 therefore e y 2 x

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Unformatted text preview: )2 with probability one, and the inequality is strict whenever L∗ < 0. Example 4.10.2 Suppose (X, Y ) is uniformly distributed over the triangular region with vertices at (−1, 0), (0, 1), and (1, 1), shown in Figure 4.24. (a) Find and sketch the minimum MSE estimator v 1 u !1 0 1 Figure 4.24: Support of fX,Y . of Y given X = u, g ∗ (u) = E [Y |X = u], for all u such that it is well defined, and find the resulting minimum MSE for using g ∗ (X ) = E [Y |X ] to estimate Y. (b) Find and sketch the function L∗ (u) used for minimum MSE linear estimation of Y from X , and find the resulting MSE for using L∗ (X ) to estimate Y. 166 CHAPTER 4. JOINTLY DISTRIBUTED RANDOM VARIABLES Solution: (a) We will first solve this by going through all the usual steps, first finding fX,Y and fX in order to identify the ratio of the two, which is fY |X (v |u). Then E [Y |X = u] is the mean of the conditional density fY |X (v |u) for u fixed. It will then be explained how the answer could be deduced by inspection. The support of fX,Y is the triangular region, which has area 0.5. (Use the formula one half base...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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