Wrap up functions of one or more random variables the

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Unformatted text preview: ution on an interval of length 1−|u| , which is (1−|u|) . Averaging 2 48 over u using the pdf fX yields that 0 MSE for g ∗ (X ) = (1 + u) −1 1 =2 0 1 (1 − |u|)2 du + 48 (1 − u) 0 (1 − |u|)2 du 48 1 (1 − |u|)2 du = . (1 − u) 48 96 (b) Finding L∗ (u) requires calculating some moments. By LOTUS: ∞ ∞ vfX,Y (u, v )dudv E [Y ] = −∞ −∞ 1 2v −1 = 2vdudv v −1 0 1 = 0 1 2 2v 2 dv = , 3 2v −1 E [XY ] = 2uvdudv v −1 0 1 1 v [(2v − 1)2 − (v − 1)2 ]dv = = 0 3v 3 − 2v 2 dv = 0 1 E [X 2 ] = 2 0 1 , 12 1 u2 (1 − u)du = . 6 1 A glance at fX shows E [X ] = 0, so Var(X ) = E [X 2 ] = 6 and Cov(X, Y ) = E [XY ] = by (4.32), 2 1/12 2u L∗ (u) = + u= + . 3 1/6 32 1 12 . Therefore, This estimator is shown in Figure 4.25(b). While the exact calculation of L∗ was tedious, the graph of L∗ could have been drawn approximately by inspection. It is a straight line that tries to be close to E [Y |X = u] for all u. To find the MSE we shall use (4.35). By LOTUS, 1 2v −1 E [Y 2 ] = 2v 2 dudv v −1 0 1 = 0 1 2v 3 dv = . 2 168 CHAPTER 4. JOINTLY DISTRIBUTED RANDOM VARIABLES and E [L∗ (X )2 ] = E...
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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 Tech.

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