Hw5sol - STAT 401 Homework 5 Solutions(Spring 2009 4.3.5 1(a P(X ≤ 1 Y ≤ 1 = 0.1 0.04 0.08 0.2 = 0.42(b A Pmf of X X=x 0 1 2 P(X = x 0.16 0.34

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STAT 401: Homework 5 Solutions (Spring 2009) 4.3.5 1. (a) P ( X 1 ,Y 1) = 0 . 1 + 0 . 04 + 0 . 08 + 0 . 2 = 0 . 42 (b) A. Pmf of X X = x 0 1 2 P ( X = x ) 0.16 0.34 0.5 B. Pmf of Y Y = y 0 1 2 P ( Y = y ) 0.24 0.38 0.38 (c) E ( X ) = 0 × 0 . 16 + 1 × 0 . 34 + 2 × 0 . 5 = 1 . 34 V ar ( X ) = E ( X 2 ) - E ( X ) 2 = 2 . 34 - 1 . 34 2 = 0 . 544 E ( Y ) = 0 × 0 . 24 + 1 × 0 . 38 + 2 × 0 . 38 = 1 . 14 V ar ( Y ) = E ( Y 2 ) - E ( Y ) 2 = 0 × 0 . 24 + 1 × 0 . 38 + 4 × 0 . 38 - 1 . 14 2 = 0 . 6004 3. (a) Dependent since conditional distribution changes as given variable changes. (b) p ( x,y ) = p Y | X ( y | x ) · p ( x ) y 0 1 x 0 0.3726 0.1674 1 0.1445 0.0255 2 0.2436 0.0464 (c) Marginal distribution of Y is, y 0 1 p(y) 0.7607 0.2393 (d) P ( X = 1 ,Y = 1) = 0 . 0255 6 = P ( X = 1) · P ( Y = 1) = 0 . 17 × 0 . 2393 = 0 . 040681 X and Y are not independent. 4. (a) Pmf of X X = x 400 500 600 P ( X = x ) 0.3 0.5 0.2 P ( Y = 1 | X = x ) = p ( x ) , P ( Y = 0 | X = x ) = 1 - p ( x ), so we can compute a conditional table for P ( Y | X = x ) y 1 0 x 400 0.1437 0.8563 500 0.1296 0.8704 600 0.0895 0.9105 The joint distribution is computed by using p ( x,y ) = P ( Y = y | X = x ) * P ( X = x ) and is given in the table below, y 1 0 x 400 0.0431 0.2569 500 0.0648 0.4352 600 0.0179 0.1821 (b) P ( Y = 1) = p (1 , 400) + p (1 , 500) + p (1 , 600) = 0 . 0431 + 0 . 0648 + 0 . 0179 = 0 . 1258, P ( Y = 0) = 1 - P ( Y = 1) = 0 . 8742. (c) E ( X ) = 400 * 0 . 3 + 500 * 0 . 5 + 600 * 0 . 2 = 490 We need to compute P ( X = x | Y = y ) = p ( x,y ) /P ( Y = y ) which is given in the table below, y 1 0 x 400 0.3426 0.2939 500 0.5151 0.4978 600 0.1422 0.2083 E ( X | Y = 1) = 400 * 0 . 3426 + 500 * 0 . 5151 + 600 * 0 . 1422 = 479 . 91 E ( X | Y = 0) = 400 * 0 . 2939 + 500 * 0 . 4978 + 600 * 0 . 2083 = 491 . 44 1
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4.4.5 1. (a) R -∞ R -∞ f ( x,y ) dxdy = 1 = R 2 0 R 3 x kxy 2 dydx = k R 2 0 ± xy 3 3 ² ² ² 3 x ³ dx == k R 2 0 h 9 x - x 4 3 i dx = k ´ 9 x 3 2 - x 5 15 ² ² ² 2 0 µ = 238 15 k = 1 = k = 15 238 = 0 . 0630 (b) f X ( x ) = R -∞ f ( x,y ) dy = R 3 x kxy 2 dy = kx · y 3 3 ² ² ² 3 x = h 9 kx - kx 4 3 i = (0 . 567 x - 0 . 0210 x 4 )(0 x 2) f Y ( y ) = R
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This note was uploaded on 10/08/2009 for the course STAT 401 taught by Professor Akritas during the Spring '00 term at Pennsylvania State University, University Park.

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Hw5sol - STAT 401 Homework 5 Solutions(Spring 2009 4.3.5 1(a P(X ≤ 1 Y ≤ 1 = 0.1 0.04 0.08 0.2 = 0.42(b A Pmf of X X=x 0 1 2 P(X = x 0.16 0.34

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