solution10

solution10 - Solutions to Homework 10 1. a. P [|X| < 5,...

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1. a. P [ | X | < 5 , Y > 2 , Z 2 2] = P [ | X | < 5] P [ Y > 2] P [ Z 2 2] = P [ - 5 < X < 5](1 - P [ Y 2])(1 - P [ - 2 < Z < 2] = ( F X (5 - ) - F X ( - 5))(1 - F Y (2))(1 - F Z ( 2 - ) + F Z ( - 2)) b. P [ X > 5 , Y < 0 , Z = 1] = P [ X > 5] P [ Y < 0] P [ Z = 1] = (1 - F X (5)) F Y (0 - ))( F Z (1) - F Z (1 - )) c. P [min( X, Y, Z ) > 2] = P [ X > 2 , Y > 2 , Z > 2] = P [ X > 2] P [ Y > 2] P [ Z > 2] = (1 - F X (2))(1 - F Y (2))(1 - F Z (2)) d. P [max( X, Y, Z ) < 6] = P [ X < 6 , Y < 6 , Z < 6] = P [ X < 6] P [ Y < 6] P [ Z < 6] = F X (6 - ) F Y (6 - ) F Z (6 - ) 2. a. Summing up the rows and columns we get the marginal pmf’s i. ii. iii. X Y - 1 0 1 P Y ( y ) - 1 1 6 0 1 6 1 3 0 0 1 3 0 1 3 1 1 6 0 1 6 1 3 P X ( x ) 1 3 1 3 1 3 1 X Y - 1 0 1 P Y ( y ) - 1 1 9 1 9 1 9 1 3 0 1 9 1 9 1 9 1 3 1 1 9 1 9 1 9 1 3 P X ( x ) 1 3 1 3 1 3 1 X Y - 1 0 1 P Y ( y ) - 1 0 0 1 3 1 3 0 0 1 3 0 1 3 1 1 3 0 0 1 3 P X ( x ) 1 3 1 3 1 3 1 b. i
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solution10 - Solutions to Homework 10 1. a. P [|X| &lt; 5,...

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