Assignment3-Solutions

Assignment3-Solutions - 1 Assignment 3, STAT220-Fall 2010...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Assignment 3, STAT220-Fall 2010 Instructor: Kamyar Moshksar Due on Friday, October 29th Problem 1 - Let X Bi(10 , 0 . 1) and Y Bi(5 , 0 . 2) be two independent random variables defined on similar sample spaces. Find P(X < Y) and P(X > Y | Y 2) . Solution- (a)- To compute P(X < Y) , we proceed as follows: P(X < Y) ( a ) = 5 X y =0 P(X < Y | Y = y ) P (Y = y ) = 5 X y =0 P(X < y | Y = y ) P (Y = y ) ( b ) = 5 X y =0 P(X < y ) P (Y = y ) ( c ) = 5 X y =1 P(X < y ) f Y ( y ) . (1) In this expression, ( a ) is by the law of total probability, ( b ) is by independence of X and Y and ( c ) is by the fact that P( X < 0) = 0 . Now, P( X < y ) = y - 1 X x =0 f X ( x ) . (2) By the above two expressions, P(X < Y) = 5 X y =1 y - 1 X x =0 f X ( x ) f Y ( y ) = 5 X y =1 y - 1 X x =0 ± 10 x ² (0 . 1) x (0 . 9) 10 - x ± 5 y ² (0 . 2) y (0 . 8) 5 - y . (3)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 (b)- To compute P(X > Y | Y 2) , P(X > Y | Y 2) = P(X > Y , Y 2) P(Y 2) . (4) To compute the numerator, P(X > Y , Y 2) = 10 X x =0 P(X > Y , Y 2 | X = x )P(X = x ) = 10 X x =0 P(X > Y , Y 2 | X = x ) f X ( x ) = 10 X x =0 P( x > Y , Y 2 | X = x ) f X ( x ) = 10 X x =0 P(2 Y < x ) f X ( x ) = 10 X x =3 P(2 Y < min { x, 6 } ) f X ( x
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/26/2012 for the course ECON 401 taught by Professor Burbidge,john during the Fall '08 term at Waterloo.

Page1 / 6

Assignment3-Solutions - 1 Assignment 3, STAT220-Fall 2010...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online