STAT 415 - Homework 1 Solution

STAT 415 - Homework 1 Solution - Homework 1 Solution 4.5-9...

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Homework 1 Solution 4.5-9 a). ( ) () 2 3/4 2 12 0 6 333 (max ) ( ) ( ) 3 444 729 4096 i P X PX x <= == b). 1 2 0 3 () () 3 4 EX x xd x = i 1 22 2 2 0 3 3 5 x = i [] 2 2 1 1 3 () ( ) () 80 Var X Var X E X E X −= So 3 ( ) 2 33 ( ) 2 80 40 EY Var Y Var X Var X =+= × = 4.5-11 a). 123 1 2 3 2 4 5 7 (2 , 2 , 5 ) ) ) (5 ) 461 2 1/ 2 1/ 2 1/ 3 2/3 1/ 6 5/ 6 0.0035 225 X X === ×= ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ b). ( ) ( ) ( ) 1 2 3 11 1 2 8 23 6 EXXX EX EX EX × × × × × = c). 1 2 ( ) ( ) 4 12 7 4 6 33 3 XX X Var Y Var X Var X σσ =+ = + = + = × × + × × = 4.6-4
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4.6-5 4.6-7 4.6-12
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4.6-13 a). 234 2 6 23456789 1 0 11 () ( ) ( ) 46 1 (2344432 ) 24 tttt tt t wX Y tttttttt t M tM t M t e eee e e e eeeeeeee e == + + + + + + =+ + + + + + + + " b). First W can be 2,3,4…10. If W=2, this means that X=1 and Y=1. So the probability that W is one is equal to (1 ,1 )(1 ) ( 1 ) 1 / 2 4 PX Y PY = = = = × = . If W=3, this means that X=1, Y=2 or X=2, Y=1. So the probability that W is 2 is equal to
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STAT 415 - Homework 1 Solution - Homework 1 Solution 4.5-9...

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