Solution to HW 6 - 22 0.9 23 0.9 24 0.9 × × × × ×...

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Solutions to EE351K homework assignment 6 2. X has a sample space {0,1,2,3,4}, and we list the PMF as in the following table. x 0 1 2 3 4 ( ) X P x 4 1 ( ) 2 4 1 4 ( ) 2 × 4 1 6 ( ) 2 × 4 1 4 ( ) 2 × 4 1 ( ) 2 a. 2 2 4 4 2 4 2 4 2 2 4 1 1 1 1 1 ( ) ( ) 0 ( ) 1 4 ( ) 2 6 ( ) 3 4 ( ) 2 4 ( ) 2 2 2 2 2 X x E X x P x = = × + × × + × × + × × + × = 5 4 4 4 4 2 4 1 1 1 1 1 ( ) ( ) 0 ( ) 1 4 ( ) 2 6 ( ) 3 4 ( ) 2 4 ( ) 2 2 2 2 2 X x E X xP x = = × + × × + × × + × × + × = 2 2 2 ( ) ( ) ( ) 5 4 1 Var X E X E X = - = - = hence 1 x σ = b. 7 (2 1 2 1) (1 3) (1) (2) (3) 8 X X X P X P X P P P - + = = + + = 3. N has the sample space {0,1,2} n 0 1 2 ( ) N P n 0.2 0.7 0.1 a. ( ) ( ) 0 0.2 1 0.7 2 0.1 0.9 N n E N nP n = = × + × + × = b. 2 2 2 2 2 ( ) ( ) 0 0.2 1 0.7 2 0.1 1.1 N n E N n P n = = × + × + × = c. 2 2 2 ( ) ( ) ( ) 1.1 0.9 0.29 Var N E N E N = - = - = d. ( ) 0.29 0.54 N Var N σ = = =
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4. a 1 ( ) (1 ) n N P n p p - = - n is any positive integer b. 21 | ( ) (1 ) n N B P n p p - = - n is any positive integer bigger than 20 c. 21 | 20 20 ( | ) ( ) (1 ) n N B n n E N B nP n n p p - > > = = - 0 1 2 3 21 0.9 0.1 22 0.9 0.1 23 0.9 0.1 24 0.9 0.1 ... = × × + × × + × × + × × + = 0 1 2 3 0.1 (21 0.9 22 0.9 23 0.9 24 0.9 ...) × × + × + × + × + Let’s denote 0 1 2 3 21 0.9 22 0.9 23 0.9 24 0.9 ... S
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Unformatted text preview: 22 0.9 23 0.9 24 0.9 ...) × × + × + × + × + Let’s denote 1 2 3 21 0.9 22 0.9 23 0.9 24 0.9 ... S = × + × + × + × + Multiple both sides by 0.9 1 2 3 4 0.9 21 0.9 22 0.9 23 0.9 24 0.9 ... S = × + × + × + × + Subtraction the second equation from the first, we get 1 2 3 4 0.1 21 0.9 0.9 0.9 0.9 0.9 ... S = × + + + + + 0.9 21 0.9 30 1 0.9 = × + =-Hence ( | ) 0.1 30 E N B S = × = (Tips on sum of series: you might practice on computing b n n a S nr = = ∑ , where a, b, r are constants. Think about different situations when r>1, r=1, r<1; what about b is finite or positive infinity? ) See the next page for solutions to problem 2.16, 2.17...
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