solution08

# solution08 - Solutions to Homework 08 1 inf ty E[X = k=0...

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Solutions to Homework 08 1. E [ X ] = inf ty X k =0 kP [ X = k ] = inf ty X k =0 k e - α k ! = α inf ty X k =1 α k - 1 e - α ( k - 1)! = α inf ty X i =0 α i e - α i ! = α 2. Y = A cos( ωt ) + c E [ Y ] = E [ A cos( ωt ) + c ] = Z -∞ ( A cos( ωt ) + c ) f A ( a ) da = cos( ωt ) Z -∞ Af A ( a ) da + c Z -∞ f A ( a ) da = cos( ωt ) E [ A ] + c = m cos( ωt ) + c V AR [ Y ] = E [( Y - E [ Y ]) 2 ] = Z -∞ ( A cos( ωt ) + c - m cos( ωt ) - c ) 2 f A ( a ) da = Z -∞ ( A - m ) 2 cos 2 ( ωt ) f A ( a ) da = cos 2 ( ωt ) Z -∞ ( A - m ) 2 f A ( a ) da = cos 2 ( ωt ) V AR [ A ] = σ 2 cos 2 ( ωt ) 3. Y = ba X E [ Y ] = E [ ba X ] = X i =0 ba x e - α α x x ! 1

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= be - α X i =0 ( ) x x ! = be - α e = be α ( a - 1) 4. a. E [ X ] = Z -∞ xf X ( x ) dx = Z 1 - 1 x 3 x 2 2 dx = 3 2 Z 1 - 1 x 3 = 3 2 " x 4 4 # 1 - 1 = 0 b. E [ X 2 ] = Z -∞ x 2 f X ( x ) dx = Z 1 - 1 x 2 3 x 2 2 dx = 3 2 Z 1 - 1 x 4 = 3 2 " x 5 5 # 1 - 1 = 3 5 c. V AR [ X ] = E [ X 2 ] - E [ X ] 2 = 3 5 - 0 = 3 5 d. standard deviation = q V AR [ X ] = s 3 5 = 0 . 7746 5. The call duration is exponentially distributed with parameter
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## This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue.

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solution08 - Solutions to Homework 08 1 inf ty E[X = k=0...

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