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# midsoln2 - 332:321 Probability and Stochastic Processes...

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Unformatted text preview: 332:321 Probability and Stochastic Processes Examination 2 November 22, 2004 SOLUTION (version 1) You have 80 minutes to answer the following questions in the notebooks provided. You are permitted 1 double-sided sheet of notes. Make sure that you have included your name, Rutgers netid and signature in each book used. (5 points) Read each question carefully. All statements must be justified. Computations should be simplified as much as possible. The following table may be helpful: x . . 25 . 5 . 75 1 . 1 . 25 1 . 50 1 . 75 2 . 2 . 25 2 . 50 8( x ) . 50 . 599 . 692 . 773 . 841 . 894 . 933 . 960 . 977 . 988 . 994 1. 40 points SHORT ANSWER: Each part is a separate problem. (a) Y is a Gaussian (μ = 5 , σ = 10 ) random variable. Calculate P [ Y > 20]. P [ Y > 20] = 1- P [ Y ≤ 20] = 1- P Y- 5 10 ≤ 20- 5 10 = 1- 8( 1 . 5 ) = 1- . 933 = . 067 (b) X is a Gaussian (μ = , σ = 1 ) random variable. Z is a Gaussian ( , 4 ) random variable. X and Z are independent. Let Y = X + Z . Find the correlation coefficient ρ of X and Y . Are X and Y independent? Independence of X and Z implies Var[ Y ] = Var[ X ] + Var[ Z ] = 1 2 + 4 2 = 17 . Since E [ X ] = E [ Y ] = , the covariance of X and Y is Cov [ X , Y ] = E [ XY ] = E [ X ( X + Z ) ] = E X 2 + E [ X Z ] . Since X and Z are independent, E [ X Z ] = E [ X ] E [ Z ] = . Since E [ X ] = , E X 2 = Var[ X ] = 1 . Thus Cov [ X , Y ] = 1 . Finally, the correlation coefficient is ρ X , Y = Cov [ X , Y ] √ Var[ X ] √ Var[ Y ] = 1 √ 17 = . 243 . Since ρ X , Y 6= , we conclude that X and Y are dependent. (c) X 1 , X 2 and X 3 are independent identically distributed (iid) continuous uniform random vari- ables. Random variable Y = X 1 + X 2 + X 3 has expected value E [ Y ] = 0 and variance σ 2 Y = 4. What is the PDF f X 1 ( x ) of X 1 ? Each X i has PDF identical to a random variable X. First we observe that E [ Y ] = 3 E [ X ] = , implying E [ X ] = . Second, we observe that since the X i are independent, Y has variance Var[ Y ] = Var[ X 1 ] + Var[ X 2 ] + Var[ X 3 ] = 3 Var[ X ] . Thus Var[ X ] = σ 2 Y / 3 = 4 / 3 . Finally, since X is a continuous uniform random variable, we need to find the parameters ( a , b ) which satisfy E [ X ] = b + a 2 = , Var[ X ] = ( b- a ) 2 12 = 4 3 . Thus b + a = and b- a = 4 , implying b = 2 and a = - 2 . The PDF of X 1 is thus f X 1 ( x ) = f X ( x ) = 1 / 4- 2 ≤ x ≤ 2 , otherwise . (d) X is a continuous uniform ( , 1 ) random variable. Given X = x , Y is conditionally a continuous uniform ( , 1 + x ) random variable. What is the joint PDF f X , Y ( x , y ) of X and Y ? This problem is mostly about translating words to math. From the words, we learned that f X ( x ) = 1 ≤ x ≤ 1 , otherwise , f Y | X ( y | x ) = 1 /( 1 + x ) ≤ y ≤ 1 + x , otherwise ....
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midsoln2 - 332:321 Probability and Stochastic Processes...

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