307_assignment_112208KEY

307_assignment_112208KEY - Problem Assignment Math 307/607...

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Unformatted text preview: Problem Assignment Math 307/607 due Thursday December 4 KEY 1. Pedestrian Review: the random variables X , Y have joint density of the form f X ,Y ( x, y ) = c( x + y ), 0 < x, 0 < y, x + 2 y < 2, where c is an unspecified constant. a) Determine c. ANS: 1= ∫ 0 1 2−2 y ∫ 0 c ( x + y ) dxdy = c ∫ 0 1 (2 − 2 y) 2 2 + y ( 2 − 2 y ) dy = c ∫ 2 − 2 ydy = c. Therefore 0 1 c = 1. b) Find the marginal densities f X , fY . ANS: f X ( x ) = ( 2− x ) ∫ 0 2 2 ⎛ 2 − x ⎞ 1 ⎛ 2 − x ⎞ 1 x 3x x + y dy = x ⎜ +⎜ =+− , 0 < x < 2. ⎟ ⎟ 22 8 ⎝ 2 ⎠ 2⎝ 2 ⎠ 2 2 fY ( y ) = 2−2 y ∫ 0 (2 − 2y) x + y dx = 2 + y ( 2 − 2 y ) = 2 − 2 y, 0 < y < 1. c) Find μ X , μY , σ X , σ Y . x x 2 3x3 435 21 ANS: From b) μ X = ∫ + − dx = 1 + − = . Further μY = ∫ 2 y − 2 y 2 dy = 1 − = . Also 22 8 326 33 0 0 x 2 x3 3x 4 4 12 14 211 E(X ) = ∫ + − dx = + 2 − = . And E (Y 2 ) = ∫ 2 y 2 − 2 y 3 dy = − = . So 22 8 3 5 15 326 0 0 2 2 1 2 1 43 1 14 ⎛ 5 ⎞ 168 − 125 43 1 ⎛1⎞ 1 2 . Finally σ Y = − ⎜ ⎟ = so that σ Y = . and σ X = σ = −⎜ ⎟ = = 180 18 15 ⎝ 6 ⎠ 180 180 6 ⎝ 3 ⎠ 18 2 2 2 X d) Find the covariance ANS: E ( XY ) = ∫∫ 0 0 1 2−2 y x y + xy dxdy = ∫ 2 2 0 1 (2 − 2y) 3 3 y (2 − 2y) + 2 2 y2 1 dy = . 5 Therefore Cov ( X , Y ) = 1 5 1 18 − 25 7 −⋅= = − . 5 63 90 90 e) Find the correlation ρ ( X , Y ) . ANS: ρ ( X , Y ) = Cov ( X , Y ) σ XσY =− 7 90 7 10 =− = −.67514. 43 1 5 43 180 18 f) Find the conditional expectation E ( X | Y = .5) . ANS: f X |Y ( x | .5 ) = 1 f X ,Y ( x,.5 ) fY (.5 ) = x + .5 = x + .5, 0 < x < 1. Therefore 1 11 7 E ( X | Y = .5 ) = ∫ x 2 + .5 x dx = + = . 3 4 12 0 NOTE: you should be able to do a)‐f) without conceptual difficulty. They may be tedious, but you should not be stumped at all. 2. Suppose we have random variables X , Y with μ X = 5, μY = 3, σ X = 2, σ Y = 3, and correlation ρ ( X , Y ) = −.6. Find the mean and standard deviation of W = X + 2Y + 1. ANS: μW = E (W ) = E ( X ) + 2 E (Y ) + 1 = 5 + 2 ⋅ 3 + 1 = 12. Var (W ) = Var ( X + 2Y ) = Var ( X ) + 4Cov ( X , Y ) + 4Var (Y ) = 4 + 4 ( −.6 ) ( 2 )( 3) + 36 = 25.6 So σ W = 25.6. 3. If the random variable Z has the normal distribution with mean 0, standard deviation 1, then the moment generating function mZ ( t ) = exp t 2 2 . Taking this as known, use this to find 2 3 4 2 () E ( Z ) , E ( Z ) , E ( Z ) , E ( Z ) . Find Var ( Z ) . ( ) ( ′ ′′ ANS: Since mZ ( t ) = exp t 2 2 we have mZ ( t ) = t exp t 2 2 = tmZ ( t ) , mZ ( t ) = 1 + t 2 mZ ( t ) , ( ′′′ mZ ( t ) = ( 3t + t 3 ) mZ ( t ) , mZ4 ) ( t ) = ( 3 + 6t 2 + t 4 ) mZ ( t ) . Since mZ ( 0 ) = 1, we get ) ( ) ( ′ ′′ ′′′ E ( Z ) = mZ ( 0 ) = 0, E ( Z 2 ) = mZ ( 0 ) = 1, E ( Z 3 ) = mZ ( 0 ) = 0, E ( Z 4 ) = mZ4) ( 0 ) = 3. It follows that 2 σ Z = 3 − 12 = 2. 2 4. If the random variable X has the Poisson distribution with mean λ > 0, then the moment generating function of X is mX ( t ) = exp λ et − 1 . Use this fact to show: if V , W are (( )) independent Poisson random variables with means λV , λW respectively, then V + W is Poisson with mean λV + λW . ANS: mV +W ( t ) = mV ( t ) mW ( t ) = exp λV et − 1 exp λW et − 1 = exp ( λV + λW ) et − 1 . Since this last is the mgf of the Poisson with mean λV + λW , the assertion is proved. 5. Suppose X 1 , X 2 ,… , X 1000 are independent random variables, each uniformly distributed on the unit interval [ 0,1] . Let S = 1000 i =1 (( )) (( )) ( ( )) ∑ X . i (a) Find E ( S ) and Var ( S ) . ANS: By straightforward calculus, the mean and variance of the uniform(0,1) distribution are 1 2 and 1 12 respectively. Since the X i ' s are independent we see easily that E ( S ) = 1000 (1 2 ) = 500 and Var ( S ) = 1000 (1 12 ) = 1000 . 12 (b) Explain why S has an approximately normal distribution. ANS: The central limit theorem tells us that we can calculate probabilities for the sample mean S by using the normal distribution. Since S = 1000 x we can therefore use the normal 1000 distribution to work out probabilities for S. x= (c) Find the middle 95% of likely outcomes of S. ANS: The middle 95% of outcomes of a normal random variable fall in the range μ ± 1.96σ . Applying this to S we place the middle 95% of outcomes as 500 ± 1.96 1000 = 500 ± 17.9. 12 ...
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