Week_3_Practice_Problems_Solutions

# Week_3_Practice_Problems_Solutions - Week 3 Practice...

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Week 3 Practice Problems Solution (Review continuous random variables and their properties + likelihoods for discrete variates) 1. Suppose Y is an exponential random variable with pdf f ( y ) = exp{ y /2}/2, y > 0. Find a) (0 1) PY << exp{ y /2}/2 dy =− exp{ y /2} 0 1 = 0 1 1 exp{ 1/2} 0.3935 b) for any () ( ) Sy PY y => 0 y > P(Y > y) = 1-F(y) = 1-(exp{-y/2}) = exp{-y/2} c) Suppose are 5 independent copies of Y and let 1 ,..., YY 5 Y 15 min{ ,. .., } VY = . Find the pdf of V by first finding PV v > P(V > v)= P (min{ Y 1,. ..., Y 5 } > v ) = P ( Y i > v ) = exp{ v /2} = exp{ ( v /2} i = 1 5 i = 1 5 i = 1 5 = exp{-5v/2} = S(v) F(v) = 1- S(v) = 1 – exp{-5v/2} f ( v ) = F '( v ) = 5exp{ 5 v /2}/2 2. Suppose . ~( 0 , 1 ) ZG a) Find and (1 . 8 7 PZ > ) (| | 2) > P( Z > 1.87) = 0.0307 P( |Z| > 2) = P( Z < -2) + P ( Z > 2) = 0.0228 + 0.0228 0.0456 b) If , find 32 YZ =+ (2 4) < < and (| 2 | 1) > P( 2 < Y < 4) =P ( 2 < 3 + 2Z < 4) = P( -1/2 < Z < 1/2 ) = 0.6915 – 0.3085 0.383 P( |Y-2| > 1) = P( Y-2 > 1) + P( Y-2 < -1) = P( Y > 3) + P( Y < 1) P( Y > 3) = P ( 3 + 2Z > 3) = P( Z > 0) = 0.5 P( Y < 1) = P ( 3 + 2Z < 1) = P( Z < -1 ) 0.1587 P( |Y-2| > 1) = 0.5 + 0.1587 0.6587 c) What is the distribution of 24 * WY = ? Because W is multiplier and scale of Y and Y is multiplier and scale of Z, W is a Gaussian distribution. E(W) = E ( 2 - 4Y) = E( 2 – 4(3 + 2Z)) = E( -10 – 8Z) = -10 – 8E(Z) = -10 Var(W) = Var( 2 – 4(3 + 2Z)) = Var( -10 – 8Z) = -10 +64Var(Z) = -10+64 = -54 W ~ G(-10, 54 )

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Week_3_Practice_Problems_Solutions - Week 3 Practice...

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