ECE302HW3Soln_Sp10

# ECE302HW3Soln_Sp10 - ECE 302 Homework 3 Prof Saul Gelfand...

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Unformatted text preview: ECE 302: Homework 3 Prof. Saul Gelfand Spring 2010 Due February 19, 2010 Problem 1 a. First we find the constant c by noting that integraldisplay ∞ −∞ f X ( x ) dx = integraldisplay ∞ − 1 ce − 2 x dx = c 2 e 2 = 1 . Therefore c = 2 e − 2 and f X ( x ) = 2 e − 2 x − 2 . Now since Y = X 2 ⇒ X = √ Y , we compute We then have that the CDF F Y ( y ) = Pr [ Y < y ] = Pr [ X 2 < y ] = Pr [- √ y < X < √ y ] for the case that y < 1, we have F Y ( y ) = Pr [- √ y < X < √ y ] = integraldisplay √ y − √ y f X ( x ) dx = integraldisplay √ y − √ y 2 e − 2 x − 2 dx = e 2 √ y − 2- e − 2 √ y − 2 . For the case that y ≥ 1, we have F Y ( y ) = Pr [- 1 < X < √ y ] = integraldisplay √ y − 1 f X ( x ) dx = integraldisplay √ y − 1 2 e − 2 x − 2 dx = 1- e − 2 √ y − 2 . 1 Therefore F Y ( y ) = braceleftbigg e 2 √ y − 2- e − 2 √ y − 2 , < y < 1 1- e − 2 √ y − 2 , y ≥ 1 ⇒ f Y ( y ) = d dy F Y ( y ) = , y ≤ exp(2 √ y − 2) √ y + exp( − 2 √ y − 2) √ y , < y < 1 exp( − 2 √ y − 2) √ y , y ≥ 1 We find Pr [ Y > 2 X ] = Pr [ X 2 > 2 X ] = Pr [ X > 2] as an integral of the PDF f X ( x ) as follows Pr [ X > 2] = integraldisplay ∞ 2 f X ( x ) dx = integraldisplay ∞ 2 2 e − 2 x dx = e − 4 b. We again find c by noting ∞ summationdisplay i = −∞ p X ( x i ) = ∞ summationdisplay x i = − 1 c parenleftbigg 1 2 parenrightbigg x i +1 = 1 ⇒ (let k = x i + 1) ∞ summationdisplay k =0 c parenleftbigg 1 2 parenrightbigg k = 2 c = 1 , so that c = 1 2 and p X ( x i ) = ( 1 2 ) x i +2 . We then have that p Y ( y ) = Pr [ Y = y ] = Pr [ X 2 = y ] We have that for y = 1, there are two values of X ( X = 1 ,- 1) such that X 2 = y = 1. Therefore p Y (1) = Pr [ Y = 1] = Pr [ X = 1] + Pr [ X =- 1] = p X (1) + p X (- 1) = parenleftbigg 1 2 parenrightbigg 3 + parenleftbigg 1 2 parenrightbigg = 5 8 . For all other values of X = 0 , 2 , 3 , 4 , . . . , we have only one value of Y which is mapped to, therefore p Y ( y ) = Pr [ X 2 = y ] = Pr [ X = √ y ] = p X ( √ y ) = parenleftbigg 1 2 parenrightbigg √ y +2 , for y = 0 , √ 2 , √ 3 , . . . . 2-b-a a b-b b X Y = g(X) Figure 1: Problem 2 Problem 2 We have the limiter function Y = g ( X ) plotted as shown in Figure 1. We let a = 1 and b = 2 in this problem....
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