Therefore f y y d dy f y y f x y f x y 1 2 π e y 2 1

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Therefore,FY(y) =ddyFY(y) =fX(y) +fX(-y) =12πe-y2+12πe-y2=r2πe-y2.Thus,EY=Z0yr2πe-y2dy=r2πZ0e-udu=r2πh-e-uflflfl0i=r2π,whereu=y22.EY2=Z0y2r2πe-y2dy=r2π"-ye-y2flflfl0+Z0e-y2dy#=r2πrπ2= 1.This was done using integration by part withu=yanddv=ye-y2dy. ThenV ar(Y) = 1-2π.2.12 We have tanx=y/d, thereforetan-1(y/d) =xandddytan-1(y/d) =11+(y/d)21ddy=dx. Thus,fY(y) =2πd11 + (y/d)2,0< y <.This is the Cauchy distribution restricted to (0,), and the mean is infinite.2.14Z0(1-FX(x))dx=Z0P(X > x)dx=Z0ZxfX(y)dydx=Z0Zy0dxfX(y)dy=Z0yfX(y)dy=EX,where the last equality follows from changing the order of integration.3
2.16 From Exercise 2.14,ET=Z0[ae-λt+ (1-a)e-μt]dt=-ae-λtλ-(1-a)e-μtμflflflflfl0=aλ+1-aμ.2.17 a.Rm03x2dx=m3set=12m= (12)1/3=.794b. The function is symmetric about zero, thereforem= 0 as long as the integral is finite.1πZ-∞11 +x2dx=1πtan-1(x)flflflflfl-∞=1π(π2+π2) = 1.This is the Cauchy pdf.2.23 a. Use Theorem 2.1.8 withA0= 0,A1= (-1,0) andA2= (0,1). Theng1(x) =x2onA1andg2(x) =x2onA2. ThenfY(y) =12y-1/2,0< y <1.b.EY=R10yfY(y)dy=13EY2=R10y2fY(y)dy=15V arY=15-(13)2=445.2.25 a.Y=-Xandg-1(y) =-y. ThusfY(y) =fX(g-1(y))flflflddyg-1(y)flflfl=fX(-y)| -1|=fX(y) foreveryy.b. To show thatMX(t) is symmetric about 0 we must show thatMX(0 +²) =MX(0-²) for all² >0.MX(0 +²)=Z-∞e(0+²)xfX(x)dx=Z0-∞e²xfX(x)dx+Z0e²xfX(x)dx=Z0e²(-x)fX(-x)dx+Z0-∞e²(-x)fX(-x)dx=Z-∞e-²xfX(x)dx=Z-∞e(0-²)xfX(x)dx=MX(0-²).

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