Second Edition2-5Therefore,FY(y) =ddyFY(y) =fX(y) +fX(-y) =1√2πe-y2+1√2πe-y2=2πe-y2.Thus,EY=∞0y2πe-y2dy=2π∞0e-udu=2π-e-u∞0=2π,whereu=y22.EY2=∞0y22πe-y2dy=2π-ye-y2∞0+∞0e-y2dy=2ππ2= 1.This was done using integration by part withu=yanddv=ye-y2dy. Then Var(Y) = 1-2π.2.12 We have tanx=y/d, therefore tan-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.2.13P(X=k) = (1-p)kp+pk(1-p),k= 1,2, . . .. Therefore,EX=∞k=1k[(1-p)kp+pk(1-p)]=(1-p)p∞k=1k(1-p)k-1+∞k=1kpk-1=(1-p)p1p2+1(1-p)2=1-2p+ 2p2p(1-p).2.14∞0(1-FX(x))dx=∞0P(X > x)dx=∞0∞xfX(y)dydx=∞0y0dxfX(y)dy=∞0yfX(y)dy=EX,where the last equality follows from changing the order of integration.
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