已知随机变量X~N(0,1),则随机变量Y=2X 1的概率密度fy(Y)?
X~N(0,1), 随机变量Y=2X 1也服从正态分布, EY=2EX 1=1,DY=4DX=4, 所以,fY(y)=[1/√(8π)]e^[-(y-1)^2/8],(-∞ 又解: P(Y≤y)=P(2X 1≤y) =P[X≤(y-1)/2] =[1/√(2π)]∫[-∞,(y-1)/2]e^(-t^2/2)dt 两边对y求导,得, fY(y)=[1/√(8π)]e^[-(y-1)^2/8],(-∞ 倒数第2步或用变量代换,令t=(u-1)/2,得 P(Y≤y)=P(2X 1≤y) =P[X≤(y-1)/2] =[1/√(2π)]∫[-∞,(y-1)/2]e^(-t^2/2)dt =...全部
X~N(0,1), 随机变量Y=2X 1也服从正态分布, EY=2EX 1=1,DY=4DX=4, 所以,fY(y)=[1/√(8π)]e^[-(y-1)^2/8],(-∞ 又解: P(Y≤y)=P(2X 1≤y) =P[X≤(y-1)/2] =[1/√(2π)]∫[-∞,(y-1)/2]e^(-t^2/2)dt 两边对y求导,得, fY(y)=[1/√(8π)]e^[-(y-1)^2/8],(-∞ 倒数第2步或用变量代换,令t=(u-1)/2,得 P(Y≤y)=P(2X 1≤y) =P[X≤(y-1)/2] =[1/√(2π)]∫[-∞,(y-1)/2]e^(-t^2/2)dt =[1/√(8π)]∫[-∞,y]e^[-(u-1)^2/8]du fY(y)=[1/√(8π)]e^[-(y-1)^2/8],(-∞ 收起