yjh3210
幼苗
共回答了20个问题采纳率:100% 举报
since Xhas a uniform distribution pdf of fX(x)is fX(x)=1 x∈[0,1] fX(x)=0 otherwise FY(y)=p(Y≤y)=P(aX+b≤y)if a>0, then FY(y)=p(X≤(y-b)/a)=FX((y-b)/a)differentiating FY(y)yields fY(y):fY(y)=dFY(y)/dy=dFX((y-b)/a)/dy=1/a×fX((y-b)/a)=1/a (y-b)/a∈[0,1] →y∈[b,b+a]fY(y)=0 otherwiseifa<0,then FY(y)=p(X≥(y-b)/a)=1-p(x<(y-b)/a)=1-FX(y-b)/a)differentiation in this case gives fY(y)=dFY(y)/dy=-1/a×fX((y-b)a)=1/|a| (y-b)/a∈[0,1]→y∈[b-a,b]fY(y)=0 otherwiseso random variable Y has uniform distribution.
1年前
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