已知随机变量ξ=0,1,2且P(ξ=0)=m/3,P(ξ=1)=m/3,则Dξ的最大值是?25/36

Question

已知随机变量ξ=0,1,2且P(ξ=0)=m/3 ,P(ξ=1)=m/3 ,则Dξ的最大值是?
25/36

Answer 1

由题意知,P(ξ=2)=1-2m/3所以Eξ=0*m/3+1*m/3+2*(1-2m/3)=2-m所以Dξ=(2-m)^2*m/3+(1-m)^2*m/3+m^2*(1-2m/3)=m^3/3-4m^2/3+4m/3+m^3/3-2m^2/3+m/3+m^2-2m^3/3=-m^2+5m/3=-(m-5/6)^2+25/36由于m可以取到5/6所以Dξ的...