概率论,均值方差

X$\sim$G(0.1),即其概率函数为p${(1-p)}^{x}$,x=0,1,2,$\cdots$,EX=$\cfrac{1-p}{p}$,VarX=$\cfrac{1-p}{{p}^{2}}$,矩母函数$\cfrac{p}{1-(1-p){e}^{t}}$
Y,${f}_{Y}$(0)=0.3,${f}_{Y}$(x)=$\cfrac{7}{9}$${f}_{X}$(x),x=1,2,3,$\cdots$
求Y的均值和方差
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Math001

赞同来自:

EY=$\sum_{i=1}^{\infty}$i${f}_{Y}$(i)=$\cfrac{7}{9}$$\sum_{i=1}^{\infty}$i${f}_{X}$(i)=$\cfrac{7}{9}$EX=7
VarY=E[${Y}^{2}$] - ${[EY]}^{2}$
E[${Y}^{2}$]=$\sum_{i=1}^{\infty}$${i}^{2}$${f}_{Y}$(i)=$\cfrac{7}{9}$$\sum_{i=1}^{\infty}$${i}^{2}$${f}_{X}$(i)=$\cfrac{7}{9}$E[${X}^{2}$]
VarX=E[${X}^{2}$] - ${[EX]}^{2}$
E[${X}^{2}$]=90+81=171
E[${Y}^{2}$]=133
VarY=84

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