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Question Number 50855 by peter frank last updated on 21/Dec/18
For the random variable x show that a)Var(x)=E^2 (x)−[E(x)]^2 b)Var(ax+b)=a^2 Var(x)
Fortherandomvariablexshowthata)Var(x)=E2(x)−[E(x)]2b)Var(ax+b)=a2Var(x)
Answered by peter frank last updated on 21/Dec/18
var(x)=((Σf(x−x^− )^2 )/N) var(x)=((Σf[x^2 −2xx^− +(x^− )^2 ])/N) var(x)=((Σfx^2 )/N)−2x^− ((Σfx)/N)+(x^− )^2 var(x)=((Σfx^2 )/N)−2x^− x^− +(x^− )^2 var(x)=((Σfx^2 )/N)−2(x^− )^2 +(x^− )^2 var(x)=((Σfx^2 )/N)−(x^− )^2 x^− =E(x) var(x)=E(x^2 )−[E(x)]^2 shown from var(x)=E(x^2 )−[E(x)]^2 var(ax+b)=E(ax+b)^2 −[E(ax+b)]^2 ((Σf(a^2 x^2 +2abx+b^2 ))/N)−[a^2 [E(x)]^2 −2abE(x)+b^2 ] simplify a^2 E(x^2 )−a^2 [E(x)]^2 a^2 var(x) shown
var(x)=Σf(x−x−)2Nvar(x)=Σf[x2−2xx−+(x−)2]Nvar(x)=Σfx2N−2x−ΣfxN+(x−)2var(x)=Σfx2N−2x−x−+(x−)2var(x)=Σfx2N−2(x−)2+(x−)2var(x)=Σfx2N−(x−)2x−=E(x)var(x)=E(x2)−[E(x)]2shownfromvar(x)=E(x2)−[E(x)]2var(ax+b)=E(ax+b)2−[E(ax+b)]2Σf(a2x2+2abx+b2)N−[a2[E(x)]2−2abE(x)+b2]simplifya2E(x2)−a2[E(x)]2a2var(x)shown

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