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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)

$${F}\mathrm{or}\:{the}\:{random}\:{variable} \\ $$$${x}\:{show}\:{that} \\ $$$$\left.{a}\right){Var}\left({x}\right)={E}^{\mathrm{2}} \left({x}\right)−\left[{E}\left({x}\right)\right]^{\mathrm{2}} \\ $$$$\left.{b}\right){Var}\left({ax}+{b}\right)={a}^{\mathrm{2}} {Var}\left({x}\right) \\ $$$$ \\ $$

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

$${var}\left({x}\right)=\frac{\Sigma{f}\left({x}−\overset{−} {{x}}\right)^{\mathrm{2}} }{{N}} \\ $$$${var}\left({x}\right)=\frac{\Sigma{f}\left[{x}^{\mathrm{2}} −\mathrm{2}{x}\overset{−} {{x}}+\left(\overset{−} {{x}}\right)^{\mathrm{2}} \right]}{{N}} \\ $$$${var}\left({x}\right)=\frac{\Sigma{fx}^{\mathrm{2}} }{{N}}−\mathrm{2}\overset{−} {{x}}\frac{\Sigma{fx}}{{N}}+\left(\overset{−} {{x}}\right)^{\mathrm{2}} \\ $$$${var}\left({x}\right)=\frac{\Sigma{fx}^{\mathrm{2}} }{{N}}−\mathrm{2}\overset{−} {{x}}\overset{−} {{x}}+\left(\overset{−} {{x}}\right)^{\mathrm{2}} \\ $$$${var}\left({x}\right)=\frac{\Sigma{fx}^{\mathrm{2}} }{{N}}−\mathrm{2}\left(\overset{−} {{x}}\right)^{\mathrm{2}} +\left(\overset{−} {{x}}\right)^{\mathrm{2}} \\ $$$${var}\left({x}\right)=\frac{\Sigma{fx}^{\mathrm{2}} }{{N}}−\left(\overset{−} {{x}}\right)^{\mathrm{2}} \\ $$$$\overset{−} {{x}}={E}\left({x}\right) \\ $$$${var}\left({x}\right)={E}\left({x}^{\mathrm{2}} \right)−\left[{E}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${shown} \\ $$$$ \\ $$$${from}\:{var}\left({x}\right)={E}\left({x}^{\mathrm{2}} \right)−\left[{E}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${var}\left({ax}+{b}\right)={E}\left({ax}+{b}\right)^{\mathrm{2}} −\left[{E}\left({ax}+{b}\right)\right]^{\mathrm{2}} \\ $$$$\frac{\Sigma{f}\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{abx}+{b}^{\mathrm{2}} \right)}{{N}}−\left[{a}^{\mathrm{2}} \left[{E}\left({x}\right)\right]^{\mathrm{2}} −\mathrm{2}{abE}\left({x}\right)+{b}^{\mathrm{2}} \right] \\ $$$${simplify} \\ $$$${a}^{\mathrm{2}} {E}\left({x}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} \left[{E}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} {var}\left({x}\right) \\ $$$${shown} \\ $$

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