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Question Number 192375 by Spillover last updated on 16/May/23

Show that E(Z)=0 and Var(Z)=1 where Z is the standard normal variable

$${Show}\:{that}\:{E}\left({Z}\right)=\mathrm{0}\:\:\:{and}\:{Var}\left({Z}\right)=\mathrm{1}\:{where} \\ $$$${Z}\:{is}\:{the}\:{standard}\:{normal}\:{variable} \\ $$

Answered by mehdee42 last updated on 16/May/23

We know ∵ Z=((x−μ)/σ) & E(kx)=kE(x) & E(x+k)=E(x)+k ; k∈R & Var(kx)=k^2 Var(x) & Var(x+k)=Var(x) ⇒E(Z)=E(((x−μ)/σ))=(1/σ)E(x−μ)=(1/σ)(E(x)−μ)=0 ✓ & Var(Z)=Var(((x−μ)/σ))=(1/σ^2 )Var(x−μ)=(1/σ^2 )Var(x)=1 ✓

$${We}\:{know}\:\because\:\:\:{Z}=\frac{{x}−\mu}{\sigma}\:\:\: \\ $$$$\&\:\:{E}\left({kx}\right)={kE}\left({x}\right)\:\:\&\:\:{E}\left({x}+{k}\right)={E}\left({x}\right)+{k}\:\:;\:{k}\in\mathbb{R} \\ $$$$\:\&\:{Var}\left({kx}\right)={k}^{\mathrm{2}} {Var}\left({x}\right)\:\:\:\&\:\:{Var}\left({x}+{k}\right)={Var}\left({x}\right) \\ $$$$\Rightarrow{E}\left({Z}\right)={E}\left(\frac{{x}−\mu}{\sigma}\right)=\frac{\mathrm{1}}{\sigma}{E}\left({x}−\mu\right)=\frac{\mathrm{1}}{\sigma}\left({E}\left({x}\right)−\mu\right)=\mathrm{0}\:\checkmark \\ $$$$\&\:\:{Var}\left({Z}\right)={Var}\left(\frac{{x}−\mu}{\sigma}\right)=\frac{\mathrm{1}}{\sigma^{\mathrm{2}} }{Var}\left({x}−\mu\right)=\frac{\mathrm{1}}{\sigma^{\mathrm{2}} }{Var}\left({x}\right)=\mathrm{1}\:\checkmark \\ $$

Commented by Spillover last updated on 17/May/23

thanks

$${thanks} \\ $$

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