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Question Number 85859 by jagoll last updated on 25/Mar/20
$$\:\mathrm{Is}\:\mathrm{the}\:\mathrm{Var}\left(\mathrm{aX}+\mathrm{b}\right)\:=\:\mathrm{a}^{\mathrm{2}} \:\mathrm{Var}\left(\mathrm{X}\right)\:+\:\mathrm{b}? \\ $$
Answered by Joel578 last updated on 25/Mar/20
![nope Assume a and b is constant, then var(b) = E[b^2 ] − {E[b]}^2 = b^2 − b^2 = 0 var(aX + b) = var(aX) + var(b) = a^2 var(X)](Q85863.png)
$${nope} \\ $$$$\mathrm{Assume}\:{a}\:\mathrm{and}\:{b}\:\mathrm{is}\:\mathrm{constant},\:\mathrm{then} \\ $$$$\mathrm{var}\left({b}\right)\:=\:{E}\left[{b}^{\mathrm{2}} \right]\:−\:\left\{{E}\left[{b}\right]\right\}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{var}\left({aX}\:+\:{b}\right)\:=\:\mathrm{var}\left({aX}\right)\:+\:\mathrm{var}\left({b}\right)\:=\:{a}^{\mathrm{2}} \mathrm{var}\left({X}\right) \\ $$
Commented by jagoll last updated on 26/Mar/20
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{agree} \\ $$
Answered by jagoll last updated on 26/Mar/20
