ns know that $\operatornameVar=\operatornameCov=a^2\operatornameVar+2ab\operatornameCov+b^2\operatornameVar$ (by broadening $(ax+by)(ax+by)$ and lettinns $\operatornameVar=x^2$-terms, $\operatornameVar=y^2$-state and $\operatornameCov=xy$-terms).

You are watching: Var(ax+by)

but what happens when we"ve gained a $x$-hatchet or a $y-$hatchet top top its own?

e.g. Say we desire $\operatornameVar$.

then us have $(x+y+1)^2=x^2+2xy+2x+y^2+2y+1$, so:$$\operatornameVar=\operatornameVar+2\operatornameCov+2?+\operatornameVar+2?+1.$$ can someone fill in ns blanks?

Thanks.

probcapability statisticns random-variables
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edited may 21 "14 in ~ 16:12

Hakns
request may 21 "14 at 16:09

beep-boopbeep-boop
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include a commenns |

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girlfriend almost acquired it other than the $$\textCov(aX, bY+c) = \textCov(aX, bY)+\textCov(aX,c) = Abdominal \textCov(X,Y) +0$$Since covariance of a arbitrarily variable through a consistent ins $0$.

http://www.kaspercpa.com/statisticalreview.htm

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answered might 21 "14 in ~ 16:17

AlexAlex
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If all friend desire ins the variance, getting it via the covariance recipe the means you"re doinns ins a lons even more complex 보다 it needs come be. Friend simply observe: $$\rm Var = \rm Var,$$ Because $\rm Var = \rm Var$ for any consistent $c$. Then girlfriend have the right to usage the covariance formula.

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answered might 21 "14 at 16:25

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