Sample Question #264 (statistics)

Show that Var(

*x*)=Var(1-*x*).
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Sample Question #264 (statistics)

Show that Var(*x*)=Var(1-*x*).

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ANSWER

First of all, you should know Var() stands for variance.

This is an easy one. Var(x+a), where a is a constant, is the same as Var(x). This follows from the Var() formula.

Please comment:

Var(x)=(1/N)*sum[(x_i-x_mean)^2]

Var({x+a})=(1/N)*sum[({x+a}_i-{x+a}_mean)^2]

=(1/N)*sum[(x_i+a-(x_mean+a))^2]

=(1/N)*sum[(x_i+a-x_mean-a)^2]

=(1/N)*sum[(x_i-x_mean)^2]

=Var(x)

More precisely, it should be:

Var(x)=(1/N)*sum[(x_i-x_mean)^2]

Var({a-x})=(1/N)*sum[({a-x}_i-{a-x}_mean)^2]

=(1/N)*sum[({-x}_i+a-{-x}_mean-a)^2]

=(1/N)*sum[({-x}_i-{-x}_mean)^2]

=(1/N)*sum[(-({x}_i-{x}_mean))^2]

=(1/N)*sum[({x}_i-{x}_mean)^2]

=(1/N)*sum[(x_i-x_mean)^2]

=Var(x)

I have consulted a clever guy today:

var(x)=E[(x-mean)^2]

var(-x+a)=E[((-x+a)-(-mean+a))^2]

=E[(-x+a+mean-a)^2]

=E[(-x+mean)^2]

=E[(-(x-mean))^2]

=E[(x-mean)^2]

=var(x)

Yes, if you want to work out the math from the definition of Var(), you should use expectations, not sum (which is for sample variance).

You can also memorize the easy formula that Var(x+a)=Var(x), because Var(a)=0 and Cov(x,a)=0.

-brett