# Statistics - proof. Please help?

Prove mathematically what happens with the sample variance (one of the measures pf the variability) when multiplying all values of the data by a nonnegative constant b.

### 2 Answers

- MichaelLv 74 years agoBest Answer
Well,

edit : thanks for Randy's answer, nevertheless he is not correct

Sum [ (x_i)^2 ] = sum [ b^2 x_i^2 = b^2 sum x_i^2.

expresses :

VarX = Sum [ (x_i)^2 ]

which is only correct if E(X) = 0

_______________________________________________

first of all, welcome to this forum from (maybe the only) french guy in the zone !! ;-)

by definition the variance of a RV X is :

VarX = E( [ X - E(X) ]^2 <------------------ E(X) is the mean of X

as the E( . ) is linear :

it can easily been proven that, after expansion we obtain the theorem :

VarX = E(X^2) - E(X)^2

now let's examine Y=bX then,

VarY = E( [bX]^2) - E( [bX] )^2

= E( b^2 X^2 ) - E( bX )^2

= b^2 E(X^2) - [ E(bX) ]^2

= b^2 E(X^2) - [b* E(X) ]^2

= b^2 E(X^2) - b^2* E(X)^2

= b^2 (E(X^2) - E(X)^2 )

= b^2 VarX

conclusion :

Var(bX) = b^2 VarX

comment : the condition b >= 0 is not necessary

hope it' ll help !!

PS: if you want good answers, please do not forget to give Best Answers to one of those who answer.....

to me, or anybody else, or course provided the answer deserves one !

this is to encourage people to answer !! ;-)

- Randy PLv 74 years ago
Apply the definition of variance.

Sum(b*x_i) = b sum(x_i).

Sum [ (b*x_i)^2 ] = sum [ b^2 x_i^2 = b^2 sum x_i^2.

You're in an advanced math course, you shouldn't need help writing down those facts. Now write down the expression for variance of bX and apply those facts.