Mathematical Explanation of The Square Root Law—on page 139 in the Notes
In class I gave an intuitive explanation of why SE of the sum of n draws = n x (SD of the box)
Now here’s the mathematical proof:
!
The SE of the sum is a measure of how far on the average, the sum of n draws is likely to be from
what you’d expect.
The sum of n draws from a box will be around the expected value (n times the average of
the box) give or take some chance error. The likely size of the chance error is called the
SE.
The Square Root Law says that
the SE of the sum of n draws with replacement = n (SD of the box)
Why?
2
First remember that the SD of the box = sqrt
! of the average of the deviations
Now suppose you draw 2 tickets from the box, x1 and x2
x1 = x + "x 1 and x2= x + "x 2 where x means the average of the box, and "x is shorthand for deviation from the average.
!
!
So x1 + x2 =
! !
2 x + "x 1
!
!
+ "x 2
! words,
In other
the
!
! error in a particular instance of the sum is just the sum of the errors
that go into it.
We want to describe the typical size of the sum of the errors, but since the errors can be
positive or negative, taking the average is not helpful, instead we square the errors, then
take the average of the squares, and then take the square root of the average.
The SE of the sum of 2 draws then, is the sqrt of the average of ( "x 1 + "x 2)2. (Here
average means that you’re taking the sum of 2 draws many times and averaging the
squared sum of the error over many times.)
2
Now let’s see what the average of ( "x 1 + "x 2) is.
( "x1 + "x2 )
!
2
2
!
!
2
= ( "x1) + ( "x 2 ) + 2"x1"x 2
!
!
Here’s the key point. Because x1 and x2 are independent the average of Δx1Δx2 is
zero. When x1 happens to come out higher than its mean, that tells you nothing at all
about whether x2 is higher or lower. You know the average of Δx1 is zero (since the
deviations from the average must sum to 0), so the average of Δx1Δx2 is zero. (If the
draws were not independent, this wouldn’t generally be true.)
So now we know
2
2
2
average of ("x1 + "x 2 ) = average of ("x1 ) + average of ("x 2 ) = 2 (SD of the box)2
2
since the average over many draws of ( "x1) 2 = average over many draws of ("x 2 ) =
(SD of the box)2
!
Taking the sqrt of both sides of!the equation above we get:
SQRT of the average of ( "x 1 + "x 2)2! = 2 (SD of the box)
So in general the SE of n draws = n (SD of the box)
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