Divide the sum of squares by N and work with the averages. Let's use the notation:
<X> for the average of X. E.g.:
<X> = ∑(X)/N = Mean
<X^2 - 2X<X> + <X>^2> =
<X^2> - <X>^2
Note that <a X> = a <X> for a constant factor a. In an average like <X <X>>, the inner <X> is a constant when carrying out the outer average, so you can take it out of the outer average sign. So, you have <X <X>> = <X>^2. The average of a constant is, of course, the same constant so e.g. <<X>^2> = <X^2> because once the inner average is carried out it is a constant w.r.t. the outer average.
If you work with averages and use these rules then you can derive the desired result in just one line. If you use summations, you'll tend to re-derive these rules in every step you make, so you'll get a complicated mess.
<(X - <X>)^2> =
<(X - m + m - <X>)^2> =
<(X-m)^2> + <(m - <X>)^2>
+ 2 <X-m><m-<X>>
The last term is zero if the average of X equals the average of the Model.
I don't follow this:
<<X>^2> = <X^2>
Of course, if k is constant and x is variable:
<kx> = k<x>
<k> = k
<k^2> = k^2
<x^2> != <x>^2
Sorry, that was a typo.
I meant to write:
<<X>^2> = <X>^2
I don't follow this at all:
<(X - m + m - <X>)^2> = <(X-m)^2> + <(m - <X>)^2> + 2 <X-m><m-<X>>
Trying to follow your logic, for the left side:
<(X - m + m - <X>)^2>
= <(X - <X>)^2>
= <x^2> - <x>^2
For SSM + SSE:
<(x - m>^2> + <(m - <x>)^2>
= <x^2 - 2xm + m^2> + <m^2 - 2m<x> + <x>^2>
= <x^2> - 2<xm> + <m^2> + <m^2> - 2<m><x> + <x>^2>
= 2<x^2> + 2<m^2> - 2<xm> - 2<m><x>
And I'm stuck...
Answer this Question
Math - ∑(x+y) c. ∑(x+∑(y)) d. ∑x+ ∑y e. ∑(x...
statistics - I have a simple set of 10 data points My ten Data Points 2 3 3 4 5 ...
Calculus - Find a series ∑a_n for which ∑(a_n)^2 converges but ...
Probability - Let N be a geometric r.v. with mean 1/p; let A1,A2,… be a sequence...
Statistics - 3. The formula for finding sample standard deviation is ...
Statistics - Where can I find a proof for: SST = SSM + SSE
Math - Mathematical Induction - 3. Prove by induction that∑_(r=1)^n▒...
Calculus - If a_n >0 and b_n >0 and series ∑ sqrt( (a_n)^2 +(b_n)^2...
Calculus - If a_n does not equal zero for any n>=1 and ∑a_n converges ...
calculus - Given that ∑(n=1 to inf) 1/(n^2) = (pi^2)/6, find the value of...