Maths - Standard Deviation

anpanman SG

I want to know the formula for Standard Deviation derived. Please help.

Let E represent the sign "sigma", and x- be the sign for average since I can't insert the horizontal line on top of x.

I will exclude the square root in this case, to simplify matters.

E(x - (x-))^2 / N = Ex^2 /N - (Ex / N)^2 (prove this)

Here goes:

E(x^2 - 2x(x-) + (x-)^2) / N

= Ex^2 /N - E 2x(x-) / N + E(x-)^2 / N

= Ex^2 / N - 2(x-)(Ex/N) + N(x-)^2 / N

=Ex^2 - 2(x-)^2 + (x-)^2

=Ex^2 / N - (Ex / N)^2

I don't understand why the E(x-) is changed to N(x-) over at the 3rd statement. This is my teacher's explanation. Appeal to all for help because I really want to know about this! Thanks!

c0mplex SG

wow, your notations got me confused at first... i was thinking your 'sigma' means 'standard deviation'... after staring at the formulae, i realised it 'big sigma' instead of 'small sigma'... anyway, below is the explanation, using your notations...

E(x-)^2/N=[(x-)^2E(1)]/N since x- can be 'taken out'

= [(x-)^2][N]/N since E(1)=N

anpanman SG

E(x-)^2/N=[(x-)^2E(1)]/N since x- can be 'taken out'

= [(x-)^2][N]/N since E(1)=N

why is it that E(1) = N?

By the way, can you tell me what the (1) means? i'm kind of confused by this all.

c0mplex SG

why is it that E(1) = N?

By the way, can you tell me what the (1) means? i'm kind of confused by this all.

ohh... the (1) is just the integer 1...

so E(1) means you are summing the integer 1 from n=1 to n=N, or whatever your dummy variable is... since you sum the integer 1 N times, the total will be N.