Why are there 2 variance formulas equal?

Question

I have found two separate definitions for variance, listed below. Could you please explain why they are equivalent?

i) Variance of y $= \displaystyle \sum_{i=1}^n p_i(y_i - \mu)^2 $

ii) Variance of y $= \displaystyle \left(\sum_{i=1}^n p_i y_i^2\right) - \mu^2$

sum_{i=1}^{n} is supposed to be sigma notation with i=1,2....n — user95087, Sep 16 '13 at 03:16
Hint: $(a-b)^2 = a^2 + b^2 - 2ab$ and $\mu_y = \sum_i p_iy_i$. — Dilip Sarwate, Sep 16 '13 at 03:24
I expanded the first equation but I cannot get it to look like the second one. — user95087, Sep 16 '13 at 03:35

Michael Hardy · Accepted Answer · 2013-09-16T17:53:21.977

3

$$ \begin{align} \sum_{i=1}^n p_i(y_i - \mu)^2 & = \sum_i p_i(y_i^2 - 2 \mu y_i + \mu^2) \\[12pt] & = \left(\sum_i p_i y_i^2\right) -2\mu\left(\sum_i p_i y_i\right) + n\mu^2 \\[12pt] & = \left(\sum_i p_i y_i^2\right) -2\mu(n\mu) + n \mu^2 \end{align} $$ Now do some routine algebraic simplifications.

edited Sep 16 '13 at 17:53

answered Sep 16 '13 at 04:08

Michael Hardy

1

How come $p_i$ did not distribute to 2$uy_i$? – user95087 Sep 16 '13 at 13:21
Typo. Fixed. ${{{}}}$ – Michael Hardy Sep 16 '13 at 17:53

Why are there 2 variance formulas equal?

1 Answers1