Question

Brain fart:

I really don't know why I'm getting the wrong answer when doing the algebra. Can somebody please show the work for this equivalency? http://upload.wikimedia.org/math/3/0/e/30eb0964923e7b3594319b939dae9afa.png

You can skip steps. I am probably just making a silly mistake (aware that mu = sigma(x_i)/n).

Answer 1

what are we doing exactly?

Answer 2

oh

Answer 3

you want to show that equality is true?

Answer 4

ya. i'm getting the wrong answer when expanding the left side

Answer 5

\[\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2 \\ \frac{1}{N} \sum_{i=1}^N(x_i^2-2x_i \mu+\mu^2)\]
are you cool with this so far?

Answer 6

I expanded the square thingy

Answer 7

(A-B)^2=A^2-2AB+B^2

Answer 8

you know since (A-B)^2=(A-B)(A-B)

Answer 9

yes. so far so good

Answer 10

\[\frac{1}{N} \sum_{i=1}^N x_i^2 -\frac{1}{N} \sum_{i=1}^{N}(2 x_i \mu +\mu^2) \]
so it looks like we need to somehow show
the following:
\[\frac{-1}{N} \sum_{i=1}^N(2x_i \mu +\mu^2) \text{ is equal to } -\mu^2 \]

Answer 11

this is what i keep getting: \[\frac{ \sum_{i = 1}^{N} (X_i^2 - \mu^2)}{ N }-2\mu ^2\]

Answer 12

my bad. should be x_i^2 + mu^2 in bracket

Answer 13

recall the following:
\[\mu=\frac{1}{N} \sum_{i=1}^{N}x_i\]

Answer 14

yes

Answer 15

mistake in:
\[\frac{1}{N} \sum_{i=1}^N x_i^2 -\frac{1}{N} \sum_{i=1}^{N}(2 x_i \mu +\mu^2)\]
oops I made a mistake earlier in this distribution
so we should have:
\[\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2 \\ \frac{1}{N} \sum_{i=1}^N(x_i^2-2x_i \mu+\mu^2) \\ \frac{1}{N} \sum_{i=1}^{N}x_i^2 -\frac{1}{N} \sum_{i=1}^N 2 x_i \mu + \frac{1}{N} \sum_{i=1}^N \mu^2 \\ \frac{1}{N} \sum_{i=1}^N x_i^2 -2 \mu \frac{1}{N} \sum_{i=1}^N \mu +\frac{1}{N} \sum_{i=1}^N \mu^2 \]
guess what we can rewrite that middle term

Answer 16

\[\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2 \\ \frac{1}{N} \sum_{i=1}^N(x_i^2-2x_i \mu+\mu^2) \\ \frac{1}{N} \sum_{i=1}^{N}x_i^2 -\frac{1}{N} \sum_{i=1}^N 2 x_i \mu + \frac{1}{N} \sum_{i=1}^N \mu^2 \\ \frac{1}{N} \sum_{i=1}^N x_i^2 -2 \mu \frac{1}{N} \sum_{i=1}^N \mu +\frac{1}{N} \sum_{i=1}^N \mu^2 \\ \frac{1}{N} \sum_{i=1}^N x_i^2-2 \mu \cdot \mu +\frac{1}{N} \cdot N \mu^2 \]
I went ahead and rewrote the last term as well

Answer 17

can you simplify it from there

Answer 18

oops that one thing in the sum is x_i
type-o

Answer 19

but yeah the last line is correct
the line before it i accidentally put mu in the sum instead of x_i

Answer 20

thank you!

my mistake was that that sigma mu = N*mu . i left it as mu

Answer 21

\[\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2 \\ \frac{1}{N} \sum_{i=1}^N(x_i^2-2x_i \mu+\mu^2) \\ \frac{1}{N} \sum_{i=1}^{N}x_i^2 -\frac{1}{N} \sum_{i=1}^N 2 x_i \mu + \frac{1}{N} \sum_{i=1}^N \mu^2 \\ \frac{1}{N} \sum_{i=1}^N x_i^2 -2 \mu \frac{1}{N} \sum_{i=1}^N x_i +\frac{1}{N} \sum_{i=1}^N \mu^2 \\ \frac{1}{N} \sum_{i=1}^N x_i^2-2 \mu \cdot \mu +\frac{1}{N} \cdot N \mu^2 \]

Answer 22

yeah pretend c is a constant
(and mu is in this situation so mu^2 is also)
\[\sum_{i=1}^{n}c=nc\]

Answer 23

but anyways np :)

Answer 24

I can show you why that is if you like :)
pretend f(i)=c
\[\sum_{i=1}^{n}c=\sum_{i=1}^{n}f(i)=f(1)+f(2)+f(3)+f(4)+\cdots + f(n) \\ =c+c+c+c + \cdots +c =nc \\ \text{ since there are a } n \text{ amount of } c \text{ 's }\]

Answer 25

anyways peace