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January 13th, 2015, 11:31 AM   #1
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Standard Deviation Formulas - Short Proof

Today I learnt that:

$\displaystyle Variance=\frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } =\frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 }$

Considering that the formula for the mean in the United Kingdom is:

$\displaystyle \overline { x } =\frac { \Sigma x }{ n } $

Now, I wasn't told why the formulas posted above were equal to one another. I was just supposed to accept such truths without any evidence.

So, to satisfy such claims - I've made a short little proof as to why the two formulas at the top of your screen ARE in fact equal to one another.

If you have any better proofs, then please feel free to post them on this thread. For the meantime, this is what I've achieved...

----------------------------------------------------

PROOF:

Say that:

$\displaystyle For\quad convenience,\\ \\ d=a+b+c,\\ \\ \therefore \quad \frac { \Sigma x }{ n } =\frac { a+b+c }{ n } =\frac { d }{ n } =\overline { x } \\ \\ However,\quad \frac { d }{ n } =p,\quad \therefore \quad \overline { x } =p,\quad d=np\quad and\quad n=\frac { d }{ p } .\\ \\ Note\quad that:\quad n=3\quad in\quad this\quad case.$

Also note that:

$\displaystyle { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=g,\\ \\ \therefore \quad \frac { \Sigma { x }^{ 2 } }{ n } =\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ n } =\frac { g }{ n } \\ \\ However,\quad \frac { g }{ n } =q,\quad \therefore \quad g=nq\quad and\quad n=\frac { g }{ q } .$

Now, with these set of rules, we should reach the conclusion that:

$\displaystyle \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 }=\frac { g }{ n } -{ \left( \frac { d }{ n } \right) }^{ 2 }=q-{ p }^{ 2 }$

We can also discover that:

$\displaystyle \frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } \\ \\ =\frac { { \left( a-p \right) }^{ 2 }+{ \left( b-p \right) }^{ 2 }+{ \left( c-p \right) }^{ 2 } }{ n } \\ \\ =\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+3{ p }^{ 2 }-2ap-2bp-2cp }{ n } \\ \\ =\frac { g+3{ p }^{ 2 }-2p\left( a+b+c \right) }{ n } \\ \\ =\frac { g+3{ p }^{ 2 }-2pd }{ n } \\ \\ =\frac { g }{ n } +\frac { 3{ p }^{ 2 } }{ n } -\frac { 2pd }{ n } \\ \\ \boxed { But,\quad n=3. } \\ \\ =q+{ p }^{ 2 }-2p\cdot \frac { d }{ n } \\ \\ =q+{ p }^{ 2 }-2{ p }^{ 2 }\\ \\ =q-{ p }^{ 2 }$

So, as:

$\displaystyle q-{ p }^{ 2 }=q-{ p }^{ 2 }\\ \\ \frac { \Sigma { \left( x-\overline { x } \right) }^{ 2 } }{ n } =\frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 }$

I came up with this proof, as I wasn't told why these two formulas are equal to one another. Hopefully this will be able to help some maths students come to terms with why it's the case.

If you have any better proofs, please post them below. Thankyou for your attention.

Last edited by perfect_world; January 13th, 2015 at 11:33 AM.
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January 13th, 2015, 02:01 PM   #2
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$\displaystyle \sum \frac{(x-m)^2}{n}=\sum (\frac{x^2}{n}-2\frac{mx}{n}+\frac{m^2}{n})$
$\displaystyle \sum \frac{mx}{n} = m\sum \frac{x}{n} = m^2$
$\displaystyle \sum \frac{m^2}{n} = m^2 \sum \frac{1}{n} =m^2$

Put it all together to get result.
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January 13th, 2015, 03:59 PM   #3
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Quote:
Originally Posted by mathman View Post
$\displaystyle \sum \frac{(x-m)^2}{n}=\sum (\frac{x^2}{n}-2\frac{mx}{n}+\frac{m^2}{n})$
$\displaystyle \sum \frac{mx}{n} = m\sum \frac{x}{n} = m^2$
$\displaystyle \sum \frac{m^2}{n} = m^2 \sum \frac{1}{n} =m^2$

Put it all together to get result.
I understand that you multiplied the binomial in the first line, but I have no idea how you got to the second line.
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January 14th, 2015, 12:44 PM   #4
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In $\displaystyle \sum \frac {mx}{n} $, m is constant, so take it outside summation.
By definition: $\displaystyle \sum \frac{x}{n} = m $.
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January 15th, 2015, 10:32 AM   #5
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To @mathman,

In your proof I presume that n=1, as the data set for x consists of x alone.

It's funny how that summation sign can be manipulated around in specific cases.

How would I define this data set mathematically?

Would it be:

x={x}

Therefore n=1.

Last edited by perfect_world; January 15th, 2015 at 10:35 AM.
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January 15th, 2015, 10:51 AM   #6
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Can you let me know if I'd be able to manipulate the sigma symbol in such a manner...

$\displaystyle \frac { \Sigma { \left( x-m \right) }^{ 2 } }{ n } \\ =\sum { \frac { { \left( x-m \right) }^{ 2 } }{ n } } \\ \\ =\sum { \frac { { x }^{ 2 }-2mx+{ m }^{ 2 } }{ n } } \\ \\ =\sum { \left( \frac { { x }^{ 2 } }{ n } -2\frac { mx }{ n } +\frac { { m }^{ 2 } }{ n } \right) } \\ \\ =\sum { \frac { { x }^{ 2 } }{ n } } -2\sum { \frac { mx }{ n } } +\sum { \frac { { m }^{ 2 } }{ n } } \\ \\ =\frac { \Sigma { x }^{ 2 } }{ n } -2{ m }^{ 2 }+{ m }^{ 2 }\\ \\ =\frac { \Sigma { x }^{ 2 } }{ n } -{ m }^{ 2 }\\ \\ =\frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 }$

I'm just experimenting with it, and seeing if it's possible to create proofs in this way.
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January 15th, 2015, 11:47 AM   #7
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Quote:
Originally Posted by EvanJ View Post
I understand that you multiplied the binomial in the first line, but I have no idea how you got to the second line.
@EvanJ,

I believe that these truths can be derived from the information we have at our disposal.

$\displaystyle x=\left\{ x \right\} \Rightarrow n=1\\ \\ \frac { \Sigma x }{ n } =\frac { x }{ n } =\sum { \frac { x }{ n } } \\ \\ ---------------------------\\ \\ m=\frac { \Sigma x }{ n } ,\\ \\ \therefore \quad { m }^{ 2 }=m\frac { \Sigma x }{ n } =\frac { mx }{ n } =m\sum { \frac { x }{ n } } =\frac { \Sigma mx }{ n } \\ \\ ----------------------------\\ \\ \frac { \Sigma { x }^{ 2 } }{ n } =\frac { { x }^{ 2 } }{ n } =x\cdot \frac { x }{ n } =x\frac { \Sigma x }{ n } =x\sum { \frac { x }{ n } } \\ \\ -----------------------------\\ \\ \frac { { m }^{ 2 } }{ n } ={ m }^{ 2 }\sum { \frac { 1 }{ n } } =\frac { { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } }{ n } ={ \left( \frac { \Sigma x }{ n } \right) }^{ 2 },\quad (as\quad n=1)\\ \\ \\ \\ \\ \\ \\ \\ $
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January 15th, 2015, 05:31 PM   #8
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I believe that in the original expression $\displaystyle x$ was meant to be $\displaystyle x_i$ (one sample) where i went from 1 to n. My answers were made on that assumption.
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