How do I prove that $n \sum i = 1 {(x_{i} - μ)}_{i}^{2} = n \sum i = 1 {(x_{i})}_{i}^{2} - n μ^{2}$ $sum_(i=1)^n(x_i-mu)^2=sum_(i=1)^n(x_i)^2-nmu^2$ ?

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How do I prove that $n \sum i = 1 {(x_{i} - μ)}_{i}^{2} = n \sum i = 1 {(x_{i})}_{i}^{2} - n μ^{2}$ $sum_(i=1)^n(x_i-mu)^2=sum_(i=1)^n(x_i)^2-nmu^2$ ?

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Answer 1 · 2018-08-07T05:11:41

Start with:

$n \sum i = 1 {(x_{i} - μ)}_{i}^{2}$ $sum_(i=1)^n (x_i-mu)^2$

Within the sum, distribute the square:

$= n \sum i = 1 (x_{i} - μ) (x_{i} - μ)$ $=sum_(i=1)^n (x_i-mu)(x_i-mu)$

$= n \sum i = 1 (x_{i}^{2} - 2 x_{i} μ + μ^{2})$ $=sum_(i=1)^n (x_i^2-2x_i mu + mu^2)$

Distribute the sum notation to each term:

$= n \sum i = 1 x_{i}^{2} - n \sum i = 1 2 x_{i} μ + n \sum i = 1 μ^{2}$ $=sum_(i=1)^n x_i^2- sum_(i=1)^n 2x_i mu + sum_(i=1)^n mu^2$

Factor out the $2 μ$ $2mu$ from the middle sum (since $2 μ$ $2mu$ doesn't depend on $i$ $i$ ) and notice that the last term is just $underset (n" times") underbrace(mu^2 + mu^2 + ... + mu^2):$

$=sum_(i=1)^n x_i^2- 2musum_(i=1)^n x_i + nmu^2$

Using $mu = 1/n sum_(i=1)^n x_i,$ rearrange to get $sum_(i=1)^n x_i = nmu,$ then sub this in:

$=sum_(i=1)^n x_i^2- 2mu(nmu) + nmu^2$

$=sum_(i=1)^n x_i^2- 2nmu^2 + nmu^2$

Combine like terms:

$=sum_(i=1)^n x_i^2- nmu^2$

Done!

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How do I prove that $n \sum i = 1 {(x_{i} - μ)}_{i}^{2} = n \sum i = 1 {(x_{i})}_{i}^{2} - n μ^{2}$ $sum_(i=1)^n(x_i-mu)^2=sum_(i=1)^n(x_i)^2-nmu^2$ ?

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How do I prove that $n \sum i = 1 {(x_{i} - μ)}_{i}^{2} = n \sum i = 1 {(x_{i})}_{i}^{2} - n μ^{2}$ $sum_(i=1)^n(x_i-mu)^2=sum_(i=1)^n(x_i)^2-nmu^2$ ?