The question is:

Let $Y_1,....,Y_n$ be independent random variables each with the distribution $N\left ( \mu ,\sigma ^2 \right )$.


$\overline{Y} = \frac{1}{n}\sum_{i=1}^{n}Y_i$

$S^2 = \frac{1}{n-1}\sum_{i=1}^{n}\left ( Y_i - \overline{Y} \right )^2$

Show that: $S^2 = \frac{1}{n-1}\left [ \sum_{i=1}^{n}\left ( Y_i-\mu \right )^2 - n\left ( \overline{Y}-\mu \right )^2 \right ]$

This is as far as I've being able to get:

$S^2 = \frac{1}{n-1}\sum_{i=1}^{n}\left ( Y_i - \overline{Y} \right )^2$

$= \frac{1}{n-1}\left [ \sum_{i=1}^{n}\left ( Y_i-\overline{Y} \right )\sum_{j=1}^{n}\left ( Y_j -\overline{Y} \right ) \right ]$

I intend the first double summation to be for when $i=j$ and the second for when $i\neq j$...

$= \frac{1}{n-1}\sum_{i=1}^{n}\sum_{j=1}^{n}\left ( Y_i-\overline{Y} \right ) \left ( Y_j-\overline{Y} \right )+\frac{1}{n-1}\sum_{i=1}^{n}\sum_{j=1}^{n}\left ( Y_i-\overline{Y} \right ) \left ( Y_j-\overline{Y} \right )$

$= \frac{1}{n-1}\sum_{i=1}^{n}\left ( Y_i-\overline{Y} \right )^2 +\frac{1}{n-1}\sum_{i=1}^{n}\sum_{j=1}^{n}\left ( Y_i-\overline{Y} \right ) \left ( Y_j-\overline{Y} \right )$

Would anyone know how to finish the question off from here or any mistakes I've made thus far?

The question is exercise 1.4 from this textbook.

Thank you.

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    $\begingroup$ This is an FAQ, mainly because this question is usually stated exactly the same way each time--but it's in mathematical notation, which is difficult to search! I am confident there are at least 20 exact duplicates here and would therefore be grateful to anyone who could locate one. $\endgroup$ – whuber Jul 12 '17 at 13:29
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    $\begingroup$ BTW, this question has nothing to do with independence, normality, or even random variables: it's a purely algebraic result. Demonstrating it is merely an exercise in using summation notation. If that's not clear, write it out explicitly for the case $n=2$: the pattern should become apparent. $\endgroup$ – whuber Jul 12 '17 at 13:43

As whuber points out, this is simply an exercise in algebra that has nothing to do with random variables or normality; all that is needed is the definition that $\bar{Y} \triangleq \frac 1n \sum_{i=1}^n Y_i$ which can be manipulated into $$0 = \left(\sum_{i=1}^n Y_i\right) - n\bar{Y} = \sum_{i=1}^n (Y_i -\bar{Y}) .$$ We have that \begin{align} \require{cancel} \sum_{i=1}^n (Y_i-\mu)^2 &= \sum_{i=1}^n (Y_i-\bar{Y} + \bar{Y} - \mu)^2\\ &= \sum_{i=1}^n \left[(Y_i-\bar{Y})^2 + (\bar{Y}-\mu)^2 + 2(Y_i-\bar{Y})(\bar{Y}-\mu)\right]\\ &= \sum_{i=1}^n (Y_i-\bar{Y})^2 + \sum_{i=1}^n (\bar{Y}-\mu)^2 + \sum_{i=1}^n 2(Y_i-\bar{Y})(\bar{Y}-\mu)\\ &= \sum_{i=1}^n (Y_i-\bar{Y})^2 + n\cdot (\bar{Y}-\mu)^2 + 2(\bar{Y}-\mu)\sum_{i=1}^n (Y_i-\bar{Y})\\ &= \sum_{i=1}^n (Y_i-\bar{Y})^2 + n\cdot (\bar{Y}-\mu)^2 +2 (\bar{Y}-\mu)\cancelto{0}{\sum_{i=1}^n (Y_i-\bar{Y})}\\ &= \sum_{i=1}^n (Y_i-\bar{Y})^2 + n\cdot (\bar{Y}-\mu)^2\\ &\Downarrow\\ \frac{1}{n-1}\sum_{i=1}^n (Y_i-\bar{Y})^2 &= \frac{1}{n-1}\left[\sum_{i=1}^n (Y_i-\mu)^2 - n\cdot (\bar{Y}-\mu)^2\right]. \end{align} Note that the proof works for all choices of number $\mu$; there is no requirement that $\mu$ equal the expected value of anything.

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  • $\begingroup$ Thank you very much! However, how did you know to start off with (Yi−μ)? Is there any other way to obtain μ from the original equation for S^2? $\endgroup$ – Tejay Lovelock Jul 12 '17 at 23:25

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