if I have $\displaystyle \frac {x}{n} + x - \frac{n}{n-1}$ how can I transform it into $\displaystyle x + nx - \frac{n}{n-1}$?
I can't seem to do it without changing the n/(n-1) term into n^2/(n-1)
Please give the whole problem. The expression $\displaystyle \dfrac{x}{n}+x - \dfrac{n}{n-1}$ is a different expression from $\displaystyle x+nx - \dfrac{n}{n-1}$. I would need the full context to be able to help further.
If $\displaystyle \dfrac{x}{n} + x - \dfrac{n}{n-1} = 0$ then $\displaystyle x + nx - \dfrac{n^2}{n-1} = 0$ as you suggested. Or, $\displaystyle \dfrac{x}{n} + x - \dfrac{n}{n-1} = \dfrac{1}{n}(x + nx) - \dfrac{n}{n-1}$. Without more information, I don't know how to help you.
sorry about that. It's a statistics problem that becomes an algebra problem.
I have as an estimator of the variance of the population$\displaystyle \frac{\sum(y_i - \bar{y})^2}{n} - \frac{n}{n-1} + \sum(y_i - \bar{y})^2$ and I'm trying to find out if it's biased or not, meaning if I take the expectation whether or not the expression remains the same. When I take the expectation I get
$\displaystyle \sum(y_i - \bar{y})^2 - \frac{n}{n-1} + n \sum(y_i - \bar{y})^2$ (I hope that's right, if not the whole thing here is wrong) and the question is how would I make this unbiased, which I interpret to mean that I have to find some way of algebraically making the second statement equivalent to the first. In the above problem I let $\displaystyle E( \sum (y_i - \bar{y})^2 )$= VARIANCE(Y) = x so as to simplify the algebra.
It could be that my statistics up to this point are wrong and perhaps this is why the algebra is not working.
Ahh, my knowledge of statistics is rather limited. But, would this help with the algebra? Rewriting your original expression, you have
$\displaystyle \dfrac{\sum{(y_i-\overline{y})^2}}{n} - \dfrac{n}{n-1} + \sum{(y_i - \overline{y})^2} = \dfrac{n+1}{n}\sum{(y_i - \overline{y})^2} - \dfrac{n}{n-1}$
If you find the expected value of the RHS expression, do you get $\displaystyle (n+1)\sum(y_i-\overline{y})^2 - \dfrac{n}{n-1}$? If so, then your algebra looks correct to me, and maybe the estimator is biased.