Show that the co-variance is zero between $\bar{Y}$ and $Y_i - \bar{Y}$ I have a quick question about how to mathematically show this result. Here is the question and my thoughts.
Let $Y_1, \ldots, Y_n$ be i.i.d. R.Vs with mean $\mu$ and variance $\sigma^2$ and let $\bar{Y}$ be the mean of those R.Vs.  Show that the co variance is zero between $\bar{Y}$ and $Y_i - \bar{Y}$. 
Since we are told that $Y_1, \ldots, Y_n$ are i.i.d. R.V.s then we also know that $\bar{Y}$ also has the same distribution but is i.i.d correct? Would that not imply that the co-variance is zero between $\bar{Y}$ and $Y_i - \bar{Y}$ since this is simply a linear combination of R.V.s that are already independent? I am confused about how to mathematically show this however and help would be greatly appreciated! Thanks :) 

EDIT: 
While doing the problem, another question arose. It appears that $E(\bar{Y})$ = $\bar{Y}$ but how can that be if $\bar{Y}$ = $ (1\div n)$ $\sum_{i=1}^n Y_i$ ? If we take the expectation of this second value don't we get $\mu$? But how could $\mu$ = $\bar{Y}$ ?

EDIT #2:
So I have been solving the problem with various updates and here is where I am stuck now:
We start here:
$$\text{cov} (\bar{Y}, Y_i - \bar{Y})$ = $E[(\bar{Y}-E(\bar{Y})) * (Y_i-\bar{Y}-E(Y_i-\bar{Y}))]$$
so simplifying this expression according to the fact that $E(\bar{Y})$ = $\mu$ gives the following:
$$\bar{Y}\mu - \bar{Y^2} - \mu^2 + \mu\bar{Y}$$ 
I am confused whether or not this equals zero. I would also greatly appreciate an explanation of how $\mu$ and $\bar{Y}$ relate in practical purposes. Thanks so much!
 A: Starting from the formula for Covariance.
$$
Cov(x,y) = \mathrm{E}\left[\left(X-\mathrm{E}(X)\right)\left(Y-\mathrm{E}(Y)\right)\right]
$$
For your case
$$
Cov(x,y) = \mathrm{E}\left[\left(\bar{Y}-\mathrm{E}(\bar{Y})\right)\left(Y-\bar{Y}-\mathrm{E}(Y-\bar{Y})\right)\right] = \mathrm{E}\left[0\cdot\left(Y-\bar{Y}\right)\right] = 0 
$$
Here I used the fact that $E(\bar{Y})= \bar{Y}$ since it is a constant value.
Does that help?
A: Remark that $\text{cov}(\bar Y, Y_i - \bar Y) $ does not depend of $i$,
$$ \forall i, \ \sum_{j=1}^n \text{cov}(\bar Y, Y_j - \bar Y) = n \ \text{cov}(\bar Y, Y_i - \bar Y).  $$
The covariance is bilinear, so
$$ \frac 1 n \sum_{j=1}^n \text{cov}(\bar Y, Y_j - \bar Y) = \text{cov}(\bar Y, \frac 1 n \sum_{j=1}^n Y_j - \bar Y) = \text{cov}(\bar Y,0) = 0.$$
A: Regarding the question posed in the edit, it is not true that $E[\bar Y] = \bar Y$.  Instead, $\bar Y$ is an estimator of $E[Y]$. It is a linear combination of $n$ i.i.d. random variables. $E[\bar Y]=E[\frac{1}{n}\sum_{i=1}^n Y_i]=\frac{1}{n}\sum_{i=1}^n E[Y_i]=\frac{1}{n}\cdot n\mu = \mu$.
To see the computation of the covariance, replace the first $\bar{Y}$ with its definition and apply linearity to get
$$\text{Cov}[\bar{Y}, Y_i - \bar{Y}] = \frac1n\sum_j \text{Cov}[Y_j,Y_i] - \frac1n \sum_j \text{Cov}[Y_j,\bar{Y}]$$
Note that $\text{Cov}[Y_j,Y_i]$ is zero for $j\neq i$ and is $\text{Var}[Y_i] = \sigma^2$ for $j=i$, so the first sum is just $\frac1n \sigma^2$. So we now have
$$\text{Cov}[\bar{Y}, Y_i - \bar{Y}] = \frac1n \sigma^2 - \frac1n\sum_j \text{Cov}[Y_j,\bar{Y}]$$
Now do it again, working on the second instance of $\bar{Y}$. Replace the second $\bar{Y}$ with its definition and apply linearity again to get
$$\text{Cov}[\bar{Y}, Y_i - \bar{Y}] = \frac1n \sigma^2 - \frac1n \sum_j \left( \frac1n \sum_k \text{Cov}[Y_j,Y_k]\right)$$
Just as before, $\text{Cov}[Y_j,Y_k]$ is zero for $k\neq j$ and is $\text{Var}[Y_j] = \sigma^2$ for $k=j$, so we now have
$$\text{Cov}[\bar{Y}, Y_i - \bar{Y}] = \frac1n \sigma^2 - \frac1n \sum_j \left( \frac1n \sigma^2\right)$$
Since each of the $n$ inner summands is the same, we have
$$\text{Cov}[\bar{Y}, Y_i - \bar{Y}] = \frac1n \sigma^2 - \frac1n  (n \cdot \frac1n \sigma^2) = \frac1n \sigma^2 - \frac1n \sigma^2 = 0$$
as desired.
