Relating a sum to the standard deviation I have a summation that looks like
$$\sum_{j=0}^{n-1} S_j(v_{j} - \bar{v})$$
The $S_j$ terms all have values in $[0,1]$. $v_j$ is a data value and $\bar{v}$ is the mean of all $v_j$.
This expression is clearly related to the standard deviation. If $\sigma=0$ than all $v_j=\bar{v}$ so the result is $0$. But what I would like to do is relate the change in value of this expression to $\sigma$ for all values of $\sigma$. Is this possible? I have tried without luck but think that perhaps there is some property of $\sigma$ which I do not know which might help here? 
 A: Your original expression can be positive or negative.  
Here is an upper bound in terms of the standard deviation, which is greater than or equal to the mean absolute deviation about the mean.
$$\sum_{j=0}^{n-1} S_j(v_{j} - \bar{v}) \le \frac12 \sum_{j=0}^{n-1} \left| v_{j} - \bar{v} \right|  = \frac{n}{2} {\displaystyle\sum_{j=0}^{n-1} \left| v_{j} - \bar{v} \right|}{ / n}  \le \frac{n}{2} \sqrt{{\displaystyle\sum_{j=0}^{n-1} \left( v_{j} - \bar{v} \right)^2}{ / n}} = \frac{n}{2} \sigma_v$$
To show this is tight, take the example of $v_0 = S_0 = 0$ and $v_1 = S_1=1$.  
A: An inequality
Let $S = (S_0 - \bar{S}, S_1- \bar{S}, \ldots, S_{n-1}- \bar{S})$ and $r = (v_0 - \bar{v}, v_1 - \bar{v}, \ldots, v_{n-1} - \bar{v})$ be the mean-centered $n$-vectors of coefficients and data.  Let $\sigma_v = ||r||_2/n$ be the standard deviation of the $v_j$. The Cauchy-Schwarz Inequality (which basically asserts the cosine of the angle between two vectors lies between $-1$ and $1$, reaching those extremes if and only if the vectors are parallel) implies
$$\sum_j S_j(v_j - \bar{v}) = \sum_j (S_j- \bar{S})(v_j - \bar{v}) = S \cdot r \le ||S||_2 ||r||_2 =   n ||S||_2 \sigma_v.$$
The first equality exploits the sum-to-zero property of the residuals $\sum_j (v_j-\bar{v}) = 0$, the second is the definition of the dot product, and the last follows from the definition of the standard deviation.  The inequality becomes an equality if and only if the vector $r$ is a multiple of the vector $S$.
Assuming all $S_j$ are either $0$ or $1$, this can further be simplified by writing $k$ for the number of nonzero $S_j$, for then $$||S||_2 = \frac{\sqrt{k(n-k)}}{n}.$$
Putting the two results together gives
$$\sum_j S_j(v_j - \bar{v}) \le \sqrt{k(n-k)}\sigma_v.$$
The right hand side is largest when $k=n/2$ or $k=(n-1)/2$ (depending on whether $n$ is even or odd, respectively); in either case the right hand side does not exceed $n/2$ times the standard deviation (and for odd $n$ it is strictly less).
Effects of changing the data
To determine the effect of changing the data $v_0, \ldots, v_{n-1}$ on $f(v_0, \ldots, v_{n-1})$ = $\sum_j S_j(v_j - \bar{v}),$ take the partial derivatives:
$$\frac{\partial f(v_0, \ldots, v_{n-1})}{\partial v_j} = S_j - \frac{1}{n}\sum_j S_j = S_j - \frac{k}{n}.$$
The gradient of $f$ is just the mean-centered vector $S$.  This does not involve the data at all, and particularly it's independent of their standard deviation $\sigma_v$.
When the $S_j$ can take other values in the interval $[0,1]$, all the results that do not involve $k$ continue to hold, which includes the two major ones: the implication of the C-S Inequality and the lack of dependence of the gradient of $f$ on $\sigma_v$.
