Solve $\frac{\Delta y}{y}=\sum_{i}s_{i}\frac{\Delta x_{i}}{x_{i}}$ Colleague here but rusty and confused. I have to get this
done, I just must, must, must.
[Note: as far as I can tell, the problem is in part calculus (sums
and integrals) and in part sequences or statistics, and it can be
worked out in discrete or continuous time. Hence all the tags to this question. All this is throwing me
off. Feel
free to change notations or restate the problem if it helps]
Problem. For now the problem is best stated in discrete time. Solve for $y$ $$\frac{\Delta y}{y}=\sum_{i=1}^{n}s_{i}\frac{\Delta x_{i}}{x_{i}}$$
where every variable is a function of time. and the ratios denote
rates of change (all time subscripts are omitted throughout for simplicity).
The term $s_{i}=\frac{p_{i}x_{i}}{\sum_{i}p_{i}x_{i}}$ is a share
(or proportion in total), with $0<s_{i}<1$. It is not immediately clear what the $p_i$'s are, and they may be related to the $x_i$'s. The equation states that
the rate of change of $y$ is an arithmetic mean of the rates of change
of the $x_{i}$'s when weighted by $s_{i}$. 
Question. How would you solve this? I can provide more background.

Alternatively, what do you think about my starting point:
Switch to continuous time
$$\int_{t}\frac{dy}{y}\,dt=\int_{t}\int_{i}f(x_{i})\frac{dx_{i}}{x_{i}}\,dt$$
where $f(x_{i})=s_{i}=\frac{p_{i}x_{i}}{\int_{i}p_{i}x_{i}}$. Again, the $p_i$'s may be related to the $x_i$'s. 
The left-hand side is immediate:
$$\ln y+C_{1}=\int_{t}\int_{i}f(x_{i})\frac{dx_{i}}{x_{i}}\,dt$$
But now the right-hand side is bothering me. Best I can think of is
integrate by parts, possibly twice. But then I don't know what I'm
doing or if it's correct.
 A: Assuming that $y$ and $x_k$ are functions $\mathbb N \to \mathbb R$, and that $\Delta y$ means tthe forward finite difference
$$
\Delta y\;:\;\Delta y(n) = y(n + 1) - y(n)
$$
then we can write
$$
\Delta y(n) = y(n + 1) - y(n) = \left( {{{y(n + 1)} \over {y(n)}} - 1} \right)y(n)
$$
and
$$
{{\Delta y(n)} \over {y(n)}} = \left( {{{y(n + 1)} \over {y(n)}} - 1} \right)
$$
So your sum becomes
$$
{{y(n + 1)} \over {y(n)}} - 1 = \sum\limits_i {s_{\,i} \left( {{{x_{\,i} (n + 1)} \over {x_{\,i} (n)}} - 1} \right)}
  = \sum\limits_i {s_{\,i} {{x_{\,i} (n + 1)} \over {x_{\,i} (n)}}}  - \sum\limits_i {s_{\,i} } 
$$
and if the weights sum to $1$
$$
{{y(n + 1)} \over {y(n)}} = \sum\limits_i {s_{\,i} {{x_{\,i} (n + 1)} \over {x_{\,i} (n)}}} 
$$
Now, for the expression above we have that 
$$
z(n) = {{y(n + 1)} \over {y(n)}}\quad  \Leftrightarrow \quad \prod\nolimits_{\,n\, = \,a}^{\;b} {z(n)}  = {{y(b)} \over {y(a)}}
$$
that is that $y(n)$ is the  multiplicative primitive of $z(n)$ or equivalently that
$$
\ln z(n) = \Delta \ln y(n)
$$
--- addendum  ---
Some additional notes and clarifications in reply to your comment.
The conclusion of the above is that, if the weights $s_i$ sum to $1$ we can pass from 
a relation between $\Delta y /y, \; vs. \;  \Delta x_i / x_i$ to a relation between $y(n+1)/y(n) \; vs. \; x_i(n+1) / x_i(n)$.
That is not a big improvement, if not for the fact that the correspondence $z(n) \Leftrightarrow  y(n+1)/y(n)$ is more "standard"
and available for many functions as you can see in the given reference.
In any case, it must be clear that, either in your formulation (discrete / continuous) and in the one above you cannot reach to any 
further significant conclusion if you do not know which specific functions of $n$ the $x_i$ are.
It is easy to see that changing the interpretation of the $\Delta$ from "forward" to "backwards" bring to the same relations
with $y(n-1)/y(n)$ in place of $y(n+1)/y(n)$, and analogously for the $x_i$.
I used $n$ as a variable integer, understanding $y, \, x_i$ to be functions $\mathbb N \to \mathbb R$.
But if they are $\mathbb R \to \mathbb R$, the formulation does not change : $y(x+1)/y(x)$ is well "extendable" also 
to real, and even complex, variables. Of course you can change $x$ to $t$ or any other symbol.
