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I have two time-varying variables X and Y. The product of which is Z(= X*Y). A curve Z' has been fit on Z. So Z'can be described as a product of some new X' and Y'. Where X' = X + dX and Y' = Y + dY. How do I get these new X' and Y'? It looks like a regression problem to me. But what are the rules I will have to follow to get X' and Y'?

Yakku
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    I assume that there is some common independent variable $t$ which you didn't mention. Your problem is undeterminate as there is an infinity of ways to solve $X'(t)Y'(t)=Z'(t)$. You can address that by fitting another data set such as $X_i$ or $Y_i$ or $X_i/Y_i$..., if available; otherwise, there is nothing you can do. –  Jan 09 '17 at 13:10
  • That is what I meant by time-varying variables. So it is X(t) and Y(t). For brevity, I just called them X and Y.

    What if I use a rule, to arrive at a solution which has the least mean squared error (sqrt(dX^2) + sqrt(dY^2)).

     – Yakku Jan 09 '17 at 13:34
  • This just amounts to fitting $S_i:=X_i+Y_i$, indirectly mentioned in my comment. Then you solve $X'Y'=Z', X'+Y'=S'$. –  Jan 09 '17 at 13:41

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$Z'=X'Y'$

The problem here is that there are (infinitely) many choices of $X'$ and $Y'$ that will give the product you desire.

$Z'=(X+dX)(Y+dY)$

$Z'=XY+YdX+XdY+dXdY$

You could decide that you want the variations $dX$ and $dY$ to be equal.

In that case set $dX=dY=\delta$

Then $Z'=XY+Y\delta+X\delta+\delta^2$

You can solve the resulting quadratic equation to find $\delta$.

If $\bar X= k \bar Y$ then you might prefer to set $dX=k dY$.

Your relationship is then: $Z'=XY+Yk\delta+X\delta+k\delta^2$

tomi
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