Challenge. Solve for $y$ the expression $$\Delta \log\frac{y}{x_{1}}=s.\Delta \log\frac{x_{2}}{x_{1}}$$ That's it, that's my question. But you will probably need some background.

Background. This is from my research in economics, but my math is rusty so I am posting here. More precisely, $y$ measures how much income (or GDP) there is in an economy at a given time. I am interested in finding an equation for $y$, which if I am correct can be derived from the equation above. Some further points:

1. My notations are not the best. $\Delta:=\frac{d}{dt}$, and $\Delta \log=\frac{dx/dt}x$ i.e. the rate of change over time. Furthermore, every variable $y,x_{1},x_{2}$ as well as $s$ are all indexed in time. Feel free to rewrite the equation above if needed.
2. In addition I know that the following relationship holds true: $\frac{x_{2}}{x_{1}}=\frac{s}{1-s}\frac{p_{1}}{p_{2}}$

What I've done. Below I provide some steps I have already taken, hoping that they are correct:

1. First I integrated both with respect to time, i.e. $$\int\Delta \log\frac{y}{x_{1}}dt=\int s.\Delta \log\frac{x_{2}}{x_{1}}dt$$

2. Then I note that I have a product of two functions in the right-hand side integral, so I proceeded by integrating by parts. Let $u=s, du=ds$ and $dv=\Delta \log\frac{x_{2}}{x_{1}}, v=\log\frac{x_{2}}{x_{1}}$. Again every variable is indexed in time. The integration by part follows $\int u.dv = u.v -\int v.du$. This leads to $$\log\frac{y}{x_{1}}+C_{1}=s\log\frac{x_{2}}{x_{1}}-\int \log\frac{x_{2}}{x_{1}}ds$$

3. This is already a nice and very interesting first step for my research. But I am not done as I still have to calculate the integral $\int \log\frac{x_{2}}{x_{1}}ds$. As I understand it, I have to integrate $\log\frac{x_{2}}{x_{1}}$ with respect to $s$. This is where my problem starts, as $\log\frac{x_{2}}{x_{1}}$ is not an expression in $s$. But I can use the relationship in point #2 above, which relates $\frac{x_{2}}{x_{1}}$ to an expression in $s$ times $\frac{p_{1}}{p_{2}}$. Using this substitution I get $$\log\frac{y}{x_{1}}+C_{1}=s\log\frac{x_{2}}{x_{1}}-\int \log\left(\frac{s}{1-s}\frac{p_{1}}{p_{2}}\right)ds$$ or equivalently: $$\log\frac{y}{x_{1}}+C_{1}=s\log\frac{x_{2}}{x_{1}}-\int \log\frac{s}{1-s}ds-\int \log\frac{p_{1}}{p_{2}}ds$$

4. Now I have an expression consisting of four terms. The first three do not cause any problem. It is the last term I have trouble with. Specifically, what is $$\int \log\frac{p_{1}}{p_{2}}ds$$ equal to? Maybe I should end with some more background. $\frac{p_{1}}{p_{2}}$ does not seem equal to anything useful. Those two $p$'s are the prices of two products, and the ratio of those two prices is what it is. There is no expression I can substitute for this ratio, except using the relationship in point #2 above again, which probably means going backwards, not forwards.

5. Maybe there is something to do with $s$ in that last integral? From relationship #2 again, we find that $s=\frac{p_{2}x_{2}}{p_{1}x_{1}+p_{2}x_{2}}$, which means that the last integral is $$\int \log\frac{p_{1}}{p_{2}}.d\left(\frac{p_{2}x_{2}}{p_{1}x_{1}+p_{2}x_{2}}\right)$$ ... but this is far from helping me.

So I will conclude with one word: help. Help with solving the very first equation directly, or better yet, help with solving the expressions in #6 or #7.

• It's more how fast money rotates, but interesting question nevertheless. – mathreadler Apr 5 '18 at 20:46
• $\dfrac{du}{dt} = \dfrac{ds}{dt}$, so playing with notation a little $du=\dfrac{ds}{dt}\cdot dt$. Presumably you can compute the derivative of s with respect to time. – Andy Walls Apr 5 '18 at 20:58
• if $p_1$ and $p_2$ do not depend on $s$, your last integral (#6) will simply be $s\cdot (\log p_1-\log p_2)+C$ where $C$ is a constant. – Vasya Apr 5 '18 at 21:02
• @AndyWalls: so you're saying that I should rewrite equation #6 as $$\int log\frac{p_{1}}{p_{2}}ds=\int\left(log\frac{p_{1}}{p_{2}}.\frac{ds}{dt}\right)dt$$ ? If that's the case, then I get $$\int\left(log\frac{p_{1}}{p_{2}}.\Delta s\right)dt$$. I just don't know what to do with that :/ – frencho Apr 5 '18 at 21:23
• @frencho I was thinking the integral in step 4, where you have $ds$. My comment shows you how to write the integral as $dt$, but you need to be able to express $ds/dt$ analyticly to do that. You havent provided an expression for s in terms of t though. – Andy Walls Apr 5 '18 at 21:29

Start with the original question, to solve $$\Delta log \frac{y}{x_1} = s \Delta log \frac{x_2}{x_1}.$$ We can re-write that as $$\frac{d}{dt}[log\ y -log\ x_1]=s\frac{d}{dt}[log\ x_2 - log\ x_1]$$ or $$\frac{d}{dt}[log\ y ]=s\frac{d}{dt}[log\ x_2 - log\ x_1] +\frac{d}{dt}log\ x_1$$ or $$\frac{d}{dt}log\ y =s\frac{d}{dt}log\ x_2 +(1-s)\frac{d}{dt}log\ x_1$$

Now we recall the useful fact, which is the point of the log formulation, that $$\frac{d}{dt}log\ y = \frac{1}{y}\frac{dy}{dt}$$ thus the derivative of the log is the percentage change. In econ jargon the original problem is a question about the relation of two elasticities.

So now we have a differential equation of the following form (I stick with the one variable case to make the logic more clear): $$\frac{1}{y}\frac{dy}{dt}=k\frac{1}{x}\frac{dx}{dt}$$ and using separation of variables, that reduces to $$\frac{1}{y}dy=k\frac{1}{x}dx.$$ Integrating both sides gives the solution $log \ y=k\ log\ x$ or $$y=x^k.$$

Now by analogy, the solution to the original problem becomes $$y=x_1^{1-s}x_2^s$$ which looks a lot like a Cobb-Douglas production function. It pays to remember that $x_1, x_2$ and $y$ are all functions of $t$. This says nothing about the prices $p_1,p_2$, but you can substitute out for either one of the x's and just have a few more constants floating around.

Separately, is this a growth accounting exercise? Have you looked at chapter 12 of Glaister's Mathematical Methods of Economists? or the More advanced The Theory of Equilibrium Growth by Dixit?

• @turl Your long response is much appreciated. No I haven't looked at those books and thanks for the reference. Indeed, bingo, the solution to my original problem (very first equation above) is a some sort of Cobb-Douglas production function, agreed. But here's the gist of my paper/idea. In virtually all papers on economic growth, the factor shares (parameters $s$ and $1-s$) are considered constant over time. Yet this is no longer true empirically (i.e. we observe falling labor shares $1-s$ for instance). – frencho Apr 17 '18 at 14:49
• When $s$ is constant over time, economists usually arrive at something like $$y=A.x_{1}^{1-s}.x_{2}^{s}$$ But when we lift the assumption of $s$ being constant over time, who knows what happens, and this is what I'm trying to find out. So to restate my problem more accurately, I have to solve for $y$ the expression $$\Delta\log\frac{y_{t}}{x_{1t}}=s_{t}.\Delta\log\frac{x_{2t}}{x_{1t}}$$ where I made the time subscripts explicit. My intuition is that when $s$ is allowed to vary with time, you would get a solution of the form $$y=B.x_{1}^{1-s}.x_{2}^{s}$$. I need to find out what $B$ is. – frencho Apr 17 '18 at 15:00
• @frencho: you might look at www4.ncsu.edu/~jjseater/sharealteringtechprogress.pdf – Trurl Apr 17 '18 at 17:30