# Forming first-order difference equation

• Jun 7th 2012, 08:30 AM
econolondon
Forming first-order difference equation
Hi,

I was wondering if anyone can help with this mathematical economics problem:

The multiplier-accelerator model of growth has the following three equations

St = aYt
It+1 = b(Yt+1 - Yt)
St = It

a) Briefly interpret the above three equations and combine them as a first-order difference equation.

b) Assume that output at time zero is Y0 and solve the difference equation. Which are the determinants of output growth in the model? Explain intuitively.

It's clear from b) that the difference equation must refer to Y(t). But I can't work out how to get a single first-order difference equation that includes the three equations, that is set up to refer to Y(t).
• Jun 7th 2012, 09:04 AM
HallsofIvy
Re: Forming first-order difference equation
I assume "Yt+1" is $Y_{t+1}$, not " $Y_t+ 1$" and that " $It+1$" is $I_{t+1}$. So the middle equation is the difference equation: $Y_{t+1}- Y_t= \frac{I_{t+1}}{b}$

Since you are given that $I_t= S_t$ and $S_t= aY_t$, it follows that $I_t= aY_t$ so that $I_{t+1}= aY_{t+1}$. Putting that into your equation, $Y_{t+1}- Y_t= \frac{a}{b}Y_{t+1}$ which is the same as $\left(1- \frac{a}{b}\right)Y_{t+1}= Y_t$. I would write that as $Y_{t+1}= \frac{b}{a- b}Y_t$ is relatively easy to solve.
• Jun 9th 2012, 08:15 AM
econolondon
Re: Forming first-order difference equation
Excellent, thanks