# Calculating rate of growth of per capita income

Given this question:

National income is increasing by 1.5% a year and population by 2.5% a year. What is the rate of growth of per capita income?

Attempt:

Since per capita income is GDP/ population. I divided 1.5 by 2.5 and got 0.6. Is this right? Thanks.

Consider: If national income is increasing at a slower rate than population growth, then intuitively per capita income will be falling. Here is a set-up for the rate of decline in per capital income.

$$\text{per capita income}_t = \frac{\text{GDP}(1.015)^t}{\text{population}(1.025)^t} \text{per capita income}_{t-1}$$

• Thanks for answering but I don't understand. How did you get 1.015 and 1.025. And please whats your answer? I don't know the formula you used Nov 9, 2015 at 20:40
• I used the same equation you did, just treating your 1.5% and 2.5% as "interest rates" that show the growth of GDP and pop Nov 9, 2015 at 21:11
• Alright. Thanks but what formula did you use to change it and Is the final answer 0.99? Nov 9, 2015 at 21:14
• 1.) Lookup how to setup compound interest. 2.) If you are looking at one period (year), then per capita income shrinks to ~0.9902th of it's original value, or in other words, falls by ~1% 3.) If you are looking over multiple periods, you will have a different story. I have edited the equation to make it a little clearer. Nov 9, 2015 at 23:29
• Oh. Alright. I understand now. Thank you very much. Really appreciate it. Please can you recommend a textbook for me to learn more of this type of questions on per capita income. I'm an economics major, first year college. Nov 10, 2015 at 6:35

This is not correct. It will be a bit complicated to derive this formally, so I will provide a rule of thumb. For a time dependent variable that is a function of a ratio of two other time dependent variables: $$y(t)=\frac{A(t)}{B(t)}$$ the growth rate of $y(t),$ call it $\tilde{y}=\tilde{A}-\tilde{B}$ where the RHS is the difference between the growth rates of $A(t)$ and $B(t)$. You can apply this to your question.

• Alright. Thanks so the answer is 1.5% - 2.5%. If its not. Can you please explain more. I don't understand the formula you used. Nov 9, 2015 at 20:41
• Yes its correct. Nov 9, 2015 at 20:42
• Thank you. Can you recommend any macro textbook or any textbook for me that explains this topic well. I'm in first year college, economics major. Nov 9, 2015 at 20:52

Here is the calculus that you can make. I note $\tau_{GDPc}$ the rate of change of the per capita GDP, $\tau_{pop}$ the one for the population and $\tau_{GDP}$ for the GDP rate of change.

A rate of change for any variable A is

$$\tau _A = \frac{A(t)-A(t-1)}{A(t-1)}$$

t being the moment considered, $(t-1)$ the previous moment for the calculus of the rate of change. So knowing $GDPc = \frac{GDP}{pop}$ you can write: $$\tau_{GDPc} = \frac{\frac{GDP(t)}{pop(t)} - \frac{GDP(t-1)}{pop(t-1)}}{ \frac{GDP(t-1)}{pop(t-1)}}$$ With the definition of the rate of change, you write $$pop(t)= pop(t-1)*(\tau_{pop}+1)$$ You can then simplify the $pop(t-1)$ term in the previous equation and you get

$$\tau_{GDPc} = \frac{1}{\tau_{pop}+1}\left[\frac{GDP(t)-GDP(t-1)}{GDP(t-1)}-\tau_{pop}\right]$$ $$\tau_{GDPc} = \frac{1}{\tau_{pop}+1}\left[\tau_{GDP}-\tau_{pop}\right]$$

It is indeed different than Brandon marcus answer.

We can make a try with simple numbers:

• let's consider at t-1 a population of 100 for a GDP of 100. The per capita income is then 1.
• At t, you will have a population of 102,5 fo a GDP of 101,5, that is a per capita of 0,99024.

The rate of chage of the per capita income will be $\frac{0,99024-1}{1} \simeq -0,975 \%$.

With the formula I gave you:

$$\tau_{GDPc} = \frac{1}{1,025}(0,015 - 0,025) = -0,975 \%$$ Seems to work.

• This is the best answer! Sep 21, 2017 at 8:56

Rate of growth of per capita GDP is defined as the difference between the rate of growth of GDP and the rate of growth of population as Per Capita GDP = GDP/Population. So, the growth rate of per capita GDP = 1.5% - 2.5% = -1.0%

• This formula is only an approximation and works poorly if the growth rates are not small numbers. Jun 2, 2017 at 4:28