Average employment length

Question

From Macroeconomics 7th edition by Gregory Mankiw, p166.

Here’s a numerical example. Suppose that 1 percent of the employed lose their jobs each month ($s$ = 0.01). This means that on average jobs last 100 months, or about 8 years. Suppose further that 20 percent of the unemployed find a job each month ( $f$ = 0.20), so that spells of unemployment last 5 months on average. Then the steady-state rate of unemployment is

$\frac U L = \frac {0.01} {0.01 + 0.20} = 0.0476.$

The rate of unemployment in this example is about 5 percent.

Well, here I don't get this part.

This means that on average jobs last 100 months

To me, the fact that 0.01 times 100 months comes to be 1, looks very plausible in a sense. But I don't get the process in it. My guess was like as follows. If anyone among 100 people loses their job monthly,

(suppose they can never be reemployed, which is the opposite in the text book; I'm not sure that this matters on the issue here)

then the first guy loses after 1 month, second guy 2 months, ... So we can add up all their work period 1+2+3+...+100 which is 5050, and divide it with 100 people, then we get 50.50 months per person. What am I missing?

Answer 1

This is simply the geometric distribution.

A job lasts 1 month with probability $s$.

A job lasts 2 months with probability $(1-s)s$.

A job lasts 3 months with probability $(1-s)^2s$.

$\vdots$

A job lasts $n$ months with probability $(1-s)^{n-1}s$.

Hence, the average length of a job is: $$\begin{align*} & 1\cdot s+2\left(1-s\right)s+3\left(1-s\right)^{2}s+\dots+n\left(1-s\right){}^{n-1}s+\dots\\ = & \left[s+\left(1-s\right)s+\left(1-s\right)^{2}s+\dots\right]+\left[\left(1-s\right)s+\left(1-s\right)^{2}s+\dots\right]+\dots\\ = & \frac{s}{1-\left(1-s\right)}+\frac{1-s}{1-\left(1-s\right)}+\frac{\left(1-s\right)^{2}}{1-\left(1-s\right)}+\dots=\frac{s}{s}+\frac{1-s}{s}+\frac{\left(1-s\right)^{2}}{s}+\dots\\ = & \frac{1}{s}\left(s+s\left(1-s\right)+s\left(1-s\right)^{2}+\dots\right)=\frac{1}{s}\frac{s}{1-\left(1-s\right)}=\frac{1}{s}. \end{align*}$$

In the specific numerical example given by Mankiw, we have $s=0.01$ and hence the average length of a job is $1/s=100$.