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Math Help - Help with steady-state unemployment

  1. #1
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    Help with steady-state unemployment

    Hello,

    I am a first year econ student, trying to work through Mankiw's Macroeconomics 7th edition textbook. However, I have run into a problem that I cannot seem to find where to start.

    The question is: The steady-state rate of unemployment is U/L = s/(s+f). Suppose that the unemployment rate does not begin at this level. Show that unemployment will evolve over time and reach this steady state. (Hint: Express the change in the number of unemployed as a function of s, f, and U. Then show that if unemployment is above the natural rate (or the steady-state rate), unemployment falls, and if unemployment is above the natural rate unemployment rises.)

    If anyone can point me in the right direction, I would greatly appreciate it.

    Thanks.
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  2. #2
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    That should not be too hard but first we'll need to be clear on definitions. Please define U, L, f and s. In particular, but not exclusively, what is the Unemployment Rate? U or U/L?
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  3. #3
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    U is the number of unemployed.
    L is the total Labor Force (Employed + Unemployed, or E+U)
    s is the rate at which those who are employed leave their job. (known as the job separation rate)
    f is the rate at which those who are unemployed find a job. (the job finding rate)


    Hope this helps and thanks for the quick response.
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  4. #4
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    Using the hint, we have the Change in Unemployment (for some period)

    Employed Separating = (L-U)*s -- This increases the unemployment rate
    UnEmployed Finding = (U)*f -- This decreases the unemployment rate

    Change In Unemployment = (L-U)*s - (U)*f

    This assumes the workforce does not change size.

    Given the Steady State Rate, without proof, we see with a little algebra that:

    s_{steady}\;= \;\frac{U\cdot f}{L-U}

    and

    f_{steady}\;=\;s\cdot\frac{L-U}{U}

    Adding an increase as our starting place, we have \frac{U}{L}\;=\;\frac{s}{s+f}\;+\;\Delta for some \Delta\;>\;0

    Solving again for s and f gives:

    s\;= \;\frac{(U\cdot f) - L\Delta f}{(L-U) + L\Delta}\;<\;s_{steady} so the Separation Rate decreases

    and

    f\;=\;s\cdot\frac{(L-U) + L\Delta}{(U) - L\Delta}\;>\;f_{steady} so the Finding Rate increases.

    It is rather clear, then, that unemployment forces certainly take action that would suggest the desired result. What is not yet shown is that the action causes a return to the Steady State. Does it?
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  5. #5
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    It struck me as I was headed for work...

    I'm a little concerned that EITHER response of f or s will correct the Unemployment Rate back to the steady state. I'm not quite sure what the implications are for BOTH responding. Will it overshoot, maybe become oscillatory? There's more thinking ahead. Personally, I would want to explore this prior to submitting the response for marking. Perhaps, since each is a function of the other, it is necessary to consider only one response.

    Great, TheUncle, now you have me thinking about it. How am I going to get any work done for my actual employer? :-)
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  6. #6
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    Haha, sorry about that TK.

    Thanks for the help. I think things are starting to become more clear now. I was definitely overthinking the question the first few times I read it. My professor sent this hint to me though:

    If we say that the unemployment rate is not in steady
    state, then it must be changing over time. If the unemployment rate is
    changing over time, then the number of unemployed individuals must be
    changing over time. So the change in unemployment between any two periods
    must be the difference between those separating from jobs and those who
    are unemployed that have found jobs. In other words

    U[t+1] - U[t] = sE[t] - fU[t]

    Now in steady state, this expression equals zero, but not when we are out
    of the steady state. The next step is to use the information you know
    about the unemployment rate and derive an expression.
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  7. #7
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    Exactly. Take my last expression for either s or f, not both, and drop it into the excess expression for U/L. See what it does to the augmentation in one iteration.

    Note: This is a rather unsatisfactory solution. The single-step correction would be thought of in the real world as rather magic. Reversion to a steady state might take a while, perhaps even an asymptotic tendency.

    Indeed, plenty to study on this one question.
    Last edited by TKHunny; March 31st 2010 at 06:55 PM.
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