Log-level and Log-log transformations in Linear Regression ...home.wlu.edu/~gusej/econ398/notes/ and Log-log transformations in Linear Regression Models ... A “Log-level” Regression Specification ... From previous slide the marginal effect is

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  • Log-level and Log-log transformations inLinear Regression Models

    A. Joseph Guse

    Washington and Lee University

    Fall 2012, Econ 398 Public Finance Seminar

  • Level-Level

    A Level-level Regression Specification.

    y = 0 + 1x1 +

    This is called a level-level specification because rawvalues (levels) of y are being regressed on raw values of x .

  • Level-Level

    A Level-level Regression Specification.

    y = 0 + 1x1 +

    This is called a level-level specification because rawvalues (levels) of y are being regressed on raw values of x .

    How do we interpret 1?

  • Level-Level

    A Level-level Regression Specification.

    y = 0 + 1x1 +

    This is called a level-level specification because rawvalues (levels) of y are being regressed on raw values of x .

    How do we interpret 1?

    Differentiate w.r.t. x1 to find the marginal effect of x on y . Inthis case, IS the marginal effect.

    dydx

    =

  • Log-Level

    A Log-level Regression Specification.

    log(y) = 0 + 1x1 +

    This is called a log-level specification because the naturallog transformed values of y are being regressed on rawvalues of x .

  • Log-Level

    A Log-level Regression Specification.

    log(y) = 0 + 1x1 +

    This is called a log-level specification because the naturallog transformed values of y are being regressed on rawvalues of x .

    You might want to run this specification if you think thatincreases in x lead to a constant percentage increase in y .(wage on education? forest lumber volume on years?)

  • Log-Level Continued

    How do we interpret 1?

  • Log-Level Continued

    How do we interpret 1?First solve for y .

    log(y) = 0 + 1x1 +

    y = e0+1x1+

  • Log-Level Continued

    How do we interpret 1?First solve for y .

    log(y) = 0 + 1x1 +

    y = e0+1x1+

    Then differentiate to get the marginal effect:

    dydx1

    = e0+1x1+ = 1y

    So the marginal effect depends on the value of y , while itself represents the growth rate.

    1 =dydx1

    1y

    For example, if we estimated that 1 is .04, we would saythat another year increases the volume of lumber by 4%.

  • Log-Log

    A Log-Log Regression Specification.

    log(y) = 0 + 1 log(x1) +

    You might want to run this specification if you think thatpercentage increases in x lead to constant percentagechanges in y . (e.g. constant demand elasiticity).

  • Log-Log

    A Log-Log Regression Specification.

    log(y) = 0 + 1 log(x1) +

    You might want to run this specification if you think thatpercentage increases in x lead to constant percentagechanges in y . (e.g. constant demand elasiticity).To caculate marginal effects. Solve for y ...

    log(y) = 0 + 1 log(x1) + y = e0+1 log(x1)+

    ... and differentiate w.r.t. x

    dydx1

    =1

    x1e0+1 log(x1)+ = 1

    yx1

  • Log-Log Continued.

    Log-Log Regression Specification Continued...

    log(y) = 0 + 1 log(x1) +

    From previous slide the marginal effect is

    dydx1

    = 1yx1

  • Log-Log Continued.

    Log-Log Regression Specification Continued...

    log(y) = 0 + 1 log(x1) +

    From previous slide the marginal effect is

    dydx1

    = 1yx1

    Solving for 1 we get

    1 =dydx1

    x1y

    Hence 1 is an elasticity. If x1 is price and y is demand andwe estimate 1 = .6, it means that a 10% increase in theprice of the good would lead to a 6% decrease in demand.

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