I have estimated the following dynamic panel data model using GMM:

          $\ln Y_{it}=\beta_0+\beta_1\ln Y_{it-1}+\beta_2\ln X_{it-1}+\epsilon_{it} $

where $Y$ is employment and $X$ is productivity.

Suppose that the estimated coefficient on $\ln X_{it-1}$ is $\hat{\beta_2}=0.3$. If I understand correctly, because of the dynamic structure of the model, it can be considered that the dependent variable is the growth rate of $Y$. Hence, is the following interpretation of the estimated coefficient correct: any $1\%$ increase in productivity would be expected to increase the employment growth rate by $0.3$ percentage points?

If not, what would be the correct interpretation?

  • 1
    $\begingroup$ If $\beta_1\not =1$ then you really do not have the growth rate. On the other hand, if you do then $\frac{Y_{it}}{Y_{it-1}} -1\approx \ln Y_{it}-\ln Y_{it-1}= \beta_0+\beta_2\ln X_{it}+\epsilon_{it}$ and a $1\%$ higher $X_{it}$ could indeed increase the projected growth rate by about $\beta_2 /100$ $\endgroup$
    – Henry
    May 18, 2021 at 19:37
  • $\begingroup$ Thank you. In the case where $\beta_1\not =1$, would you have any advice on how $\beta_2$ should be interpreted in the presence of the lagged dependent variable? $\endgroup$
    – gkr
    May 18, 2021 at 19:53
  • 1
    $\begingroup$ That situation is where economics starts head towards incomprehensibility so you should ask an economist rather than a statistician. Essentially if $\beta_0+\beta_2 \ln X_{it} = (1-\beta_1) \ln Y_{it-1}$ would keep employment stable subject to the $\epsilon_{it}$ term, and if that equality does not apply then the remainder of the $\beta_2 \ln X_{it}$ (positive or negative) would then change employment $\endgroup$
    – Henry
    May 18, 2021 at 22:38
  • 1
    $\begingroup$ If you subtract $\ln Y_{it-1}$ on both sides and diff with respect to $\ln X_{it-1}$ you get $\partial (\ln Y_{it}-\ln Y_{it-1})/\partial \ln X_{it-1} = \beta_2$ which supports the interpretation you are suggesting - irrespective of the value of $\beta_1$. $\endgroup$ May 19, 2021 at 19:29

1 Answer 1


tldr If $\mathbb{E}[\varepsilon_{i,t}|X_{i,t-1}, Y_{i,t-1}] = 0$ then the coefficient $\beta_2$ is equal to: $$ \frac{\partial \mathbb{E}[\ln Y_{i,t}|X_{i,t-1}= x_{i,t-1}, Y_{i,t-1} = y_{i,t-1}]}{\partial \ln x_{i,t-1}} $$ If $\varepsilon_{i,t}$ is independent of $X_{i,t-1}$ and $Y_{i,t-1}$ (which is a stronger condition) then it is also equal to: $$ \frac{\partial \ln \mathbb{E}[Y_{i,t}|X_{i,t-1}= x_{i,t-1}, Y_{i,t-1} = y_{i,t-1}]}{\partial \ln x_{i,t-1}} $$ In both cases, they give an estimate for the elasticity of employment with respect to productivity yesterday (conditional on employment yesterday), i.e. what is the (expected) percentage point change in employment (today) due to a 1 percentage increase in productivity (yesterday) given the level of employment yesterday.

estimating elasticities

If $$ \ln y = \alpha + \beta \ln x, $$ Then the $x$-elasticity of $y$ is given by: $$ \frac{\partial \ln y}{\partial \ln x} = \beta. $$ It measures the percentage change in $y$ due to a 1 percent increase in $x$.

Now let's go to the stochastic framework: $$ \ln Y = \alpha + \beta \ln X + \varepsilon $$ And assume, as usual, that $\mathbb{E}[\varepsilon] = 0$.

The elasticity of $Y$ with respect ot $X$ is not really defined as $Y$ is not a function of $X$ (i.e. both are random variables).

There are two natural generalizations of elasticity in this setting:

  1. $\dfrac{\partial \ln \mathbb{E}[Y|X = x]}{\partial \ln x}$, which is the elasticity of the conditional mean function $\mathbb{E}[Y|X = x]$.
  2. $\dfrac{\partial \mathbb{E}[\ln Y|X = x]}{\partial \ln x}$, which is the derivative of the conditional mean function $\mathbb{E}[\ln Y|X = x]$.

Because the log is a non-linear transformation, the two are not necessarily the same.

However, if $X$ and $\varepsilon$ are independent then they are identical and equal to $\beta$. To see this, notice that for 1. $$ Y = e^\alpha X^\beta e^\varepsilon $$ Then taking conditional expecations and using the independence between $\varepsilon$ and $X$: $$ \mathbb{E}[Y|X = x] = e^\alpha x^\beta \mathbb{E}[e^\varepsilon|X = x] = e^\alpha X^\beta \mathbb{E}[e^\varepsilon],\\ \to \ln \mathbb{E}[Y|X = x] = \alpha + \beta \ln x + \ln \mathbb{E}[e^\varepsilon]. $$ Taking the partial derivative with respect to $\ln(x)$ gives: $$ \frac{\partial \ln \mathbb{E}[Y|X = x]}{\partial \ln x} = \beta. $$

For 2. we immediately have: $$ \mathbb{E}[\ln Y|X = x] = \alpha + \beta \ln x + \mathbb{E}[\varepsilon|X = x] = \alpha + \beta \ln x. $$ So: $$ \frac{\partial E[\ln Y|X = x]}{\partial \ln x} = \beta. $$

If only $\mathbb{E}[\varepsilon|X] = 0$ but $X$ and $\varepsilon$ are not independent, then only the derivation of 2 is valid and 1 and 2 are not necessarily equal.

If you want to estimate the elasticity of the growth rate, then the right regression is indeed: $$ \ln(Y_{i,t}/Y_{i,t-1}) = \beta_0 + \beta_1 \ln X_{i,t} + \varepsilon_{i,t}. $$

  • $\begingroup$ Thank you. I have a couple of questions: 1. In the tldr, did you perhaps mean a 1% change in employment (rather than a percentage point change)? 2. Does conditioning on the previous level of employment impact calculations in practice? For example, if I want to use the estimated result of $\hat{\beta_2}=0.3$ to estimate the impact of employment change on productivity for every year of a ten year period, do I just apply the $0.3$ figure to the employment change figures? $\endgroup$
    – gkr
    May 24, 2021 at 9:01

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