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$X$ is democracy level

$Y$ is infant mortality rate (IMR)

In my model, 1 unit increase in $X$ predicts a $20\%$ decrease in $Y$

Does this mean 5 unit increase in $X$ will eradicate IMR? In other words, can I conclude that 5 unit increase in democracy predicts $100\%$ decrease in IMR?

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The regression models of positive outcomes usually consider multiplicative decrease in outcome and additive change in predictor. Assuming model $\log Y=\beta X+\epsilon$. Then an additive change in $X$ by one unit changes $Y$ by a multiplicative factor $e^\beta$, which can be translated into percent change with $(e^\beta-1)\cdot 100$.

Increasing $X$ by five units i.e. $X=x_0+5$ gives us $Y=y_0 \cdot \exp(\beta)^5$ with $y_0=\epsilon \exp(\beta x_0)$. Assuming that

1 unit increase in X predicts a 20% decrease in Y

then $\exp(\beta)= 1-20/100=.8$ and for 5 units increase in $X$, $Y$ decreases by a factor $\exp(\beta)^5=0.8^5=0.33$. That is approx. $67\%$ decrease.

It is sometimes recommended to approximate the logarithm with $\log(z)=z-1$ which is what You implicitly do when You claim that the percantage change is equal to $\beta$. The approximation is only valid for small changes in $Y$. Thus, with a change in $X$ by one unit, the approximation gives $-19.5$ for a true value of $(e^\beta-1)\cdot 100=-17.71\%$ which is reasonably accurate. However, when $X$ changes by five units, the approximation fails. Approximation suggests a change by $-97.5\%$ while the true value is $(e^{5\beta}-1)\cdot 100=-62.28\%$. When in doubt avoid the approximation and perform the exact calculation.

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  • $\begingroup$ Hi Matus, thanks for answering. could you explain a bit more please. i am trying to find the relationship between democracy level and infant mortality rate (IMR). I log-transformed the dependent variable (IMR). my equation is now: log Y= 4.107-0.195*democracylevel. The interpretation is: one unit increase in democracy level will predict 19.5% reduction in IMR. but how do i find the effect if democracy level is increased by 5 units? $\endgroup$ – Hans Apr 18 '17 at 18:34
  • $\begingroup$ @Hans I added more details. Let me know if additional clarification is needed. $\endgroup$ – matus Apr 18 '17 at 23:41

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