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The demand function for air travel between the U.S. and Europe has been estimated to be $$\ln Q = 2.737 - 1.247 \ln P +1.905 \ln I$$ where $Q$ denotes number of passengers (in thousands) per year, $P$ the (average) ticket price and $I$ the U.S. national income.

Determine the price elasticity and income elasticity of demand.

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Here's a hint on the price elasticity. Totally differentiate that equation $\ln Q = a + b \ln P + c \ln I$ with respect to $P$ and solve for $b$. Compare to the usual formula for elasticity. – Dimitriy V. Masterov Sep 18 '13 at 2:46
@Spring23874 Can you show us any attempts you have made at this problem or where you are stuck in trying to solve it? – user25658 Sep 18 '13 at 2:47
Unsure how to start to solve without more data (like a table to run a regression with in excel); which was not provided. Identified the answer on another site: 1.247 ln P = -5.0710 and 1.905 ln I = 7286. But trying to learn how to calculate. – Spring23874 Sep 18 '13 at 3:52
Everything you need to answer the question is in that equation. – Dimitriy V. Masterov Sep 18 '13 at 9:43
I know very little about economics but: is this a statistical question in any way? @DimitriyV.Masterov, or OP, perhaps you can clarify? If not, I'm not sure where this would be on topic since the economics SE has closed ( maybe?). – Macro Sep 18 '13 at 15:01

The estimated equation is \begin{equation}\ln Q = a + b \cdot \ln P + c \cdot \ln I.\end{equation} Take the derivative of both sides with respect to $P$. Very roughly, the estimated parameters $a,b,c$ are constants and $I$ is an exogenous variable (we've assumed the price change is so small that it does not change national income). Variables $Q$ and $P$ are functions of $P$. This gives you \begin{equation} \frac{1}{Q} \cdot \frac{\partial Q}{\partial P} = 0 + b \cdot \frac{1}{P} \cdot \frac{\partial P}{\partial P} + 0.\end{equation} Rearranging terms to get the price elasticity: \begin{equation}\varepsilon_{Q,P} = \frac{\partial Q}{\partial P} \cdot \frac{P}{Q} = b\end{equation}

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Although the answer is correct, the logic is reversed: you solve for the elasticity in order to discover that the answer is $b$, not the other way around! – whuber Sep 18 '13 at 19:33
Excellent point! Changed. – Dimitriy V. Masterov Sep 18 '13 at 22:05
+1 If you literally followed the definition in your link, you would obtain $b$ immediately as the unit change in $\log(Q)$ with respect to $\log(P)$, with no algebraic rearrangement needed: in short, the elasticities are none other than the coefficients in any log-log regression. – whuber Sep 18 '13 at 22:16

Thanks to all who have presented here on the powerful and efficient manner showing how elasticity can be computed when one has a simple OLS model relating Q to P. This may be an interesting topic for textbooks but lacks generality needed in practice. Often Q is not only a function of P but one or more lags of P and sometimes future expectations of P. Additionally there may be the need to incorporate ARIMA structure in addition to P and its lags. More generally there may be unobservable input variables in the mix which can often be proxied by level/step shifts, time trends, seasonal pulses, or one-time pulses. Other structures often needed in practice include holiday effects/day-of-week effects, promotions etc. The reason for identifying and incorporating these kinds of effects is to render the error process Gaussian, leading to valid tests of significance for all estimated coefficients.

With this in mind, a more general/robust approach to calculating an elasticity is to form a useful model and generate a forecast (baseline forecast) using an expected P (user-specified) for the next period. Take that same model and the same parameters and increase P by 1%. Compute the difference in the forecasted Q and convert the difference in the two forecasts to a percent change from the baseline forecast. In this way you have not sacrificed model form, but you can still compute elasticity. The advantage of the double log approach is speed, but this may entail using a sub-standard model.

Transformations of any kind should be viewed as possibly helpful BUT possibly harmful. The whole idea is to keep the model as simple as possible, but not too simple. Some analysts think that incorporating pulse indicators is kitchen-sink modelling, but they are only point adjustments for proven unusual values, much like the value 5 in the process {1,9,1,9,1,9,1,9,1,9,5,9,1,9,1,9,1,9,1,9}. The 5, if untreated, will bias the characterization of the process.

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