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In the log-log regression case,

$$\log(Y) = B_0 + B_1 \log(X) + U \>,$$

can you show $B_1$ is the elasticity of $Y$ with respect to $X$, i.e., that $E_{yx} = \frac{dY}{dX}(\frac{Y}{X})$?

cardinal
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    The result follows directly from the formula for elasticity in the [Wikipedia article](http://en.wikipedia.org/wiki/Elasticity_%28economics%29) because the form of the rhs shows $B_1$ is the derivative of $\log(Y)$ with respect to $\log(X)$. – whuber Apr 23 '11 at 21:54
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    BTW, I believe you need to fix the formula for $E_{y,x}$: it should be a quotient, not a product. – whuber Apr 23 '11 at 22:04

2 Answers2

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whuber has made the point in the comment.

If $\log_e(Y) = B_0 + B_1\log_e(X) + U$ and $U$ is independent of $X$ then taking the partial derivative with respect to $X$ gives $\frac{\partial Y}{\partial X}\cdot\frac{1}{Y} = B_1\frac{1}{X}$, i.e. $B_1 = \frac{\partial Y}{\partial X}\cdot\frac{X}{Y}$.

$E_{y,x} = \lim_{X \rightarrow x} \frac { \Delta Y} { y} / \frac { \Delta X} { x}$, which is the same thing. Take absolute values if you want to avoid negative elasticities.

Henry
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Responding 9 years later in case anyone is interested

If we consider $X = e^{{\log{X}}}$ then $$\beta = \frac{d \log{Y}}{d \log X} = \frac{d \log{Y}}{d Y} \frac{d Y}{d X} \frac{dX}{d \log{X}}= \frac{1}{Y} \frac{dY}{dX} \frac{X}{1}= \frac{X}{Y} \frac{dY}{dX} = \ elasticity$$

CJ.Learns.Math
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