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I am interested in two variables $x, y$. Their (observed empirical) standard deviations are $\sigma_x$, $\sigma_y$. I know the elasticity of $x$ w.r.t. $y$ is $\eta_{x,y}$.

Let $x$ be fully determined through $y$, which is the only source of randomness. Is it true that

$$ \frac{\sigma_x}{mean(X)} = \eta_{x,y} \cdot \frac{\sigma_y}{E[y]}$$

?

I can't find anything on this relationship, but it appears that appropriately scaled standard deviations of two variables should be connected through their elasticity.

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  • $\begingroup$ It seems like the end of your question got cut-off. $\endgroup$ – Ubiquitous Mar 11 '15 at 19:25
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Let $X = h(Y)$. A first-order Taylor expansion around $E(Y) = \mu_y$ gives

$$X \approx h[\mu_y] + h'[\mu_y]\cdot [Y-\mu_y]$$

This easily leads to

$$\sigma^2_x \approx \big(h'[\mu_y]\big)^2\cdot \sigma^2_y$$

Since

$$\eta_{x,y} \equiv h'\cdot \frac Yh \implies \eta_{x,y}(\mu_y) = h'(\mu_y)\frac {\mu_y}{h(\mu_y)} \implies h'(\mu_y) = \eta_{x,y}(\mu_y) \cdot \frac {h(\mu_y)}{\mu_y}$$

Substituting,

$$\sigma^2_x \approx \left(\eta_{x,y}(\mu_y) \cdot \frac {h(\mu_y)}{\mu_y}\right)^2\cdot \sigma^2_y$$

Taking the square root we are led to

$$\frac {\sigma_x}{h(\mu_y)} \approx \eta_{x,y}(\mu_y) \cdot \frac {\sigma_y}{\mu_y} $$

The difference of the above from the expression in the question is that

a) The elasticity must be evaluated at the center of the Taylor expansion used and
b) $h(\mu_y)$ is usually equal to $E(X)=E[h(Y)]=\mu_x$ only to a first approximation, due to Jensen's inequality.

Of course if the relation between $X$ and $Y$ is linear, and/or the elasticity is constant, things specialize.

In any case, it is valid as an approximation to write

$${\rm cv}_x \approx \eta_{x,y}(\mu_y) \cdot {\rm cv}_y $$

where "cv" stands for "coefficient of variation".

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