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.

  • $\begingroup$ It seems like the end of your question got cut-off. $\endgroup$ – Ubiquitous Mar 11 '15 at 19:25

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

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


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


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

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".

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.