# Obtaining demand/supply elasticity of prices

First, the general formula for price elasticity of demand or supply, the change with which demand or supply responds to price changes, is $$\frac{dQ}{dP}.\frac{P}{Q}$$. Second, if I'd instead want to express the sensitivity of prices to changes in supply/demand, then I'd presumably write it as: $$\frac{dP}{dQ}.\frac{Q}{P}$$, which is just the inverse of price elasticity of demand/supply. Now my question: if I want to know real-life responses of prices to changes in supply and demand of various goods and services, can I just take the empirical observations of their price elasticities and likewise invert these results? Or is it not that straight-forward and would such demand/supply elasticities need to be obtained from separate statistical computations?

Now my question: if I want to know real-life responses of prices to changes in supply and demand of various goods and services, can I just take the empirical observations of their price elasticities and likewise invert these results?

In principle yes. For example, if you estimate your structural model for demand, then in second stage you are in essence estimating parameters of the demand curve. For example, in simple 2SLS a second stage would look like:

$$q = \beta_0 + \beta_1 \hat{p} + \epsilon \tag{*}$$

Then once you find your $$\hat{\beta}_i$$ you still find that

$$\frac{\partial q}{\partial \hat{p}} \frac{\hat{p}}{q} = \frac{\hat{\beta}_1 \hat{p}}{\hat{\beta}_0 +\hat{\beta}_1 \hat{p}} = \left( \frac{\hat{\beta}_0+\hat{\beta_1} \hat{p}}{\hat{\beta}_1 \hat{p}}\right)^{-1} \tag{**}$$

The inside of bracket of last term of ** is $$\frac{\partial \hat{p}}{\partial q } \frac{q}{\hat{p}}$$ you can verify that by solving * for \$\hat{p} after estimating the parameters to get:

$$\hat{p} = \frac{q-\hat{\beta_0}}{\hat{\beta}_1}$$

then

$$\frac{\partial \hat{p}}{\partial q } \frac{q}{\hat{p}} = \frac{q}{q-\beta_0}$$

substituting back for $$q = \hat{\beta}_0 + \hat{\beta}_1 \hat{p}$$ you get:

$$\frac{\hat{\beta}_0+\hat{\beta_1} \hat{p}}{\hat{\beta}_1 \hat{p}}$$