I've trained a linear model to predict log(volume) to capture the elasticity of demand with respect to price difference between my product and a competitor's:

$$log(volume)= constant+elasticityCoef×(myPrice−competitorPrice)$$

I want to optimize profit, and given the log-linear form of the demand model, I assume the objective function should be: $$Objective=log(volume)+log(myPrice−costs)$$ equivalent to $$Objective=constant+ elasticityCoef×(myPrice−competitorPrice)+log(myPrice−Costs)$$

But using this model, I've noticed that for any given competitor price, the optimal price for my product remains fixed (verified using simple differentiation). This seems counterintuitive. What could I be missing?


1 Answer 1


Your objective function is additive separable between $myPrice$ and $competitorPrice$. This means that you can write: $$ Objective = f(myPrice) + g(competitorPrice). $$ For some functions $f$ and $g$. In such case, the optimal level of $myPrice$ will indeed be independent of the value of $competitorPrice$. Notice that maximizing $Objective$ will choose $myPrice$ to maximize $f$ (whatever the value of $g$).

If you want to have something non-separable, you need to change your first stage equation such that $myPrice$ and $competitorPrice$ is no longer additively separable.

  • $\begingroup$ Thanks u for your response, but can you please elaborate more on what you mean by changing my first stage equation. I estimate elasitcity coefficient via Instrumental variables that predicts the difference with competitor prices. Do you suggest to use another variable. What am i failing to see also the economic interpreation of such result $\endgroup$
    – MarcM
    Commented Aug 22, 2023 at 15:33
  • $\begingroup$ @MarcM If you want the optimal price to depend also on the price of the competitor, your regression should have a form such that $myPrice$ and $competitorPrice$ are non-separable. For example $\ln(Volume) = constant + \alpha \ln(ownPrice) + \beta \ln(competitorPrice) + \gamma \ln(ownPrice) \ln(competitorPrice)$. $\endgroup$
    – tdm
    Commented Aug 23, 2023 at 7:08

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