<p>Your objective function is additive separable between <span class="math-container">$myPrice$</span> and <span class="math-container">$competitorPrice$</span>. This means that you can write:
<span class="math-container">$$
Objective = f(myPrice) + g(competitorPrice).
$$</span>
For some functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span>. In such case, the optimal level of <span class="math-container">$myPrice$</span> will indeed be independent of the value of <span class="math-container">$competitorPrice$</span>. Notice that maximizing <span class="math-container">$Objective$</span> will choose <span class="math-container">$myPrice$</span> to maximize <span class="math-container">$f$</span> (whatever the value of <span class="math-container">$g$</span>).</p>
<p>If you want to have something non-separable, you need to change your first stage equation such that <span class="math-container">$myPrice$</span> and <span class="math-container">$competitorPrice$</span> is no longer additively separable.</p>

CC BY-SA 4.0
2023-08-22T15:33:57.373Z
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
MarcM
CC BY-SA 4.0
2023-08-23T07:08:48.210Z
@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 $\ln(Volume) = constant + \alpha \ln(ownPrice) + \beta \ln(competitorPrice) + \gamma \ln(ownPrice) \ln(competitorPrice)$.
tdm
CC BY-SA 4.0
2023-08-22T11:14:11.977Z
2023-08-22T11:14:11.977Z
tdm
<p>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:</p>
<p><span class="math-container">$$log(volume)= constant+elasticityCoef×(myPrice−competitorPrice)$$</span></p>
<p>I want to optimize profit, and given the log-linear form of the demand model, I assume the objective function should be:
<span class="math-container">$$Objective=log(volume)+log(myPrice−costs)$$</span>
equivalent to
<span class="math-container">$$Objective=constant+ elasticityCoef×(myPrice−competitorPrice)+log(myPrice−Costs)$$</span></p>
<p>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?</p>

<optimization><profit-maximization><log-linearization>
Problem with Optimizing Profit in Log-Linear Demand Model
MarcM
# Problem with Optimizing Profit in Log-Linear Demand Model

(Asked by: username_0 on 2023-08-22)

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?

**Tags:** optimization, profit-maximization, log-linearization

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## 1 answers

### Answer by username_1 on 2023-08-22. Score: 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.

> **Comments:**
> **username_0**: 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  
> **username_1**: @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 $\ln(Volume) = constant + \alpha \ln(ownPrice) + \beta \ln(competitorPrice) + \gamma \ln(ownPrice) \ln(competitorPrice)$.  

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