<p>The textbook way to derive this type of comparative statics is to use the implict function theorem.</p>
<p>The first order condition gives:
<span class="math-container">$$
\dfrac{\partial f}{\partial x}(x^\ast(\theta), \theta) - \dfrac{\partial c}{\partial x}(x^\ast(\theta)) = 0.
$$</span>
Differentiating with respect to <span class="math-container">$\theta$</span> gives:
<span class="math-container">$$
\dfrac{\partial^2 f}{\partial x^2} (x^\ast(\theta), \theta) \dfrac{d x^\ast(\theta)}{d \theta} + \dfrac{\partial^2 f}{\partial \theta \partial x}(x^\ast(\theta), \theta) - \dfrac{\partial^2 c}{\partial x^2}(x^\ast(\theta),\theta) \dfrac{d x^\ast(\theta)}{d \theta} = 0 
$$</span>
So:
<span class="math-container">$$
\dfrac{d x^\ast(\theta)}{d \theta} =  \dfrac{-\dfrac{\partial^2 f}{\partial \theta \partial x}(x^\ast(\theta), \theta)}{\dfrac{\partial^2 f}{\partial x^2}(x^\ast(\theta), \theta)- \dfrac{\partial^2 c}{\partial x^2}(x^\ast(\theta))}
$$</span>
If the objective function is strictly concave, then the denominator will be negative, so <span class="math-container">$\dfrac{d x^\ast(\theta)}{d \theta}$</span> will be positive if:
<span class="math-container">$$
\dfrac{\partial^2 f}{\partial \theta \partial x}(x^\ast(\theta), \theta) &gt; 0.
$$</span></p>

CC BY-SA 4.0
2021-12-27T08:59:13.967Z
2021-12-27T08:59:13.967Z
tdm
<p>I am thinking about the following question.</p>
<p>Let <span class="math-container">$f(x,\theta)$</span> be a strictly concave real-valued function in <span class="math-container">$x$</span> and <span class="math-container">$c(x)$</span> be a strictly convex real-valued function in <span class="math-container">$x$</span>, both <span class="math-container">$x$</span> and <span class="math-container">$\theta$</span> belong to an closed interval of <span class="math-container">$\mathbb R$</span>. Additionally, <span class="math-container">$f$</span> and <span class="math-container">$c$</span> are twicely differentiable and <span class="math-container">$\frac{\partial f}{\partial x\partial \theta}&gt;0$</span></p>
<p>Let <span class="math-container">$$x^{*}(\theta)=argmax_{x}\,\,f(x,\theta)-c(x)$$</span></p>
<p>If <span class="math-container">$x^{*}(\theta)$</span> is an interior solution, does <span class="math-container">$x^{*}$</span> increase in <span class="math-container">$\theta$</span>?</p>
<p>My answer to this is yes, and here's how I prove it.</p>
<p>By FOC, we know at the optimum, we must have <span class="math-container">$$\frac{\partial f(x^{*}(\theta),\theta)}{\partial x}-\frac{\partial c(x^{*}(\theta))}{\partial x}=0\tag{1}$$</span></p>
<p>Now if <span class="math-container">$\theta$</span> increases to <span class="math-container">$\theta'$</span>, since <span class="math-container">$\frac{\partial f}{\partial x\partial \theta}&gt;0$</span>, we know <span class="math-container">$$\frac{\partial f(x^{*}(\theta),\theta')}{\partial x}&gt;\frac{\partial f(x^{*}(\theta),\theta)}{\partial x}$$</span></p>
<p>Therefore, we know <span class="math-container">$$\frac{\partial f(x^{*}(\theta),\theta')}{\partial x}-\frac{\partial c(x^{*}(\theta))}{\partial x}&gt;0\tag{2}$$</span> Since <span class="math-container">$x^{*}(\theta')$</span> is the new maximizer, we know FOC still holds at <span class="math-container">$x^{*}(\theta')$</span>, i.e <span class="math-container">$$\frac{\partial f(x^{*}(\theta'),\theta')}{\partial x}-\frac{\partial c(x^{*}(\theta'))}{\partial x}=0\tag{3}$$</span></p>
<p>Compare equation <span class="math-container">$(2)$</span> and <span class="math-container">$(3)$</span>, and combined with the facts that <span class="math-container">$f$</span> is strictly concave and <span class="math-container">$c$</span> is strictly convex in <span class="math-container">$x$</span>, we know <span class="math-container">$x^{*}(\theta')&gt;x^{*}(\theta)$</span>. Hence we know <span class="math-container">$x^{*}$</span> strictly increases in <span class="math-container">$\theta$</span>.</p>
<p>Is my proof correct? And are there any theorems for such problems?</p>
<p>Thanks in advance!</p>

<optimization><comparative-statics>
Proof of a comparative statics result in a maximization problem
John
# Proof of a comparative statics result in a maximization problem

(Asked by: username_0 on 2021-12-26)

I am thinking about the following question.

Let $f(x,\theta)$ be a strictly concave real-valued function in $x$ and $c(x)$ be a strictly convex real-valued function in $x$, both $x$ and $\theta$ belong to an closed interval of $\mathbb R$. Additionally, $f$ and $c$ are twicely differentiable and $\frac{\partial f}{\partial x\partial \theta}>0$

Let $$x^{\*}(\theta)=argmax\_{x}\,\,f(x,\theta)-c(x)$$

If $x^{\*}(\theta)$ is an interior solution, does $x^{\*}$ increase in $\theta$?

My answer to this is yes, and here's how I prove it.

By FOC, we know at the optimum, we must have $$\frac{\partial f(x^{\*}(\theta),\theta)}{\partial x}-\frac{\partial c(x^{\*}(\theta))}{\partial x}=0\tag{1}$$

Now if $\theta$ increases to $\theta'$, since $\frac{\partial f}{\partial x\partial \theta}>0$, we know $$\frac{\partial f(x^{\*}(\theta),\theta')}{\partial x}>\frac{\partial f(x^{\*}(\theta),\theta)}{\partial x}$$

Therefore, we know $$\frac{\partial f(x^{\*}(\theta),\theta')}{\partial x}-\frac{\partial c(x^{\*}(\theta))}{\partial x}>0\tag{2}$$ Since $x^{\*}(\theta')$ is the new maximizer, we know FOC still holds at $x^{\*}(\theta')$, i.e $$\frac{\partial f(x^{\*}(\theta'),\theta')}{\partial x}-\frac{\partial c(x^{\*}(\theta'))}{\partial x}=0\tag{3}$$

Compare equation $(2)$ and $(3)$, and combined with the facts that $f$ is strictly concave and $c$ is strictly convex in $x$, we know $x^{\*}(\theta')>x^{\*}(\theta)$. Hence we know $x^{\*}$ strictly increases in $\theta$.

Is my proof correct? And are there any theorems for such problems?

Thanks in advance!

**Tags:** optimization, comparative-statics


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## 1 answers

### Answer by username_1 on 2021-12-27. Score: 3 | Accepted answer

The textbook way to derive this type of comparative statics is to use the implict function theorem.

The first order condition gives:
$$
\dfrac{\partial f}{\partial x}(x^\ast(\theta), \theta) - \dfrac{\partial c}{\partial x}(x^\ast(\theta)) = 0.
$$
Differentiating with respect to $\theta$ gives:
$$
\dfrac{\partial^2 f}{\partial x^2} (x^\ast(\theta), \theta) \dfrac{d x^\ast(\theta)}{d \theta} + \dfrac{\partial^2 f}{\partial \theta \partial x}(x^\ast(\theta), \theta) - \dfrac{\partial^2 c}{\partial x^2}(x^\ast(\theta),\theta) \dfrac{d x^\ast(\theta)}{d \theta} = 0
$$
So:
$$
\dfrac{d x^\ast(\theta)}{d \theta} = \dfrac{-\dfrac{\partial^2 f}{\partial \theta \partial x}(x^\ast(\theta), \theta)}{\dfrac{\partial^2 f}{\partial x^2}(x^\ast(\theta), \theta)- \dfrac{\partial^2 c}{\partial x^2}(x^\ast(\theta))}
$$
If the objective function is strictly concave, then the denominator will be negative, so $\dfrac{d x^\ast(\theta)}{d \theta}$ will be positive if:
$$
\dfrac{\partial^2 f}{\partial \theta \partial x}(x^\ast(\theta), \theta) > 0.
$$


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