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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!

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1 Answer 1

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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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