# Proof of a comparative statics result in a maximization problem

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?

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: $$\dfrac{\partial^2 f}{\partial \theta \partial x}(x^\ast(\theta), \theta) > 0.$$