<p>You can obtain a definite answer for <span class="math-container">$$\text{sign}\left\{\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha }\right\}$$</span> given the assumptions.</p>

<p>From</p>

<p><span class="math-container">$$S_{1}^{*} - f\left ( g\left ( S^{*}_{1}, \beta \right ), \alpha \right ) = 0 \equiv F$$</span></p>

<p>and the Implicit Function Theorem</p>

<p><span class="math-container">$$\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha } = - \frac{\partial F/\partial \alpha}{\partial F/\partial S^*_{1}} $$</span></p>

<p>we have  </p>

<p><span class="math-container">$$\frac{\partial F}{\partial \alpha} = -\frac{\partial f}{\partial \alpha}$$</span></p>

<p>and </p>

<p><span class="math-container">$$\frac{\partial F}{\partial S^*_{1}} = 1-\frac{\partial f}{\partial S_2}\cdot \frac{\partial g}{\partial S^*_{1}}$$</span></p>

<p>For these expressions we know not only the signs but also the magnitudes. The result follows.</p>

CC BY-SA 4.0
2018-10-03T15:03:10.617Z
So the effect is negative then? I don't understand why you don't have $\frac{\partial f\left (  \right )}{\partial g\left (  \right )}$ but instead have $\frac{\partial f\left (  \right )}{\partial S_{2} \left (  \right )}$.
OGC
CC BY-SA 4.0
2018-10-03T15:07:12.683Z
@OGC It is the same thing, since $g$ has the exact same position in $f$ as $S_2$. Also, it appears you are forgetting to take into account one minus sign. The effect is positive.
Alecos Papadopoulos
CC BY-SA 4.0
2018-10-03T19:17:11.620Z
Thanks a lot. It's easy to see some of these silly mistakes once someone points them to you.
OGC
CC BY-SA 4.0
2018-10-03T09:13:28.360Z
2018-10-03T09:13:28.360Z
Alecos Papadopoulos
<p>In the state of Mexas, two politicians (Mr. BO, or "Politician 1" and Mr. TC, or "Politician 2") are competing intensely for a senate seat. The two politicians spend on advertising to increase the number of supporters. A political consultant finds that the optimal advertising expenditure of Mr. BO, <span class="math-container">$S_{1}$</span>, depends on the spending <span class="math-container">$S_{2}$</span> by Mr. TC and a "likability" parameter <span class="math-container">$\alpha$</span> that influences the popularity of Mr. BO in Mexas: <span class="math-container">$$\text{Equation 1:  }S_{1} = f\left ( S_{2}, \alpha \right ),$$</span>
where <span class="math-container">$f:\mathbb{R}_{+}^{2}\rightarrow \mathbb{R}$</span> is twice continuously differentiable, <span class="math-container">$0&lt;\frac{\partial f\left ( S_{2}, \alpha \right )}{\partial S}&lt;1$</span> and <span class="math-container">$0&lt;\frac{\partial f\left ( S_{2}, \alpha \right )}{\partial \alpha}&lt;1$</span> for all <span class="math-container">$S_{2} \geq 0$</span> and all <span class="math-container">$\alpha \geq 0$</span>. </p>

<p>The optimal expenditure of politician 2 depends on the spending <span class="math-container">$S_{1}$</span> of politician 1 and a "redness" parameter <span class="math-container">$\beta$</span>, which influences how many people would stick to Mr. TC: <span class="math-container">$$\text{Equation 2:  }S_{2} = g\left ( S_{1}, \beta \right ),$$</span>
where <span class="math-container">$g:\mathbb{R}_{+}^{2}\rightarrow \mathbb{R}$</span> is twice continuously differentiable, <span class="math-container">$0&lt;\frac{\partial g\left ( S_{1}, \alpha \right )}{\partial S}&lt;1$</span> and <span class="math-container">$0&lt;\frac{\partial g\left ( S_{1}, \alpha \right )}{\partial \alpha}&lt;1$</span> for all <span class="math-container">$S_{1} \geq 0$</span> and all <span class="math-container">$\beta \geq 0$</span>. </p>

<p>The equilibrium values of <span class="math-container">$S_{1}$</span> and <span class="math-container">$S_{2}$</span> are given by the solution of the simultaneous equations (1) and (2). Suppose that there is a unique solution <span class="math-container">$S_{1}^{*}&gt;0$</span> and <span class="math-container">$S_{2}^{*}&gt;0$</span>. </p>

<p>Does an increase in <span class="math-container">$\alpha$</span>(holding <span class="math-container">$\beta$</span> constant) necessarily increase or necessarily decrease <span class="math-container">$S_{1}^{*}$</span>? Explain. </p>

<p><strong>My attempt:</strong> Here I have used Implicit Function Theorem to answer the question, as we are essentially looking for the comparative statics of <span class="math-container">$\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha }$</span>. </p>

<p>Since from equation 2, <span class="math-container">$S_{2}^{*} = g\left ( S_{1}^{*}, \beta \right )$</span>, I substituted <span class="math-container">$S_{2}^{*}$</span> in equation to obtain <span class="math-container">$$S_{1}^{*} = f\left ( g\left ( S^{*}_{1}, \beta \right ), \alpha \right ).$$</span> Then since <span class="math-container">$$S_{1}^{*} - f\left ( g\left ( S^{*}_{1}, \beta \right ), \alpha \right ) = 0 \equiv F,$$</span>
I can apply the Implicit Function Theorem (IFT) to derive <span class="math-container">$\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha }$</span>.</p>

<p>So <span class="math-container">$$\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha } = - \frac{\frac{\partial F}{\partial \alpha}}{\frac{\partial F}{\partial S_{1}}} = - \frac{\frac{\partial f\left (  \right )}{\partial \alpha}}{1 - \frac{\partial f \left (  \right )}{\partial g} \frac{\partial g \left (  \right )}{\partial S_{1}^{*}}} \frac{&gt;}{&lt;} 0.$$</span></p>

<p>Since we do not know the sign of the derivative <span class="math-container">$\frac{\partial f \left (  \right )}{\partial g}$</span>, the effect of <span class="math-container">$\alpha$</span> on <span class="math-container">$S^{*}_{1}$</span> is ambiguous. </p>

<p>Does my answer make sense?</p>

CC BY-SA 4.0
2018-10-03T08:31:30.013Z
Any thoughts to the question and my work?
OGC
<comparative-statics>
Comparative statics question with an application
OGC
# Comparative statics question with an application

(Asked by: username_0 on 2018-10-03)

In the state of Mexas, two politicians (Mr. BO, or "Politician 1" and Mr. TC, or "Politician 2") are competing intensely for a senate seat. The two politicians spend on advertising to increase the number of supporters. A political consultant finds that the optimal advertising expenditure of Mr. BO, $S\_{1}$, depends on the spending $S\_{2}$ by Mr. TC and a "likability" parameter $\alpha$ that influences the popularity of Mr. BO in Mexas: $$\text{Equation 1: }S\_{1} = f\left ( S\_{2}, \alpha \right ),$$
where $f:\mathbb{R}\_{+}^{2}\rightarrow \mathbb{R}$ is twice continuously differentiable, $0<\frac{\partial f\left ( S\_{2}, \alpha \right )}{\partial S}<1$ and $0<\frac{\partial f\left ( S\_{2}, \alpha \right )}{\partial \alpha}<1$ for all $S\_{2} \geq 0$ and all $\alpha \geq 0$.

The optimal expenditure of politician 2 depends on the spending $S\_{1}$ of politician 1 and a "redness" parameter $\beta$, which influences how many people would stick to Mr. TC: $$\text{Equation 2: }S\_{2} = g\left ( S\_{1}, \beta \right ),$$
where $g:\mathbb{R}\_{+}^{2}\rightarrow \mathbb{R}$ is twice continuously differentiable, $0<\frac{\partial g\left ( S\_{1}, \alpha \right )}{\partial S}<1$ and $0<\frac{\partial g\left ( S\_{1}, \alpha \right )}{\partial \alpha}<1$ for all $S\_{1} \geq 0$ and all $\beta \geq 0$.

The equilibrium values of $S\_{1}$ and $S\_{2}$ are given by the solution of the simultaneous equations (1) and (2). Suppose that there is a unique solution $S\_{1}^{\*}>0$ and $S\_{2}^{\*}>0$.

Does an increase in $\alpha$(holding $\beta$ constant) necessarily increase or necessarily decrease $S\_{1}^{\*}$? Explain.

**My attempt:** Here I have used Implicit Function Theorem to answer the question, as we are essentially looking for the comparative statics of $\frac{\mathrm{d} S\_{1}^{\*} }{\mathrm{d} \alpha }$.

Since from equation 2, $S\_{2}^{\*} = g\left ( S\_{1}^{\*}, \beta \right )$, I substituted $S\_{2}^{\*}$ in equation to obtain $$S\_{1}^{\*} = f\left ( g\left ( S^{\*}\_{1}, \beta \right ), \alpha \right ).$$ Then since $$S\_{1}^{\*} - f\left ( g\left ( S^{\*}\_{1}, \beta \right ), \alpha \right ) = 0 \equiv F,$$
I can apply the Implicit Function Theorem (IFT) to derive $\frac{\mathrm{d} S\_{1}^{\*} }{\mathrm{d} \alpha }$.

So $$\frac{\mathrm{d} S\_{1}^{\*} }{\mathrm{d} \alpha } = - \frac{\frac{\partial F}{\partial \alpha}}{\frac{\partial F}{\partial S\_{1}}} = - \frac{\frac{\partial f\left ( \right )}{\partial \alpha}}{1 - \frac{\partial f \left ( \right )}{\partial g} \frac{\partial g \left ( \right )}{\partial S\_{1}^{\*}}} \frac{>}{<} 0.$$

Since we do not know the sign of the derivative $\frac{\partial f \left ( \right )}{\partial g}$, the effect of $\alpha$ on $S^{\*}\_{1}$ is ambiguous.

Does my answer make sense?

**Tags:** comparative-statics
> **Comments:**
> **username_0**: Any thoughts to the question and my work?  

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## 1 answers

### Answer by username_1 on 2018-10-03. Score: 3 | Accepted answer

You can obtain a definite answer for $$\text{sign}\left\{\frac{\mathrm{d} S\_{1}^{\*} }{\mathrm{d} \alpha }\right\}$$ given the assumptions.

From

$$S\_{1}^{\*} - f\left ( g\left ( S^{\*}\_{1}, \beta \right ), \alpha \right ) = 0 \equiv F$$

and the Implicit Function Theorem

$$\frac{\mathrm{d} S\_{1}^{\*} }{\mathrm{d} \alpha } = - \frac{\partial F/\partial \alpha}{\partial F/\partial S^\*\_{1}} $$

we have

$$\frac{\partial F}{\partial \alpha} = -\frac{\partial f}{\partial \alpha}$$

and

$$\frac{\partial F}{\partial S^\*\_{1}} = 1-\frac{\partial f}{\partial S\_2}\cdot \frac{\partial g}{\partial S^\*\_{1}}$$

For these expressions we know not only the signs but also the magnitudes. The result follows.

> **Comments:**
> **username_0**: So the effect is negative then? I don't understand why you don't have $\frac{\partial f\left (  \right )}{\partial g\left (  \right )}$ but instead have $\frac{\partial f\left (  \right )}{\partial S_{2} \left (  \right )}$.  
> **username_1**: @OGC It is the same thing, since $g$ has the exact same position in $f$ as $S_2$. Also, it appears you are forgetting to take into account one minus sign. The effect is positive.  
> **username_0**: Thanks a lot. It's easy to see some of these silly mistakes once someone points them to you.  

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