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?

  • $\begingroup$ Any thoughts to the question and my work? $\endgroup$
    – OGC
    Commented Oct 3, 2018 at 8:31

1 Answer 1


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


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

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

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


$$\frac{\partial F}{\partial S^*_{1}} = 1-\frac{\partial f}{\partial S_2}\cdot \frac{\partial g}{\partial S^*_{1}}$$

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

  • $\begingroup$ 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 $\frac{\partial f\left ( \right )}{\partial S_{2} \left ( \right )}$. $\endgroup$
    – OGC
    Commented Oct 3, 2018 at 15:03
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Oct 3, 2018 at 15:07
  • 1
    $\begingroup$ Thanks a lot. It's easy to see some of these silly mistakes once someone points them to you. $\endgroup$
    – OGC
    Commented Oct 3, 2018 at 19:17

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