Overthinking a finite geometric sum


Ever have that lightbulb moment in class that stays with you? Recently in one of my ECON 611 (Macroeconomic Theory II) recitations, we explored the permanent-income life-cycle hypothesis (PILCH) – basically, how unexpected changes in income (permanent vs. temporary) affect spending habits throughout your life.

We delved into an interesting question: how much does your consumption growth react to a temporary change in your endowment (think of it as your overall financial resources)? We suspected a link between the transience (temporary nature) of the endowment shock and the change in consumption growth. Time constraints kept us from formally proving this hunch, but we did explore the extreme cases.

The cool part? The answer hinges on a straightforward concept in math – analyzing the properties of a finite geometric series. Don’t let the technical term fool you! This exercise underscored an important lesson: sometimes a step-by-step analysis of individual terms, even with technical concepts like PILCH and geometric series, can lead to a deeper understanding than jumping straight to complex solutions. After all, even the simplest problems can teach valuable lessons!

The problem

Consider the following exercise.

Problem. Let $x \in (0,1)$ and $p\in \mathbb{N}$ such that $p>1$. Then the (finite) geometric sum

\[\begin{aligned} S \equiv \sum_{i=0}^p x^i = \frac{1-x^{p+1}}{1-x}\qquad (1) \end{aligned}\]

is strictly increasing and strictly convex in $x$.

Friendly solution 1. We readily obtain that

\[\begin{aligned} \frac{d}{dx}S &= 1 + 2x + 3x^2 + \cdots + px^{p-1}\\ \frac{d^2}{dx^2}S &= 2 + 6x + \cdots + p(p-1)x^{p-2} \end{aligned}\]

which are two polynomials of degree $p-1$ and $p-2$ whose coefficients are all positive, which together with $0<x<1$ implies that both equations are strictly positive. Duh, the end. $\square$

Not-that-friendly solution 2. Let’s focus our attention on the right-hand side expression of (1). Some tedious algebra allows us to obtain that

\[\begin{aligned} \frac{d}{dx}S &= \frac{1-\left[1+p(1-x)\right]x^p}{(1-x)^2}\\ \frac{d^2}{dx^2}S &= \frac{2\left[1-(1-p)(1+p)x^p\right] - p(p+1)x^{p-1} - p(p+1)x^{p+1}}{(1-x)^3} \end{aligned}\]

Our goal is to show that the numerators of both expressions are positive.

Claim 1. For any $x\in (0,1)$ and $p\in \mathbb{N}$ with $p>2$, it is true that $1>\left[1+p(1-x)\right]x^p$.

Proof. By virtue of Bernoulli’s inequality, we find that

\[(2-x)^p \geq 1 + p(1-x)\]

Since $(1-x)^2 > 0$, it follows that $1 > x(2-x)$. Therefore $1>[x(2-x)]^p \geq [1+p(1-x)]x^p$, as desired. $\square$

This proves that $dS/dx>0$. Similarly, we argue that

Claim 2. The function $g(x) = 2\left[1-(1-p)(1+p)x^p\right] - p(p+1)x^{p-1} - p(p+1)x^{p+1}$ is strictly positive on $(0,1)$.

Proof. We readily deduce that

\[\begin{aligned} g'(x) = -p(p-1)(p+1)x^{p-2}(x-1)^2 \end{aligned}\]

which is guaranteed to be negative if the conditions on $x$ and $p$ hold. This result implies that $g$ is strictily decreasing on $(0,1)$. As $g(0) = 2$ and $g(1) = 0$, we immediately conclude that $g$ is strictly positive on $(0,1)$, as required. $\square$

While Solution 1 just relies on examining individual terms, our second approach is substantially more tiresome (yet fun). This approach highlights the value of examining terms themselves, and not just the final summation (if it exists). Don’t forget this rule of thumb: sometimes, a breakdown can lead to a breakthrough!

Implications in the context of the PILCH model with quadratic preferences

The permanent-income life-cycle hypothesis (PILCH) model in its finite-horizon, discrete version asserts that, under a quadratic utility function and the premise that $\beta(1+r) =1$ (i.e., we remove unnecessary dynamics), the optimal contingent plan is given by

\[\theta_t\Delta c_t = \frac{r}{1+r}\sum_{j=0}^{T-t} \frac{\left(\mathbb{E}_t - \mathbb{E}_{t-1}\right)y_{t+j}}{(1+r)^j}\]

which indicates that households allocate a fraction $r\theta^{-1}/(1+r)$ of the revision in permanent income to current consumption growth.

Let us suppose that the endowment process is autoregressive of order 1, i.e., AR(1). Specifically, suppose that

\[y_t = \rho y_{t-1} + \varepsilon_t\]

where $\varepsilon_t$ is a zero-mean i.i.d. innovation and $0<\rho<1$. An easy induction coupled with iterated expectations provides \(\mathbb{E}_t[y_{t+j}] = \rho^j y_t\ \forall\ j\geq 0\). This allows us to rewrite the optimal consumption path as

\[\theta_t\Delta c_t = \frac{r\varepsilon_t}{1+r}\sum_{j=0}^{T-t} \left(\frac{\rho}{1+r}\right)^j = \beta r \varepsilon_t \dfrac{1-(\beta \rho)^{T-t+1}}{1-\beta\rho}\]

But this is just the expression in our problem with $p = T-t$ and $x = \beta\rho$. We already proved that the expression above is strictly positive and convex in $x$. We conclude that the more persistent the endowment process is (i.e., the closer $\rho$ is to $1$), the stronger the changes in consumption growth to changes in the AR(1) parameter $\rho$. This makes sense, since in the presence of high persistence any small innovation in the endowment process affect all future endowments. If the endowment process is rather transitory ($\rho$ is close to 0), consumption responds to changes in permanent income, but only to a limited degree.

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