PEA 2026 PEA Q20 | ISI MSQE

PEAMCQModerate

2026 PEA Q20

$n$ agents simultaneously choose private consumption $x_i$ and public-good contribution $y_i$ to maximise

u_i = x_i\, \Big(\sum_{j=1}^n y_j\Big),

subject to $x_i + y_i = b$ , $x_i, y_i \ge 0$ . As $n$ increases, the total Nash equilibrium expenditure on the public good

Reveal answer and solution

Answer

Solution

1
Let $Y_{-i} = \sum_{j\ne i} y_j$ . Agent $i$ chooses $y_i \in [0,b]$ to maximise
2
$u_i = (b - y_i)(y_i + Y_{-i}).$
3
The FOC is
4
$-(y_i + Y_{-i}) + (b - y_i) = 0 \quad\Longrightarrow\quad y_i = \frac{b - Y_{-i}}{2}.$
5
At a symmetric equilibrium $y_i = y^*$ and $Y_{-i} = (n-1) y^*$ :
6
$y^* = \frac{b - (n-1) y^*}{2} \quad\Longrightarrow\quad (n+1) y^* = b \quad\Longrightarrow\quad y^* = \frac{b}{n+1}.$
7
Therefore total public-good expenditure is
8
$Y \;=\; n y^* \;=\; \frac{n b}{n + 1}.$
9
The map $n \mapsto \dfrac{n}{n+1}$ is strictly increasing and approaches $1$ as
10
$n\to\infty$ . Hence $Y$ monotonically increases and approaches $b$ .

Answer structure / marking notes

Per-capita contribution $b/(n+1)$ shrinks, but it shrinks slower than $1/n$ , so the aggregate grows.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2026/ISI_MSQE_PEA_2026_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.