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Suppose I have a continuum of agents $i \in [0,1] $ where each agent i takes the action $x_i$ where

$x_i = 1$ if $ \epsilon_i >-a$ and 0 otherwise

Assume that $ \epsilon_i $ has a standard normal distribution, $N(0,1)$

I want the probability that at least half the agents chose $ x_i =1$

What I have:

Let the prob that one agent chooses $ x_i =1$ be $p$, this is easily found.

We know that as n goes to infinity, the distribution of the choices of the agents goes to a normal distribution by the CLT with mean p and variance p(1-p)

Thus,the answer should be $P(X>0.5) = 1 - \Phi(\frac{0.5-p}{\sqrt{(p(1-p)}})$

Where $\Phi$ is the standard normal CDf

Thoughts? It doesn't took right.

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1 Answer

up vote 2 down vote accepted

The proportion of agents choosing action $1$ amongst $n$ agents is $X_n=\frac1n\sum\limits_{i=1}^nx_i$. The asymptotics of $\mathrm P(X_n\geqslant\frac12)$ is described by the law of large numbers or by the central limit theorem.

The random variables $(x_i)_{1\leqslant i\leqslant n}$ are i.i.d. with mean $p$ hence the law of large numbers says that $X_n\to p$ almost surely. In particular, $\mathrm P(X_n\geqslant y)\to0$ for every $y\gt p$ and $\mathrm P(X_n\geqslant y)\to1$ for every $y\lt p$. Hence:

  • If $p\lt\frac12$ (that is, if $a\lt0$), then $\mathrm P(X_n\geqslant\frac12)\to0$.
  • If $p\gt\frac12$ (that is, if $a\gt0$), then $\mathrm P(X_n\geqslant\frac12)\to1$.

If $p=\frac12$, one can use the central limit theorem. For every $p$, the common variance of the random variables $x_i$ is $\sigma^2=p(1-p)$ hence $Y_n=\frac1{\sigma\sqrt{n}}\sum\limits_{i=1}^n(x_i-p)$ converges in distribution to a standard gaussian random variable $Z$. When $p=\frac12$, $[X_n\geqslant\frac12]=[Y_n\geqslant0]$ and $\mathrm P(Y_n\geqslant 0)\to\mathrm P(Z\geqslant 0)$. Hence:

  • If $p=\frac12$ (that is, if $a=0$), then $\mathrm P(X_n\geqslant\frac12)\to\frac12$.
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