Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $\mathrm P(X_n\geqslant\frac12)\to\frac12$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.