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I have a question about the following probability:


where $u_k\sim \exp(1)$ are i.i.d. exponential random variables with mean one (also, $\frac{\sum^N_{k=1}u_k}{N}$ is gamma distributed).

I have plotted this probability for different $N$. The plot shows that as $N$ increases, this probability approaches $0.5$. Is this a well-known result? Has someone proved it already? If not, how to prove it rigorously?

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You are right. This type of result is, however, not new, and applies to a very large class of random variables of a shape similar to yours. In particular, the fact that the random variables being averaged have exponential distribution has almost nothing to do with the result.

For a proof, it is probably best to use the Central Limit Theorem. Call your random variable (the average of the $N$ exponentials) by the name $Y_N$.

Since you are taking an average of $N$ random variables with mean $1$, the random variable $Y_N$ has mean $1$. Since each of the exponentials has variance $1$, and they are independent, their sum has variance $N$, and therefore $Y_N$, which is $1/N$ times the sum, has variance $\frac{N}{N^2}$, which is $\frac{1}{N}$.

The Central Limit Theorem says that if $Y_N$ is the average of $N$ independent random variables with mean $\mu$ and variance $\sigma^2$, then $$\lim_{N\to\infty}P\left(\sqrt{N}(Y_N-\mu)\le z\right)=\Phi(z/\sigma),$$ where $\Phi$ is the cumulative distribution function of the standard normal. In our case, we have $\mu=1$ and $\sigma=1$. So we can rewrite the above result as $$\lim_{N\to\infty}P\left(Y_N\le 1+\frac{z}{\sqrt{N}}\right)=\Phi(z). \qquad(\ast)$$ Now just put $z=0$. Since $F(0)=1/2$, we get the fact that you observed.

With relatively well-behaved random variables like your mean $1$ exponentials, the approach to normality is very rapid. Thus we can, for largish $N$, remove the limit part, and use $(\ast)$ as an estimate of $P(Y_N-1\le \frac{z}{\sqrt{N}})$.

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thank you very much for such a detailed proof. It's a rigorous proof, and very clear. Thanks for these inputs. – Scholli Feb 23 '12 at 9:24
By the way, I assume that the last probability is $P(Y_N<1)<\frac{z}{\sqrt{N}}$. – Scholli Feb 23 '12 at 9:25
@Scholli: Thanks for pointing out the typo. Fixed. But it may not be the only one. – André Nicolas Feb 23 '12 at 9:44
I think the other calculations are correct ;) – Scholli Feb 23 '12 at 14:10

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