I've been facing the following problem:

Let $(X_k, Y_k)_k$ be a sequence of $2$-dimensional, independent random variables, each with uniform distribution over $ B(0,k) $

Verify if the following random variable sequence converges:

$ \frac{1}{n} \sum \limits_{k=1}^n \textbf{1}_{\{X_n < Y_n\}} $

So what I have reached so far:

If we fix $ \omega \in \Omega $, by Stolz theorem:

$ \lim\limits_{n \rightarrow \infty} \frac{a_1 + \dots + a_n}{n} = \lim\limits_{n \rightarrow \infty} a_n $, thus:

$ \lim\limits_{n \rightarrow \infty} \frac{1}{n} \sum \limits_{k=1}^n \textbf{1}_{\{X_n < Y_n\}} = \lim\limits_{n \rightarrow \infty} \textbf{1}_{\{X_n < Y_n\}} $.

And there goes the difficulty: if I am to verify if this sequence converges almost sure, I need to know what it might converge to. What might be the limit of such sequence? And if there is none, how to prove it?

Thanks in advance

  • 2
    $\begingroup$ Ok, so I got this: Let's define a new variable: $Z_n = 1 \iff X_n < Y_n $ and $ 0 $ otherwise Then, by large numbers law ($Z_n$ is iid: $ \lim\limits_{n \rightarrow \infty} \frac{1}{n}\sum\limits_{k=1}^n \textbf{1}_{\{Z_k = 1\}} = \mathbb{E}Z_1 = \frac{1}{2} $ $\endgroup$ – Jytug Jun 3 '14 at 19:24

Indeed, the key point is to apply the law of large numbers to the sequence $(Z_j)_{j\geqslant 1}:=(\chi_{\{X_j\lt Y_j\}})_{j\geqslant 1}$. Since $(X_k,Y_k)_k$ is independent, so is $(Z_j)_{j\geqslant 1}$. Since $X_j\lt Y_j\Leftrightarrow X_j/j\lt Y_j/j$ and the random variables $ X_j/j$ and $Y_j/j$ are independent and uniformly distributed on the unit interval, $Z_j$ has the same distribution as $Z_1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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