Sequences of Random variables, and Contained limits.

Question

Dear SE Mathematics Community,

I have encountered a little riddle in probability theory, that I would like to put out there.

Question:

Let $I_n=\{\frac{k}{n}; k \in \{1, 2, ..., n\}\}$. Let $X_n:\Omega \to I_n$ the random variable, such that $P(X_n = \frac{k}{n})= \frac{a_nk}{n^2+k^2}$.

1. What is the limit of the sequence $(a_n)$?
2. Prove that $lim E[X_n]=\frac{4-\pi}{2\log(2)}$.

It seems obvious that we can only prove (2.) if we know the answer to (1.). Yet, I seem to run into nowhere with my attempts to find a viable proof to (1.).

Guess:

Let $\Lambda_n$ the set of all possible values of $X_n$. Define some $\Phi\uparrow: \mathbb{N} \to \mathbb{N}$, such that $X_{\Phi(n)}$ a subsequence of $X_n$. Chose $\Phi(n)$ such that $\forall \Phi(n)>\Phi'(n), x \in \Lambda_{\Phi(n)} \implies x\in \Lambda_{\Phi'(n)}$. One possible subsequence would be defined by $\Phi(n)=2n$.

It can be proved that for such sequences $\cup_{n \in \mathbb{N}}\Lambda_{\Phi(n)} = lim_{n \to +\infty} \Lambda_{\Phi(n)}$.

Notice that $\Lambda_n=\{\frac{1}{n}, \frac{2}{n}, ..., \frac{n}{n}\}$. Thus we know that $\forall n \in \mathbb{N}, \Sigma_{k=0}^{n}\frac{a_nk}{n^2+k^2}=1$. Re-write as $a_n\Sigma_{k=0}^{n}\frac{k}{n^2+k^2}=1.$ It's straightforward that this is also true for $\Phi(n)$.

I'm admittedly stuck here, because I think the best way would be to work further with the limits of subsequences, to show some limit-behavior of the sum $\Sigma_{k=0}^{n}\frac{k}{n^2+k^2}$ that in turn provides some information about $a_n$.

Thank you for your help - please, do not spoil (2.) in any way!

Answer 1

It looks to me that this is true:

$$\forall n \in \mathbb{N}, \sum_{k=1}^{n}\frac{a_nk}{n^2+k^2}=1$$

Answer 2

Your guess part is kind of incomprehensible to me, and the actual solution is quite straightforward: $(a_n)$ are defined so that

$$ 1 = \mathbb{P}(X \in I_n) = \sum_{k=1}^{n} \mathbb{P}(X = k/n) = \sum_{k=1}^{n} \frac{a_n k}{n^2+k^2} $$

is satisfied. So $a_n$ is simply the reciprocal

$$ a_n = \left( \sum_{k=1}^{n} \frac{k}{n^2+k^2} \right)^{-1}. $$

Now identifying the sum as the Riemann sum for the function $f(x) = \frac{x}{1+x^2}$ on $[0, 1]$, you can easily deduce that

$$ \lim_{n\to\infty} a_n = \left( \int_{0}^{1} \frac{x}{1+x^2} \, dx \right)^{-1} = \frac{2}{\log 2}. $$