2
$\begingroup$

Let $\{a_n\}_{n\geq 1}$ be a decreasing sequence of strictly positive real numbers. Let $$ x_n \equiv \frac{\sum_{k=1}^n \alpha_k k}{\sum_{k=1}^n \alpha_k}. $$ Prove $x_{n+1} - x_{n}\leq 1/2$ for all $n$.

My proof is the following. Fix $n$. W.o.l.g., assume $\alpha_n=1$. Then \begin{align*} x_{n+1} - x_n &= \frac{\alpha_{n+1}}{\Big(\sum_{k=1}^n \alpha_k + \alpha_{n+1}\Big)\Big(\sum_{k=1}^n \alpha_k\Big)} \sum_{k=1}^n \alpha_k \big(n+1-k\big)\\ &\leq \frac{\alpha_{n}}{\Big(\sum_{k=1}^n \alpha_k + \alpha_{n}\Big)\Big(\sum_{k=1}^n \alpha_k\Big)} \sum_{k=1}^n \alpha_k \big(n+1-k\big)\\ &\leq \sup_{s\geq n} \frac{1}{(s+1)s} \sup_{\alpha_1\geq \cdots\geq \alpha_n=1} \sum_{k=1}^n \alpha_k (n+1-k)\\ &= \sup_{s\geq n} \frac{n(s-n)+n(n+1)/2}{s(s+1)}\\ &=1/2. \end{align*} Since this result seems very intuitive, I think there might be easier, more direct and elegant proof. Can someone help me?

Thank you!

1 Answers 1

0

(The following is not yet complete)

Put my probabilistic two cents in:

Let $U_n$ have distribution $P(U_n=i)=\alpha_i^{-1}\sum_{k=1}^n\alpha_k$ for $i\in\{1,\dots,n\}$. Then$$E(U_{n+1}|U_{n+1}\leq n)=E(U_{n}),$$ so that \begin{eqnarray*}x_{n+1}-x_n&=&E(U_{n+1})-E(U_{n+1}|U_{n+1}\leq n)\\ &=& p[n+1-E(U_{n})]\\ &\leq&1-(n+1)^{-1}, \end{eqnarray*}

since $p:=P(U_{n+1}=n+1)\leq (n+1)^{-1}$ and $EU_n\geq 1$.

  • 0
    Thanks! But I think $EU_n\leq (n+1)/2$. Indeed, you argument was my first attempt and I got stuck exactly at this step. – 2017-01-18
  • 0
    Oops! I made a temporary fix. Will examine closer later. Thank you. – 2017-01-18