Question: If $F$ is a CDF, under what conditions is $\ \lim_\limits{n\to\infty} n^{-2}\sum_\limits{k=-n}^{n}kF(k)=1/2\ ?$
I've only proved this limit analytically in one case (a double exponential distribution), but numerically it seems to hold for all the usual "named" continuous distributions on $\mathbb{R},$ as well as some others I've tested (e.g., various rational polynomial CDFs). So I'm curious just how general this behavior might be.
Context:
I was playing with this variant of Cover's "Pick the Largest Number" game: integers $(A,B)$ have been chosen uniformly at random from $\{(a,b): a\ne b,\ -n\le a\le n,\ -n\le b\le n\}$ where $n$ is some fixed unknown positive integer, and the player -- knowing only that $(A,B)$ are unequal integers -- not knowing their values nor anything else about them (not even how they were generated) -- selects one of them uniformly at random and calls it $X$. The player must then decide whether $X$ is the larger vs. smaller of the two, i.e. $``\,X=\max(A,B)"$ vs. $``\,X=\min(A,B)".$ The point is that there exists a strategy that guarantees $P[\text{correct decision}]$ to be strictly greater than $1/2:$ $$P[\text{correct decision}] \gt 1/2\\[2ex] if \\[2ex] \text{decision} = \begin{cases} ``\,\text{max}"&\text{if $F(X)\ge U$}\\ ``\,\text{min}"&\text{otherwise} \end{cases}$$ where $U$ is a number the player chooses uniformly at random from the interval $(0,1)$ and $F$ is any strictly increasing CDF the player wants to use.
But how large might this probability actually be when $n$ is "large"? Seeking the limit probability as $n\to\infty$, here's what I find: $$\begin{align}&P[\text{correct decision}]\\[2ex] &= E_{(A,B)}\,P[\text{correct decision}\mid (A,B)]\\[2ex] &= E_{(A,B)}\,\big\{P[F(X)\ge U\ \ \land\ \ X=\max(A,B)\mid(A,B)]\\ &\quad\quad\quad\quad + P[F(X)\lt U\ \ \land\ \ X=\min(A,B)\mid(A,B)]\\[2ex] &= E_{(A,B)}\,\big\{P[F(X)\ge U\mid X=\max(A,B), (A, B)]\,P[X=\max(A,B)\mid (A, B)]\\ &\quad\quad\quad\quad + P[F(X)\lt U\mid X=\min(A,B), (A, B)]\,P[X=\min(A,B)\mid (A, B)] \big\}\\[2ex] &= E_{(A,B)}\,\big\{P[F(\max(A,B))\ge U\mid X=\max(A,B), (A, B)]\,P[X=\max(A,B)\mid (A, B)]\\ &\quad\quad\quad\quad + P[F(\min(A,B))\lt U\mid X=\min(A,B), (A, B)]\,P[X=\min(A,B)\mid (A, B)] \big\}\\[2ex] &= E_{(A,B)}\,\left\{F(\max(A,B))\,{1\over 2} + (1-F(\min(A,B))\,{1\over 2}\right\}\\[2ex] &= {1\over 2} +{1\over 2}\,E_{(A,B)}\left\{F(\max(A,B))-F(\min(A,B))\right\}\\[2ex] &= {1\over 2} +{1\over 2}\,\sum_{k=-n}^nF(k)\left({n+k\over n(2n+1)}-{n-k\over n(2n+1)}\right)\\[2ex] &\quad\quad\quad\quad\text{(having inserted the PMFs of max$(A,B)$ & min$(A,B)$)}\\[2ex] &= {1\over 2} +{1\over (2n+1)n}\,\sum_{k=-n}^nkF(k).\\[2ex] \end{align}$$
Hence the reason for posting this question:
$$\lim_\limits{n\to\infty}n^{-2}\sum_{k=-n}^nkF(k)=1/2\ \ \implies\ \ \lim_\limits{n\to\infty}P[\text{correct decision}]=3/4\ \ (!)\ $$ (Not bad for consulting a random-number "oracle" to decide what seems like a 50/50 proposition!)
As $k \rightarrow -\infty$, $kF(k) =o(k)$, so that $\sum_{-n \leq k < 0}{kF(k)}=o\left(\sum_{-n \leq k < 0}{k}\right)=o(n^2)$.
Now, as $k \rightarrow +\infty$, $1-F(k) \rightarrow 0$ so that $\sum_{0 \leq k \leq n}{k(1-F(k))} = o\left(\sum_{k=0}^n{k}\right) = o(n^2)$.
It follows that $\sum_{k=-n}^n{kF(k)}=o(n^2)+\sum_{k=1}^n{k}=\frac{n(n+1)}{2}+o(n^2)=\frac{n^2}{2}(1+o(1))$, which concludes.