Suppose you repeatedly sample from continuous distribution F with convex support
Let's say you drew 2 4 3 5 in order.
Denote the biggest number at $t$th sampling by $b(t)$, second biggest number at $t$th sampling by $a(t)$
So we have
$a(4)=4$
$b(4)=5$
My question is whether
$$A=\int_{a(t)}^{b(t)}dF(x)$$
will be decreasing, at least in expectation sense, over time.
Again, we repeatedly sample from continuous distribution F with convex support.
Intuitively, this must be true. Just imagine uniform distribution, then the distance between $b(t)$ and $a(t)$ will likely shrink.
But I can't seem to prove mathematically. In fact, I don't even know how to mathematically express the concept of "second biggest".
How should I even proceed?
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My tentative approach is as follows.
If $b(t+1)>b(t)$, then $a(t+1)=b(t)$
$$A(1)=\int_{}^{b(1)}dF(x)$$ $$A(2)=\int_{b(1)}^{b(2)}dF(x)=F(b(2))-F(b(1))$$ if $b(2)>b(1)$ $$A(t)=\int_{a(t)}^{b(t)}dF(x)$$ $$A(t+1)=\int_{b(t)}^{b(t+1)}dF(x)=F(b(t+1))-F(b(t))$$
For a uniform distribution, for example on $[0,1]$, you are correct. The difference between the highest value and the second highest value of $t$ values is distributed with density $t(1-x)^{t-1}$ on $[0,1]$ and the expected difference is $\frac{1}{t+1}$, which falls towards $0$ as $t$ increases without limit
But for an exponential distribution, say with parameter $\lambda$, the difference between the highest value and the second highest value of $t$ values is also exponentially distributed with parameter $\lambda$ and so the expected difference is $\frac{1}{\lambda}$ which does not change as $t$ increases (an earlier question and answer gives more details)
And for probability distributions with a heavier right tail than an exponential distribution, such as a log-normal distribution, the expected difference between the highest value and the second highest value of $t$ values can increase as $t$ increases, and for some distributions can even be infinite.
So the truth of your idea will depend on the particular distribution