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Pick two integer sequences $d>a_n\geq a_{n-1}\geq\dots a_1\geq0$ and $d>b_n\geq b_{n-1}\geq\dots b_1\geq0$ where $d$ is an integer bound with following method:

  1. Pick $a_n$ uniformly from $[0,d]$ and rest of $i\in\{1,\dots,n-1\}$ uniformly from $[0,a_{i+1}]$.

  2. Do similar picking for $b_i$'s.

What is the distribution of $\max_i\log|a_i-b_i|$ and expected value of $\max_i\log|a_i-b_i|$ as a function of $d$ and $n$?

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  • $\begingroup$ because it is a discrete problem there is a positive probability that the a's and b's are exactly the same, and the expected value is -inf $\endgroup$ – mike Sep 16 '19 at 8:17
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The distribution of absolute difference of two uniform random variable is: $$P(|a_n-b_n| \leqslant t) = \frac{t(2d-t)}{d^2}$$ Since $\log \ \& \ \exp$ are increasing functions we have: $$P_n = P(\log|a_n-b_n| \leqslant t') = P(|a_n-b_n| \leqslant e^{t'}) = \frac{e^{t'}(2d-e^{t'})}{d^2}$$ Now notice that: $$a_m,b_m \longrightarrow 0 \Longrightarrow |a_m-b_m| \longrightarrow 0 \Longrightarrow \log|a_m-b_m| \longrightarrow -\infty$$ So the final answer must be a decreasing series $\sum_{i=n}^{0}P_i$. Since $\{P_i\}$ is a sequence that decreases by high rate, we have: $$\sum_{i=n}^{0}P_i \simeq P_n$$ And also for its expected value: $$E(\sum_{i=n}^{0}P_i) \simeq E(P_n) = \frac{e^{2d}-4de^d+2d^3+4d-1}{2d^2}$$

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(Too long for a comment.)

I doubt there's a closed-form expression, but one can obviously find the distribution recursively. Let us consider a more general scenario, where $a_n$ is chosen uniformly from $\{0,1,2,\ldots,A\}$, while $b_n$ — from $\{0,1,2,\ldots,B\}$. Write $p(A,B,n,x)$ for the probability that the maximum of $|a_i - b_i|$ is no greater than $x$. Then $$ p(A,B,n,x) = \frac{1}{(A+1)(B+1)} \sum_{i = 0}^A \sum_{j = 0}^B \mathbb{1}_{\{|i - j| \leqslant x\}} p(i, j, n-1, x) , $$ and $p(i,j,0,x) = 1$ for every $x$.

If I am not mistaken, the expectation for $A = B = n = 5$ turns out to be $\frac{48733041639733}{33592320000000}$, which does not resemble anything nice. The $n \to \infty$ limit might be more tractable.

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  • $\begingroup$ denominator factored by $2,3,5$ and numerator square free. $\endgroup$ – VS. Sep 16 '19 at 23:34
  • $\begingroup$ If anyone wants to play with it, here is a Mathematica code for evaluating $p$: p[a_, b_, 0, x_] = 1; p[a_, b_, n_, x_] := p[a, b, n, x] = Sum[p[i, j, n - 1, x], {i, 0, a}, {j, Max[0, i - x], Min[b, i + x]}]/(a + 1)/(b + 1); $\endgroup$ – Mateusz Kwaśnicki Sep 17 '19 at 8:24

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