Let $a_n$ and $b_n$ be two sequences of numbers, both uniformly distributed in $(0,1)$. Then, $$ \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}f(a_r b_r) = \int_{0}^{1}\int_{0}^{1} f(xy)dxdy $$

Now let $A_n$ and $B_n$ be two increasing sequences of positive integers such that , both the ratios $\frac{A_r}{A_n}$ and $\frac{B_r}{B_n}$, $r = 1,2,\ldots n$ approach uniform distribution in $(0,1)$ as $n \to \infty$. Then,

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}f\Big(\frac{A_r B_r}{A_n B_n}\Big) = \int_{0}^{1} f(x^2)dx $$

I found this somewhat unexpected as I would have expected both the above limits to have the same integral form since I was expecting the $a_r b_r$ to have a behavior similar to that of $\frac{A_r}{A_n}\frac{B_r}{B_n}$ but this does not seem to be the case.

Main question: What is the root cause due to which we have two different integral i.e why is the behavior of $a_r b_r$ different from that of $\frac{A_r}{A_n}\frac{B_r}{B_n}$?

Secondary question: Under what conditions or for which functions $f(x) \ne c$, is

$$ \int_{0}^1 \int_{0}^1 f(xy)dxdy = \int_{0}^1 f(x^2)dx $$


The answer to Question 1 has to do with the fact that in the first case, $a_n$ and $b_n$ are unordered, so the joint distribution of $\{(a_r, b_r)\}_{r \ge 1}$ is uniform on the unit square $(0,1)^2$. But in the second expression both $A_r/A_n$ and $B_r/B_n$ are ordered, so their joint distribution is no longer uniform on the unit square. Informally speaking, the product $$\frac{A_r B_r}{A_n B_n}$$ never has a term where $A_r/A_n$ is "small" but $B_r/B_n$ is "large" for a given index $r$, or vice versa. Each is uniform on $(0,1)$ but together, they are highly correlated--in the limit, perfectly correlated.


Take $f(x) = \ln(x)$


$\int_{[0,1]} \int_{[0,1]} f(xy)dxdy = \int_{[0,1]} \int_{[0,1]} \ln(xy) dx dy = \int_{[0,1]} \int_{[0,1]} ( \ln(x) + \ln(y)) dx dy = 2(x\ln(x) - x) |_{0}^{1} = -2 $

And $\int_{[0,1]} f(x^{2})dx = \int_{[0,1]} \ln(x^{2}) dx= \int_{[0,1]} 2\ln(x)dx = 2(x\ln(x) -x)|_{0}^{1} = -2$


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