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For any $n$, let $X_1, ..., X_n$ be i.i.d. random variables on the probability space $(\Omega, \mathcal{F}, Pr)$ where $\Omega = \mathbb{R}^d$. Define $$ \mu_n(A) = \frac{1}{n} \sum_{i=1}^n 1_{\{X_i \in A\}}, \\ \mu(A) = Pr(X_1 \in A). $$

Consider any $\mathcal{A} \subset \mathcal{F}$, and define

$$ g(X_1, ..., X_n) := \sup_{A \in \mathcal{A}} |\mu_n(A) - \mu(A)| $$

Prove that $g(X_1, ..., X_n)$ converges in probability to $0$ implies that $g(X_1, ..., X_n)$ converges almost surely to $0$.

p/s: This problem is posed as an exercise in the book "Combinatorial methods in density estimation" by Luc Devroye and Gabor Lugosi (exercise 3.2 with a hint to use the bounded difference inequality). I have tried but could not solve it. Hopefully, someone could help. Thank you.

Edit: As noted by @NickyLevering, I change "iff" to "implies" because it is well-known that a.s. convergence implies convergence in probability (also changed the title to reflect this better). I also set $\Omega = \mathbb{R}^d$ to make it clearer.

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  • $\begingroup$ Per definition of a.s. convergence and convergence in probability the a.s. convergence implies convergence in probability, thus you only have to prove the 'only if' statement. $\endgroup$ Aug 6 '20 at 11:27
  • $\begingroup$ Are you sure about the definitions of $\mu_n$ and $\mu$? One is a random variable and the other is a number? Also what is $\mathcal{A}$? $\endgroup$ Aug 6 '20 at 11:29
  • $\begingroup$ @Keen-ameteur Yes I am sure. Yes. $\mathcal{A}$ is any set of subsets of $\Omega$. $\endgroup$
    – thanhtang
    Aug 6 '20 at 11:30
  • $\begingroup$ @NickyLevering Yes, right. $\endgroup$
    – thanhtang
    Aug 6 '20 at 11:32
  • $\begingroup$ If $\mathcal{A}$ is not a sub-collection of $\mathcal{F}$, then this is not well defined. $\endgroup$ Aug 6 '20 at 11:32
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Abbreviate $g_n=g(X_1,...,X_n)$ and note, first of all, that for any $A$,

$$ \left\|\sum_{i=1}^n 1_{X_i\in A}-\sum_{i=1}^n 1_{Y_i\in A}\right\|\leq |\{i|X_i\neq Y_i\}| $$ Hence, we can apply the bounded differences inequality with $c_i=\frac{1}{n}$ and get that

$$ Pr(|g_n-\mathbb{E} g_n|\geq t)\leq 2\exp(-2tn) $$ for every $t>0$.

In particular, $$ \sum_{n=1}^{\infty}Pr(|g_n-\mathbb{E} g_n|\geq \frac{1}{\sqrt{n}})\leq 2 \sum_{n=1}^{\infty}\exp(-2\sqrt{n})<\infty, $$ so by Borel Cantelli, $g_n-\mathbb{E} g_n\to 0$ almost surely.

Thus, if $g_n\to 0$ in probability, we want to prove that $\mathbb{E}g_n$ goes to $0$. This is equivalent to proving that every subsequence $g_{n_k}$ has a subsequence $g_{n_{k_j}}$ such that $\mathbb{E} g_{n_{k_j}}\to 0$. This follows since $g_{n_k}\to 0$ in probability and hence, has a subsequence $g_{n_{k_j}}\to 0$ almost surely. Then, since $0\leq g_{n_{k_j}}\leq 1$, we can apply Dominated Convergence to get $\mathbb{E}g_{n_{k_j}}\to 0$.

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  • $\begingroup$ Of course, you're right in using Borel Cantelli and then proving that $\mathbb{E} g_n \rightarrow 0$. For proving $\mathbb{E} g_n \rightarrow 0$, I am not sure if I fully understand it for now (but you are probably right there); I just need more time to really understand it and I will approve your answer once I fully understand it. One thing I wonder is that do we really need $g_n \rightarrow 0$ in prob. for $\mathbb{E} g_n \rightarrow 0$? We have $\mathbb{E} g_n = O(1/\sqrt{n})$ (proved in Luc's book), so it approaches $0$ regardless of whether $g_n \rightarrow 0$ in probability or not? $\endgroup$
    – thanhtang
    Aug 6 '20 at 13:12
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    $\begingroup$ I mean... you can definitely convince me that $g_n$ should go to $0$ under very mild conditions. You definitely need $X_1$ to have first moment, though, right? $\endgroup$ Aug 6 '20 at 13:20
  • $\begingroup$ Nevermind. The last part is not obviously true. $\endgroup$ Aug 6 '20 at 13:21
  • $\begingroup$ Yes, right. For the result $\mathbb{E}g_n = O(1/\sqrt{n})$ to hold, it needs some conditions about the boundedness of an integral related to the covering number of $\mathcal{A}$. The condition that $Z_n \rightarrow 0$ in prob. seems like also another mild condition for $E Z_n \rightarrow 0$. I think I got it now. Many thanks :-) $\endgroup$
    – thanhtang
    Aug 6 '20 at 13:38

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