0
$\begingroup$

Let $g_i: [0, 1] \to \mathbb R$ be $L^1$ functions, equibounded in $L^1$ norm. Let $X_i$ a sequence of iid uniform random variables on $[0, 1]$. Suppose that

$$\frac{1}{n} \sum_{i = 1}^n g_i (X_i) \to 0$$

in probability as $n \to \infty$.

Question: Is it true that

$$\frac{1}{n} \sum_{i = 1}^n g_i \to 0$$

in measure as $n \to \infty$ with respect to Lebesgue measure on $[0, 1]$?

Improve this question
$\endgroup$
3
  • $\begingroup$ Maybe I am confusing the definitions, but doesn't the first assumption (by conditioning on all $X_i, i\ge 2$) imply that $g_1$ is constant almost everywhere? And the same for the rest of $g_i$, so the first and second series are just sums of constants and they converge to zero simultaneously? $\endgroup$
    – Aleksei Kulikov
    Commented Nov 25 at 13:51
  • 1
    $\begingroup$ Unless of course there are factors $\frac{1}{n}$ missing in both expressions... $\endgroup$
    – Aleksei Kulikov
    Commented Nov 25 at 13:58
  • $\begingroup$ Oh, crap my bad. @AlekseiKulikov there should be those factors of course. $\endgroup$
    – Nate River
    Commented Nov 25 at 14:03

1 Answer 1

Reset to default
1
$\begingroup$

Counterexample: $g_i(x)=x-1/2$ for all $i$ and all $x\in[0,1]$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ah, of course it’s too weak… $\endgroup$
    – Nate River
    Commented Nov 25 at 14:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .