4
$\begingroup$

I have $$X_i \sim N(0,1), \quad S_n=X_1+\cdots+X_n,$$ $$ \mathscr{S}_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum_{i=1}^{n} \left[ S_{i-1} (\omega ) + n \left( t - \frac{i-1}{n} \right) X_{i}(\omega) \right] \textbf{1}_{ \left( \frac{i-1}{n}, \frac{i}{n} \right] } (t) $$ and let $$d(f,g) := \sup_{x \in [0,1]}| f(x) - g(x) | .$$

Then $$ Y_i^{(v)} := X_i \textbf{1}_{\{ |X_i| \leq v \} }, X_i^{(v)} := \left( Y_i^{(v)} - E [ Y_1^{(v)} ] \right) $$ analogically we define $$ S_n^{(v)} = \sum_{i=1}^n X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} } - E \left[ \sum_{i=1}^n X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} } \right] $$ and $\mathscr{S}_n^{(v)}$. In the lemma 3.2 (lecture notes on Donsker's theorem Davar Khoshnevisan,p.6, https://www.math.utah.edu/~davar/ps-pdf-files/donsker.pdf) it's shown that

$$ \sup_{n \geq 1} \| d \left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) \|_2 \leq 2 \sqrt{E \left[ X_1^2; |X_1| > v \right] } \text{ for all } v>0.$$

But to prove the Donsker's theorem we also need that

$$ \lim_{v \to \infty} \sup_n P \left( d \left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) > \lambda \right)=0$$ for all $\lambda>0$ which should follow from the inequality above and Chebyshev’s inequality. I don't know how to show this.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

You have $$ \begin{aligned} &\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]} \\ &=\sup_{n\ge1}\|d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) \|_2 \leq 2 \sqrt{E \left[ X_1^2; |X_1| > v \right] } \text{ for all } v>0. \end{aligned} $$ So, by the Chebyshev/Markov inequality, $$ \begin{aligned} &\lim_{v \to \infty} \sup_n P \left( d \left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) > \lambda \right) \\ &\le\lim_{v \to \infty} \sup_n \frac{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]}{\lambda^2} \\ &\le\lim_{v \to \infty} \frac{4E \left[ X_1^2; |X_1| > v \right] } {\lambda^2} =0, \end{aligned} $$ because, by the dominated convergence theorem, $E \left[ X_1^2; |X_1| > v \right]\to0$ as $v\to\infty$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ There should be $\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]}$ right? But i have another stupid question. Why the $\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]} = \sup_{n\ge1}||{d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)}||_2$ ? $\endgroup$
    – nodis6
    Commented May 1, 2022 at 9:25
  • 1
    $\begingroup$ (i) The brackets are optional. By a standard convention, $EY^2:=E[Y^2]$. (ii) We have $\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]} =\sup_{n\ge1}\|d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) \|_2$ because, by definition, $\|d( \mathscr{S}_n^{(v) }, \mathscr{S}_n ) \|_2=\sqrt{E[d( \mathscr{S}_n^{(v) }, \mathscr{S}_n )^2]}$ or, generally, $\|Y\|_2:=\sqrt{EY^2}$. $\endgroup$
    – Iosif Pinelis
    Commented May 1, 2022 at 12:54
  • $\begingroup$ I'm so sory but i don't know one thing to understand entire proof. How doeas the the dominated convergence theorem is used here? $\endgroup$
    – nodis6
    Commented May 1, 2022 at 13:51
  • $\begingroup$ @nodis6 : Let $f_v(x):=x^2\,1(|x|>v)$ and let $\mu$ denote the probability distribution of $X_1$, so that $E[ X_1^2; |X_1| > v]=\int_{\mathbb R}f_v(x)\,\mu(dx)$ and $\int_{\mathbb R}g(x)\,\mu(dx)=E[X_1^2]<\infty$, where $g(x):=x^2$. We have $f\le g$ and $f_v(x)\to0$ for each real $x$ as $v\to\infty$. So, by the dominated convergence theorem, $E[ X_1^2; |X_1| > v]=\int_{\mathbb R}f_v(x)\,\mu(dx)\to0$ as $v\to\infty$. $\endgroup$
    – Iosif Pinelis
    Commented May 1, 2022 at 14:07
  • $\begingroup$ Okay, I'll make a note of it. Thank you very, very much. $\endgroup$
    – nodis6
    Commented May 1, 2022 at 14:14

You must log in to answer this question.

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