1
$\begingroup$

There is a worked exercise in my book, however there is a line that I am not sure sure about. I understand all of the work before and after this line to finish the proof.

Here is what we are given:

Let $X_i\sim \mathrm{Uniform}(0,1,\ldots,i-1)$, a sequence of independent discrete uniform random variables.

For reference it also uses:

$E[X_i]=\frac{i-1}{2}$

$\operatorname{Var}[X_i]=\frac{i^2-1}{12}$

$B_n^2=\sum_{i=1}^{n}\operatorname{Var}[X_i]=\frac{2n^3+3n^2-5n}{72}\approx\frac{n^3}{36}$, noting this approximation seems to be used somewhere.

For clarification Lindebergs condition is as follows in this case:

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{B_n^2}\sum_{k=1}^{n}E\big[Y_i^2\mathbb{I}_{(|Y_i|>\epsilon\cdot B_n)}\big]=0$ for all $\epsilon >0$ where $Y_i=X_i-\frac{i-1}{2}$

It then goes on to say the Lindeberg condition holds due to the following, stripping several bit's of Lindebergs condition away:

$E\big(|X_i-\frac{i-1}{2}|^2 \mathbb{I}_{\big(|X_i-\frac{i-1}{2}|>\epsilon\cdot B_n\big)}\big)=0$ for all $n\geq\frac{9}{\epsilon^2}$.

This line skipped some steps that I just do not see, and I am hoping to better understand applying Lindebergs CLT off of this example. I particularly am lost on how to deduce this value of $n$.

$\endgroup$
0
$\begingroup$

Observe that the random variable $\left\lvert Y_i\right\rvert$ is always smaller than $i$ (hence than $n$) hence the event $ \left\{\left\lvert Y_i\right\rvert\gt \varepsilon B_n\right\} $ is empty if $n\gt \varepsilon B_n$. Since $B_n$ is of order $n^3$, this certainly happen for $n$ large enough. A ranked depending on $\varepsilon$ where this starts to be true can be made explicit.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.