1
$\begingroup$

I currently have that for $X_1, \ldots, X_n$ independent Bernoulli random variables with parameter $\frac{1}{i}$ random variables,

$$ \frac{\sum_{i=1}^{n}(X_i-\frac{1}{i})}{\sqrt{\sum_{i=1}^{n}(\frac{1}{i}-\frac{1}{i^2})}} \to_{D} N(0,1) $$

This fact I established by Lyapunov's.

How can I show that if $\sum_{i=1}^n\frac1i=\log n+O(1)\qquad \ \text{and} \ \ \ \sum_{i=1}^n\frac1{i^2}=O(1)$,

Then:

$$ \frac{\sum_{i=1}^{n}X_i - \log n}{\sqrt{\log n}} \to_{D} N(0,1) $$

?

$\endgroup$
1
$\begingroup$

Since the sum from 1 to n $1/i$ = ln n +O(1) you can substitute ln n for the sum $1/i$ in the numerator of your result and since sum $1/i^2 = O(1)$. sum $(1/i -1/i^2)=$ln n+$O(1)$ and ln n can be used to replace the quantity under the square root sign in the denominator.

$\endgroup$

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.