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Suppose $x_1,x_2,...x_n$ are i.i.d. normal random variables, $a_1,a_2,...a_n$ are some positive constants. Could we have following equation $$\sum_{i=1}^n\frac{a_ix_i^2}{\sum_{i=1}^n x_i^2}=O_p(\frac{\sum^n_{i=1}a_i}{n})?$$ How to get this?

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    $\begingroup$ Hi and welcome! Can you please show some of what you've tried so far? $\endgroup$ – MattAllegro Mar 8 '15 at 10:02
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\begin{align} \sum_{i=1}^n\frac{a_ix_i^2}{\sum_{i=1}^n x_i^2} = \frac{(\sum_{i=1}^n a_ix_i^2) /n}{(\sum_{i=1}^n x_i^2)/n} \end{align}

By the strong law of large number, $(\sum_{i=1}^n x_i^2)/n$ converges to 1 almost surely, so we only need to prove $$\frac{(\sum_{i=1}^n a_ix_i^2) /n}{(\sum_{i=1}^n a_i) /n} = O_p(1)$$ i.e.

$$\frac{\sum_{i=1}^n a_ix_i^2}{\sum_{i=1}^n a_i} = O_p(1)$$

Since $E\dfrac{\sum_{i=1}^n a_ix_i^2}{\sum_{i=1}^n a_i} = 1$, we can use $$P(\frac{\sum_{i=1}^n a_ix_i^2}{\sum_{i=1}^n a_i} >M) \leq \frac{1}{M}$$ to conclude

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