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Let $(\xi_n)$ be a sequence of i.i.d. random variables with mean $0$ and variance $\sigma^2 < \infty$ and $s>0$. I need to prove that

$$\frac{1}{\sqrt{\lfloor ns \rfloor}} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i -\frac{1}{\sqrt{ ns} } \sum_{i=1}^{\lfloor ns \rfloor} \xi_i \stackrel{\mathbb{P}}{\longrightarrow} 0$$ Where the convergence above is in probability.

I tried to use the markov inequality to bound $$\mathbb{P} \left( \left|\frac{1}{\sqrt{\lfloor ns \rfloor}} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i -\frac{1}{\sqrt{ ns} } \sum_{i=1}^{\lfloor ns \rfloor} \xi_i \right|>\delta \right)$$ but I didn't have success.

Any help will be appreciated.

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    $\begingroup$ Show that expected value of your expression is 0 and its variance goes to 0. $\endgroup$ – A.S. Sep 30 '15 at 3:32
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    $\begingroup$ Can you bound $\lfloor ns \rfloor^{-1/2} - (ns)^{-1/2} $? (An easier related question: can you bound $x^{-1/2}-(x+1)^{-1/2} $?) $\endgroup$ – Ian Sep 30 '15 at 3:47
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For your specific problem, we know that $\forall n\exists \varepsilon_n: ns=\lfloor ns \rfloor+\varepsilon_n,0\leq \varepsilon_n \leq1\; $ Which makes your problem equivalent to:

$$\frac{1}{\sqrt{\lfloor ns \rfloor}} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i -\frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i \stackrel{\mathbb{P}}{\longrightarrow} 0$$

We can also pull out the sums:

$$\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)\stackrel{\mathbb{P}}{\longrightarrow} 0$$

Since we can bound $\varepsilon_n$, we can bound the absolute value of the LHS between 0 and another sequence:

$$0\leq \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)\right| \leq \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \right|=\left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right| \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right)$$

We can pull out $$\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }}$$ since it is non-negative.

We can also see that:

$$\left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \to 0^+ \;\;(\text{surely})$$

Now, lets bound the rate of convergence:

$$\left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \leq \frac{1}{\lfloor ns \rfloor}$$


Proof

Let $k_n>0$ be a monotonic increasing, unbounded sequence.

$$ k_n\left(\frac{1}{\sqrt{k_n}} - \frac{1}{\sqrt{k_n+1 }} \right) = \sqrt{k_n}-\frac{k_n}{\sqrt{k_n+1}}$$

$$\lim_{k_n\to \infty} \sqrt{k_n}-\frac{k_n}{\sqrt{k_n+1}} = 0 \implies \frac{1}{\sqrt{k_n}} - \frac{1}{\sqrt{k_n+1 }} = o\left(\frac{1}{k_n}\right)$$


Using this convergence, we can see that:

$$ \forall s>0,\exists c \in \mathbb{N}: \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right| \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \leq \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right|\frac{1}{ns}\;\forall n>c$$ Also,

$$\left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right|\frac{1}{ns} \leq \sum_{i=1}^{\lfloor ns \rfloor} \left|\xi_i\right|\frac{1}{ns}\stackrel{\mathbb{P}}{\longrightarrow} E[\left|\xi_i\right|] <\infty$$

Since $\xi_i$ have finite mean and variance.

So, we know that the sequence$\left\{\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)\right\}_n$ converges to a some finite value $T:|T|\leq\sigma^2<\infty$ in probability.


We will need to be more specific, since our first bounding was too coarse and only established that the series does not diverge in probability. Lets look at the variance of: $$G_n:=\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)$$

$$Var(G_n)=\sigma^2_n=\lfloor ns \rfloor\sigma^2\left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)^2$$

From our earlier arguments, we can see that $\sigma^2_n\to 0$.

Also, we can see that $\forall n, E[G_n]=0$. Therefore, by Chebyshev's theorem:

$$P(|G_n-0|\geq k\sigma^2_n)\leq \frac{\sigma^2_n}{k^2} \to 0 \;\;\forall k>0 \implies G_n \stackrel{\mathbb{P}}{\longrightarrow} 0$$

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    $\begingroup$ Actually, $\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} $ is of order $\frac1{n\sqrt{n}}$ hence the convergence to zero was clear from the start. $\endgroup$ – Did Oct 1 '15 at 14:52
  • $\begingroup$ @Did thanks...didn't realize that so i had to go through a two part proof, since being $o(\frac{1}{n})$ was not strong enough. $\endgroup$ – user237392 Oct 1 '15 at 14:56

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