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I am looking for some help with the chebyshev problem that I am not quite sure how to do and am stuck on. Could someone provide some advice for the following question?

  1. Let $Z_{1}$, $Z_{2}$, $Z_{3}$, . . . be independent copies of a random variable Z of expected value $\mu$ and variance $\sigma^{2}$ > 0. Let $X_{n} = \frac{Z_{1}+···+Z_{n}}{n}$. Using Chebyshev’s inequality, show that for any $a \geq 0$ with $a < \frac{1}{2}$, the random variable $Y_{n}$ defined as $n^{a}·(X_{n}−\mu)$ converges to $0$ (hint: show that for $x < 0$, $P(Y_{n} \leq x)$ converges to $0$, and for $x > 0$, $P(Y_{n} \leq x)$ converges to 1).

I dont quite understand what the hint exactly means and putting everything together. Basically, I have the fact that the $\text{Var}[Y_{n}] = \sigma^{2}*n^{2\alpha}$ and then I'm not quite sure what to do next if I plug it in to chebyshev's inequality, how exactly would I prove this because I'm not sure what happens to the $$P(|X_{n}-p|\geq \epsilon)\leq\frac{V[Y_{n}]}{\epsilon^{2}}$$. Also, I'm a bit lost on why I would substitute $x^2$ in this case. Could someone please explain?

I would greatly appreciate some explanations given.

Thank you

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closed as off-topic by Clarinetist, Silvia Ghinassi, choco_addicted, Harish Chandra Rajpoot, user91500 Mar 16 '16 at 4:33

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For $x>0$,

$$P(|n^a(X_n-\mu)|\geq x)\leq \frac{n^{2a}\sigma^2/n}{x^2}=\frac{n^{2a-1}\sigma}{x^2}\rightarrow 0,$$

since $2a-1<0$. So $|n^a(X_n-\mu)|$ converges to 0 in probability, and also almost surely since 0 is constant.

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