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Let $X_1,X_2,..,X_{100}$ be independent identical distributed random variables with $E(X_i)=0$ and $Var(X_i)=1$ ($E$ means expected value and $Var$ means variance).

Prove that $$P(|X_i| \geq 2) \leq \frac{1}{4}$$

I don't know what need to do... I try understand what notation is say but not sure:

$|X_i|$ mean how many $X_1,X_2,..X_{100}$ we concentrate at. In example they say $|X_i| \geq 2$. This mean we have two or more of these $X_i$ where $i \in [1, 100]$.

$P(|X_i| \geq 2)$ mean probability to have at least two of these $X_i$ is at least $\frac{1}{4}$.

But then I still don't understanded what is task wanted from me? Pls you can explain task when my explanation is wrong and say what is wanted to do? But more important for me is understand what task say and how solved.

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    $\begingroup$ The question doesn't seem to need more than $X_i$. $\endgroup$ – Thomas Andrews Oct 10 '17 at 23:19
  • $\begingroup$ @ThomasAndrews ty i change $\endgroup$ – roblind Oct 10 '17 at 23:20
  • $\begingroup$ The way you have written it, $X_i$ is just one of the variables...the others have no bearing on the question. If that is what you intended, then this is just Chebyshev's Inequality but maybe you meant something else? $\endgroup$ – lulu Oct 10 '17 at 23:21
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You can simply apply Chebyshev's inequality, which states that, for a random variable with mean $\mu$ and variance $\sigma^2$, we have for any real $k>0$ that$$ P(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2}. $$ In your case you have $\mu=0$ and $\sigma^2=1$. Hence, with $k=2$,$$ P(|X_i|\geq 2)\leq \frac{1}{2^2}=\frac{1}{4}. $$

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We have: $$1=\mathrm{Var}(X_i)=E(X^2)-(E(X))^2 = E(X^2)$$

You need to know:

Lemma: If $Y$ is a non-negative random variable, then $$E(Y)\geq kP(Y\geq k).$$

So if $Y=X_i^2$ and $k=4$ then:

$$1=E(X_i^2)\geq 4P(X_i^2\geq 4)=4 P(|X_i|\geq 2)$$

Proof of lemma:

IF $p_Y$ is the PDF for $Y$, then:

$$\begin{align}E(Y)&=\int_{0}^{\infty} yp_Y(y)\,dy \\ &\geq \int_{k}^{\infty} yp_Y(y)\,dy\\ &\geq \int_{k}^{\infty} kp_Y(y)\,dy\\ &=kP(Y\geq k) \end{align}$$

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