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I have a question about the Chebyshev inequality in probability. Specifically, I am concerned with the term inside the probability function.

I agree with the following: Let $Y = (X-EX)^2$. Then Y is a non-negative quantity. Therefore, Markov's inequality can be used here. I agree with the following: $$P(Y > a^2) \leq \frac{EY}{a^2}$$

Then, replacing Y inside the probability statement, $$P((X-EX)^2 > a^2) \leq \frac{EY}{a^2}$$

Note that the following four statements are true: $$ |a| > b \iff a > b \vee a < -b$$ $$ a^2 > b^2 \iff a > b \vee a < -b$$ $$ a^2 < b^2 \iff -|b| < a < |b|$$ $$ |a| < b \iff -b < a < b$$

Then, I make this statement, using the facts above (call this equation (1)): $$P((X-EX)^2 > a^2) \leq \frac{EY}{a^2} \iff P(-|X-EX| < a < |X-EX|) \leq \frac{EY}{a^2} $$

Note that the Chebyshev's Inequality usually leads to this (call this equation (2)): $$ P(|X-EX| > a) \leq \frac{EY}{a^2}$$

I am confused why my equation (1) differs from equation (2), which is found in many books.

Any help would be appreciated, thanks.

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up vote 2 down vote accepted

The inequality you gave in equation (2) is only valid for $a>0$. If you assume that $a>0$, your statement reduces to the one of your equation (2) because $-|X-EX|<a$ is always true.

Why we do this is simple, only one side of the double inequality can be relevant at a time, so there is no need to keep both. if $a<0$, then in the same way that it reduces for $a>0$, the statement $a<|X-EX|$ becomes always true.

Edit : Also, $a$ represents a distance from the expected value, so it should be positive. (Taking the negative of the distance and comparing it to a negative value is not useful)

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Ok, so if I have an inequality $A < B < C$, and I know $A < B$ is always true, I can say $A < B < C \iff B < C$ and I can replace $A<B<C$ by $B<C$. –  jrand Jul 27 '11 at 3:45
    
That is correct. $A<B<C$ is just shorthand for "$A<B$ and $B<C$". –  Vhailor Jul 27 '11 at 4:06
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