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.