Question

[Chebyshev inequality] For any random variable X and any \( a > 0 \), we have 

\( P(|X-E(X)|\geq a)\leq\frac{V(X)}{a^2} \)

Answer 1

Solution

For any random variable X and any \( a > 0 \)

[Chebyshev inequality]  :  \( P(|X-E(X)|\geq a)\leq\frac{V(X)}{a^2} \)

\( \implies P(|X-E(X)|\geq a)=P(|X-E(X)|^2\geq a^2) \)

                                              \( \leq \frac{E\left(\left|X-E\left(X\right)\right|^2\right)}{a^2}=\frac{V\left(X\right)}{a^2} \)

Therefore. \( P(|X-E(X)|\geq a)\leq\frac{V(X)}{a^2} \)

Therefore.\( P(|X-E(X)|\geq a)\leq\frac{V(X)}{a^2} \)