Question

[Markov inequality] If \( X \geq 0, i.e. X \) takes only nonnegative values, then for an \( a>0 \)  we have \\( \hspace{3mm} P(X\geq a)\leq \frac{E(X)}{a} \)

 

Answer 1

Solution

takes only nonnegative values, then for an \( a>0 \)

[Markov inequality] :\( \hspace{3mm} P(X\geq a)\leq \frac{E(X)}{a}\hspace{3mm}X\geq 0,a>0 \)

\( \implies E(X)=\sum _{x=0}^n\:xP\left(x\right)=\sum _{x=0}^a\:xp\left(x\right)+\sum \:_{x=0}^n\:xp\left(x\right) \)

                     \( \geq \sum _{x=a}^n\:xp\left(x\right)\ge a\sum _{x=a}^n\:p\left(x\right)=ap\left(x\ge a\right) \)

Therefore. \( \frac{E(X)}{a}\geq p(x\geq a) \)

Therefore. \( \frac{E(X)}{a}\geq p(x\geq a) \)