Question

Suppose that X is a random variable such that\(E\left( {{X^k}} \right)\)exists and\({\rm P}\left( {X \ge 0} \right) = 1\). Prove that for\(k > 0\)and\(t > 0\),\({\rm P}\left( {X \ge t} \right) \le \frac{{E\left( {{X^k}} \right)}}{{{t^k}}}\)

Short answer

For k>0 and t>0 \({\rm P}\left( {X \ge t} \right) = \frac{{E\left( {{X^k}} \right)}}{{{t^k}}}\)

Step-by-step solution

Given information

Let X is the random variable such that\(E\left( {{X^k}} \right)\)exist

And \({\rm P}\left( {X \ge 0} \right) = 1\)

Verifying k >0 and t >0 \({\rm P}\left( {X \ge t} \right) \le \frac{{E\left( {{X^k}} \right)}}{{{t^k}}}\)

A random variable X has the pdf\(f\left( x \right)\), then

\(\begin{array}{c}E\left( {{X^k}} \right) = \int\limits_0^\infty {{x^k}f\left( x \right)dx} \\ \ge \int\limits_t^\infty {{x^k}f\left( x \right)dx} \end{array}\)

\(E\left( {{X^k}} \right) \ge {t^k}\int\limits_t^\infty {f\left( x \right)dx} \)

\(E\left( {{X^k}} \right) = {t^k}{\rm P}\left( {X \ge t} \right)\)

\(\frac{{E\left( {{X^k}} \right)}}{{{t^k}}} = {\rm P}\left( {X \ge t} \right)\)for k>0 and t>0

Therefore for k > 0 and t>0 \({\rm P}\left( {X \ge t} \right) = \frac{{E\left( {{X^k}} \right)}}{{{t^k}}}\) is proved