an exercise about mean and probability

Question

Let $(\Omega, \mathcal{F}, P)$ be a probability space, $X : \Omega \rightarrow \mathbb{R}$ be a discrete random variable and $$\phi : [0, \infty) \rightarrow (0, \infty)$$ an increasing function (so $\phi$ is a borelian function). Suppose that $E (\phi(| X |)) =: M < \infty$ and let $c > 0$. Show that

$$P ( |X| > c ) \leq \frac{M}{\phi(c)}$$

I have no idea to solve it :(.

Thank you!

Answer 1

What your exercise asks you to prove, is a generalization of the Markov Inequality which can be done as follows $$\begin{align*}M&=E[\phi(|X|)]=\int_{-\infty}^{+\infty}\phi(|x|)f_X(x)dx\\&=\int_{|X|< c}^{}\phi(|x|)f_X(x)dx+\int_{|X|\ge c}^{}\phi(|x|)f_X(x)dx\\&\ge \int_{|X|\ge c}^{}\phi(|x|)f_X(x)dx\\&\overset{*}\ge\phi(c)\int_{|X|\ge c}^{}f_X(x)dx=\phi(c)P(|X|\ge c)\end{align*}$$ where $(*)$ is due to the monotonicity of $\phi$.

Answer 2

$$\mathbb E\phi(|X|) \geq \mathbb E\phi(|X|)1_{\{|X| >c\}} \geq \phi(c)\mathbb E1_{\{|X|>c\}}$$

Answer 3

Hint: Use the Markov Inequality here