# How should I show that for each $t>0$, $P(|X| \ge t) \le E_\phi(X) / \phi(t)$?

Suppose $$X$$ is a random variable and $$\phi:(-\infty,\infty) \to(0,\infty)$$ satisfies $$\phi(-t)=\phi(t)$$. Assume that $$\phi(\cdot)$$ is an increasing function on $$(0,\infty)$$. Show that for each $$t>0$$, $$P(|X| \ge t) \le E_\phi(X) / \phi(t)$$.

My work:

I first identified that $$\phi(\cdot)$$ is an even function. Since $$\phi(\cdot)$$ is increasing on $$(0,\infty)$$, then $$E_\phi(X) / \phi(t) \to 0$$ as $$t \to \infty$$, since the denominator is growing quite large and the numerator is a constant.

Working on the LHS of the inequality:

$$P(|X| \ge t)=P(-X \le-t)$$ and $$P(X \ge t)$$. However, I do not know where to go from here.

• can you elaborate on the meaning of $E_{\phi}(X)$? Sep 25, 2019 at 4:15
• That notation seemed new to me, too. I assume that it just means the expected value of the function $\phi(t)$ with respect to the random variable $X$. Sep 25, 2019 at 4:52
• @Edison Is it $E(\phi(X))$ instead? Sep 25, 2019 at 5:05

\begin{align} \Bbb P(|X|>t) &= 2\Bbb P(\phi(X)>\phi(t)) \; \; \;\text{(Because \phi is increasing and even)}\\ &\le 2 \frac{\Bbb E(\phi(X))}{\phi(t)} \, \; \; \text{Using Markov's inequality} \\ \end{align}
• Where do you use the fact that $\phi(\cdot)$ is an even function? Also, is the Markov inequality needed to solve this? Is there another way? I am just curious! Sep 25, 2019 at 23:40
• @Edison: Because $\phi$ is even, $\Bbb P(|X|>t) = 2 \Bbb P(X >t)= \Bbb P(\phi(X)>\phi(t))$. Else, we would have to consider the general case, $\Bbb P(|X|>t) = \Bbb P(\{X >t\} U \{ X < -t\})$. Sep 26, 2019 at 5:04
• Your answer states that $P(|X| > t) \le 2E(\phi(X))/ \phi(t)$, but your comment implies that $P(|X| > t) \le E(\phi(X))/ \phi(t)$ Oct 3, 2019 at 0:10
• $(X>0 => P(X>t) = P(\phi(X) > \phi(t))) \land (X<0 => P(-X>-t) = P(\phi(-X) > \phi(-t))= P(\phi(-X) > \phi(t))) => P(|X| > t) = P(\phi(|X|) > \phi(t) = P(\phi(X) > \phi(t))$ Oct 3, 2019 at 7:12