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.

  • 2
    $\begingroup$ can you elaborate on the meaning of $E_{\phi}(X)$? $\endgroup$ Sep 25, 2019 at 4:15
  • $\begingroup$ 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 $X$. $\endgroup$
    – Ron Snow
    Sep 25, 2019 at 4:52
  • 2
    $\begingroup$ @Edison Is it $E(\phi(X))$ instead? $\endgroup$ Sep 25, 2019 at 5:05

1 Answer 1


This is a general approach using in bounds called the Cramer Chernoff bounds

\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}

  • $\begingroup$ 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! $\endgroup$
    – Ron Snow
    Sep 25, 2019 at 23:40
  • 1
    $\begingroup$ @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, $\Bbb P(|X|>t) = \Bbb P(\{X >t\} U \{ X < -t\})$. $\endgroup$ Sep 26, 2019 at 5:04
  • $\begingroup$ Where did the coefficient of the two go? $\endgroup$
    – Ron Snow
    Oct 3, 2019 at 0:07
  • $\begingroup$ Your answer states that $P(|X| > t) \le 2E(\phi(X))/ \phi(t)$, but your comment implies that $P(|X| > t) \le E(\phi(X))/ \phi(t)$ $\endgroup$
    – Ron Snow
    Oct 3, 2019 at 0:10
  • $\begingroup$ $(X>0 => P(X>t) = P(\phi(X) > \phi(t))) \land (X<0 => P(-X>-t) = P(\phi(-X) > \phi(-t))= P(\phi(-X) > \phi(t))) => P(|X| > t) = P(\phi(|X|) > \phi(t) = P(\phi(X) > \phi(t))$ $\endgroup$
    – quester
    Oct 3, 2019 at 7:12

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