# How is $P(|X| > t) \le E(\phi(X))/ \phi(t)$?

I asked the following question on an early post:

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 $$t>0$$, $$P(|X| \ge t) \le E_\phi(X) / \phi(t)$$.

However, after looking through the answer one more time, I realize that the answer that I accepted is a bit confusing for me.

The answer states that $$P(|X| > t) \le 2E(\phi(X))/ \phi(t)$$, but a comment implies that $$P(|X| > t) \le E(\phi(X))/ \phi(t)$$, which is what we are actually trying to show. Can anyone please help me make these two connections?

Thank you.

## 3 Answers

The error on the previous question is in the fact that $$P(|X|>t) = 2 P(X>t)$$ is untrue.

It should be (if $$t>0$$) $$P (|X|>t) = P (X>t) + P (-X>t)$$

and this translates to

$$P (|X|>t) = P (\phi (X)>\phi (t) \land X>0) + P (\phi (X)>\phi (t) \land X<0) = P (\phi (X)>\phi (t))$$

note that $$P (X>t) \neq P (\phi (X)>\phi (t))$$ but instead $$P (X>t) = P (\phi (X)>\phi (t )\land X>0)$$

I guess that this is the origin of the confusion behind the factor 2.

$$[X>0 => P(X>t) = P(\phi(X) > \phi(t))] \\ \land [X<0 => P(X-t) = P(\phi(-X) > \phi(-t)) = P(\phi(-X) > \phi(t)) = P(\phi(X) > \phi(t))] \\ => P(|X| > t) = P(\phi(|X|) > \phi(t)) = P(\phi(X) > \phi(t)) \le \frac{E\phi(X)}{\phi(t)}$$

• +1 the key point is the conversion to $P (\phi (X) > \phi (t))$ but the formatting is difficult to read. Oct 3 '19 at 7:42
• I started changing your question bit then found out some error. So I reverted it (I did not want to change your question beyond editting). See my answer. Oct 3 '19 at 8:24

Let's go over this one step at a time

$$E[\phi(X)] = \int_{-\infty}^\infty \phi(x) f(x)dx\\ = \int_{-\infty}^{-t} \phi(x) f(x)dx + \int_{-t}^{t} \phi(x) f(x)dx + \int_{t}^{\infty} \phi(x) f(x)dx \\ \underbrace{\geq}_{\phi\text{ non-negative}}\int_{-\infty}^{-t} \phi(x) f(x)dx + \int_{t}^{\infty} \phi(x) f(x)dx \\ \underbrace{=}_{\text{multiply and divide by \phi(t)}} \phi(t)\int_{-\infty}^{-t} \frac{\phi(x)}{\phi(t)} f(x)dx + \phi(t)\int_{t}^{\infty} \frac{\phi(x)}{\phi(t)} f(x)dx\\ \underbrace{\geq}_{\phi\text{ increasing on }(0,\infty)\\ \text{and } \phi(-a) = \phi(a)} \phi(t)\int_{-\infty}^{-t} \frac{\phi(t)}{\phi(t)} f(x)dx + \phi(t)\int_{t}^{\infty} \frac{\phi(t)}{\phi(t)} f(x)dx$$

$$= \phi(t)\left(\int_{-\infty}^{-t} f(x)dx + \int_t^\infty f(x)dx\right)\\ =\phi(t)\left(P(X\leq -t) + P(X\geq t)\right)\\ = \phi(t) P(|X|\geq t)$$

Dividing both sides by $$\phi(t)>0$$, we obtain

$$P(|X| \geq t) \leq \frac{E[\phi(X)]}{\phi(t)}$$.