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.

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

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