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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.

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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.

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

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    $\begingroup$ +1 the key point is the conversion to $P (\phi (X) > \phi (t))$ but the formatting is difficult to read. $\endgroup$ Oct 3 '19 at 7:42
  • $\begingroup$ 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. $\endgroup$ Oct 3 '19 at 8:24
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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

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

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