Proving the bound $P(|X| \geq x) \leq (M_X(c) + M_X(-c)) e^{-cx}$ under certain conditions

Question

If the moment generating function of a random variable X, $M_X (\lambda) = E[e^{\lambda X}]$, is defined for $|\lambda| < \delta$ with some $\delta > 0$, then $P(|X| \geq x) \leq (M_X(c) + M_X(-c)) e^{-cx}$ for any $x > 0$ and $\delta > c > 0$.

I can calculate that $(M_X(c) + M_X(-c)) e^{-cx}$ $= e^{-cx} (E[e^{cX}] + E[e^{-cX}])$ $= e^{-cx} ( \int_{-\infty}^\infty e^{ct} dF(t) + \int_{-\infty}^\infty e^{-ct} dF(t))$ $= \int_{-\infty}^\infty (e^{c(t-x)} + e^{-c(t + x)}) dF(t)$, but I don't know how to say anything at all about the relationship between the moment generating function and the probability $P(|X| \geq x)$. My main problem is that, for a general random variable and function $f(t)$ that is more complicated than, say, an indicator,, I really have no idea how to relate anything of the form $\int_{-\infty}^\infty f(t) dF(t)$ to an actual probability, or really any actual number at all. I think that $P(|X| \geq x)$ $= E[\chi_{|X| \geq x}]$ $= \int_{-\infty}^\infty \chi_{|X| \geq x}(t) dF(t)$, but this seems entirely unrelated to what I wrote above, aside from both of them being an expectation/integral over $\mathbb{R}$ with regard to the cumulative function.

Answer 1

• $\mathrm e^{cx}P(X\geqslant x)\leqslant E(\mathrm e^{cX};X\geqslant x)\leqslant E(\mathrm e^{cX})=M_X(c)$

• $\mathrm e^{cx}P(X\leqslant-x)=\mathrm e^{cx}P(-X\geqslant x)\leqslant E(\mathrm e^{-cX};-X\geqslant x)\leqslant E(\mathrm e^{-cX})=M_X(-c)$

Answer 2

Hint:

$$M_X(c) \geq E e^{cX}1_{X\geq x} \geq e^{cx}P(X\geq x)$$

$$M_X(-c) \geq E e^{-cX}1_{X\leq -x} \geq e^{cx} P(X \leq -x)$$