# Probability inequalities: Jensen's?

I am trying to prove the following inequality: $$P(X \geq x) \leq e^{-tx} \phi_x(t)$$ Where $\phi_x(t) = E(e^{tx})$ is the moment generating function. $X$ is a non-negative rv and $t>0$.

Since we have the exponential inside and outside the expectation this seems to call for Jensen's inequality: $$\phi_x(t) = E(e^{tx}) \geq e^{t \ Ex}$$

In order to get rid of the expectation I have tried using Markov's inequality so for some $x>0$: $$P(X \geq x) \leq E(x)/x \implies E(x) \geq x \ P(X \geq x)$$ So then, since $t>0$ and $x>0$: $$\phi_x(t) \geq e^{t \ Ex} \geq e^{t \ x \ P(X \geq x)}$$

If I got the probability outside the exponential I would be done, but I don't think this is possible. Is there any inequality that allows this? Maybe even Markov's inequality is not the way to go. Any ideas?

Thanks

Note that you cannot use Markov in that way, since it is not assumed that $X$ is nonnegative.
To answer your original question, note that the event $\{X \ge x\}$ is the same as the event $\{e^{tX} \ge e^{tx}\}$. (Why?) Then apply Markov.
• $X$ is assumed to be non-negative in the question. Also, I don't see what do you imply by $X \geq x$ after Jensen's inequality. I should end up with some relationship of the kind$e^{t \ Ex} > P(X \geq x) e^{tx}$, right? – ebabio Oct 31 '17 at 5:01
• Now, I got the second part. Took me some time. You meant $P(X \geq x) = P(e^{tX} > e^{tx}) \leq e^{t \ Ex} / e^{tx}$ – ebabio Oct 31 '17 at 5:11
• For the record: the last comment is wrong. The idea was just $P(X \geq x) = P(e^{tX} \geq e^{tx}) \leq E(e^{tx}) / e^{tx} = \phi_x(t) / e^{tx}$ – ebabio Oct 31 '17 at 5:22