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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

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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.

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  • $\begingroup$ $X$ 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? $\endgroup$ – ebabio Oct 31 '17 at 5:01
  • $\begingroup$ Now, I got the second part. Took me some time. You meant $P(X \geq x) = P(e^{tX} > e^{tx}) \leq e^{t \ Ex} / e^{tx}$ $\endgroup$ – ebabio Oct 31 '17 at 5:11
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    $\begingroup$ @ebabio Your right-hand side is still wrong, but you are almost there. $\endgroup$ – angryavian Oct 31 '17 at 5:21
  • $\begingroup$ For the record: the last comment is wrong. The idea was just $P(X \geq x) = P(e^{tX} \geq e^{tx}) \leq E(e^{tx}) / e^{tx} = \phi_x(t) / e^{tx}$ $\endgroup$ – ebabio Oct 31 '17 at 5:22

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