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I am trying to prove the following inequality for an arbitrary random variable $X$ and a constant $c$:

$$P(X\ge c)\le \min_{s\ge 0}e^{-sc}M_X(s),$$

where $M_X(s)$ is the moment generating function of $X$.

Here is my attempt so far:

In terms of the unit step function, $u(x)$, we can express the probability as an integral:

$$P(X\ge c)=\int_c^{\infty}f_X(x)dx=\int_{-\infty}^{\infty}u(x-c)f_X(x)dx.$$

However, I am stuck at this point and would appreciate any help to proceed further.

Update:

The Markov inequality states that for any non-negative random variable $Y$ and any $a > 0$, we have:

$$P(Y \ge a) \le \frac{E[Y]}{a}.$$

Now, let's consider $Y = e^{sX}$ for some $s > 0$. Then, we have:

$$P(e^{sX} \ge e^{sc}) = P(Y \ge e^{sc}) \le \frac{E[Y]}{e^{sc}} = \frac{E[e^{sX}]}{e^{sc}} = e^{-sc}E[e^{sX}].$$

But $E[e^{sX}]$ is just the moment generating function $M_X(s)$ of $X$. So, we have:

$$P(X \ge c) = P(e^{sX} \ge e^{sc}) \le e^{-sc}M_X(s).$$

This holds for any $s > 0$, so we can take the minimum over all such $s$ to get the inequality:

$$P(X \ge c) \le \min_{s \ge 0} e^{-sc}M_X(s).$$

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    $\begingroup$ There is a trick: $P(X \ge c) = P(s X \ge sc) = P(\mathrm{e}^{sX} \ge \mathrm{e}^{sc})$ for any $s > 0$. $\endgroup$
    – River Li
    Commented Apr 24 at 12:52
  • $\begingroup$ Thanks for your hints, I updated my post, can you help to have a look whether it is a legal proof? $\endgroup$
    – prob1 yuma
    Commented Apr 24 at 14:44
  • $\begingroup$ I think It is fine. $\endgroup$
    – River Li
    Commented Apr 24 at 15:26
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    $\begingroup$ Nice, thank you again for your guidance! $\endgroup$
    – prob1 yuma
    Commented Apr 24 at 15:45

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