# Proving an Inequality Involving the Moment Generating Function

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:

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

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