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Prove that $P[X>\epsilon] \leq \dfrac{M(t)}{e^{\epsilon t}}$

Looks like Markov's inequality, it's very easy to derive for $t>0$

$P[X>\epsilon] =P[Xt>\epsilon t]=[e^{Xt}>e^{\epsilon t}]\leq \dfrac{M(t)}{e^{\epsilon t}}$

But the problem statement implies that this is true for all t. Is this true? If it is, can you please suggest a way to proceed for $t\leq0$?

Thank you for your time.

Edit: note that $M(t)=E[e^{Xt}]$

Edit2: made a mistake replaced $X$ by $\epsilon$ in three places

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  • $\begingroup$ Do you mean to say $\Pr(X>\epsilon) \leqslant M(t) \mathrm{e}^{-\epsilon t}$ perhaps? $\endgroup$ – Sasha Mar 1 '14 at 18:10
  • $\begingroup$ Yes thats what I meant to say I'll edit it now, thank you $\endgroup$ – wwbb90 Mar 1 '14 at 18:25
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In general, this inequality is not true for $t <0$. Consider the constant random variable $X=2$, and set $\varepsilon=1$. Then

$$\mathbb{P}[X>\varepsilon]=1$$

and

$$\frac{M(t)}{e^{\varepsilon t}} = \frac{e^{2t}}{e^t} = e^t.$$

Hence,

$$\mathbb{P}[X>\varepsilon] > \frac{M(t)}{e^{\varepsilon t}}$$

for $t<0$.

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