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This is a question from one of my statistics courses at uni, on which I am completely stumped (this is not something I'm being marked on, it's a question from practice problems). So here I go :)

Let $X$ be a continuous random variable, whose moment generating function $m_X(t)$ exists, i.e. $\exists$ $h>0$ such that $m_X(t)<\infty$ for $t\in(-h,h)$.

Let $a\geq0$ and $(X_i)_{i=1,\dots,n}$ is a sequence of i.i.d. random variables whose distribution follows that of $X$. We set $S_n := \sum_{i=1}^nX_i$ and $\bar{X}_n=\frac{1}{n}S_n$.

(i) Show that $\forall x\in \mathbb{R}$ and $t>0$, $I_{\{x>a\}}\leq e^{(x-a)t}$.

(ii) Show that $\mathbb{P}(S_n>a)\leq e^{-at}[m_X(t)]^n$ for $0<t<h$.

(iii) Use the facts that $m_X(0)=1$ and $m_X'(0)=\mathbb{E}(X)$ to show that if $\mathbb{E}(X)<0$ then there exists $0<c<1$ with $\mathbb{P}(S_n>a)\leq c^n$. Hint: use the mean value theorem and the intermediate value theorem.

Any help would be greatly appreciated.

Thanks in advance!


For (i), you need to prove that $e^{(x-a)t}\ge0$ for $x\le a$ (clear) and that $e^{(x-a)t}\ge1$ for $x>a$ (don't forget $t>0$).

For (ii), apply (i) noting that $P(S_n>a)=E([S_n>a])$. You need the MGF of $S_n$.

For (iii), you need a positive $t$ with $m_X(t)<1$.

  • $\begingroup$ Thanks for your reply, but how does proving this for (i), prove the inequality? $\endgroup$ – mrbing Jun 4 '17 at 8:28
  • $\begingroup$ Because $I_{[x>a]}$ is the function which is $1$ when $x>a$ and zero otherwise. $\endgroup$ – Lord Shark the Unknown Jun 4 '17 at 10:01

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