0
$\begingroup$

I'm new on Mathematics Stack Exchange, and it's awesome to be a new member of this awesome community. Please correct me if I have made any errors in terms of following the rules for asking questions here. Also, please feel free to drop down below and say hi.

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!

$\endgroup$
1
$\begingroup$

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 $m_X(t)<1$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.