Random Variables and Moment Generating functions Let $(X_i)_{i∈\Bbb{N^+}}$ be a sequence of i.i.d random variables and for $n ∈ \Bbb{N^+}$ set $S_n := \sum _{i=1}^{n} X_i$ and $Y_n := max(X_1, . . . , X_n)$. Assume that the moment generating function exists and has continuous derivatives. (the constant h is the radius of the interval containing zero for which the moment generating function exists).
1. Show that $∀x∈\Bbb{R}$ and $t>0, \,I_{\{x\le c\}}\le e^{t(c−x)} $ and hence $P(X_i \le c)\le e^{tc} m_{X_i}(−t)$ for $0 < t < h$.
I understand why the indicator function statement is true but don't know how to show the second bit. I would guess that you take probabilities of both sides and hence if $x\le c$ (from the indicator function statement) $P(X_i \le c)\le P(e^{t(c−x)}), $ pulling out terms independent of x, we have $ = e^{tc}P(e^{-tx})=e^{tc}m_{X_i}(-t). $ I feel what I have done lacks mathematical accuracy.
2. Show that $∀x∈\Bbb{R}$ and $t>0,\,I_{\{x> c\}} \le e^{t(x-c)} $ and hence $P(S_n >a)\le e^{at} [m_{X}(t)]^n$ for $0 < t < h$.
Once again, I understand why the indicator function statement is true but don't know how to show the second bit. If I follow the logic from part 1, replacing the $c$ with an $a$ and then taking probabilities of both sides we have $P(S_n>a)\le P(e^{t(x-a)})=e^{-at}P(e^{tx})=e^{-at}[m_{X}(t)]^n.$ because it's a sum from $1$ to $n$. 
3. Use the facts that $m_X(0) = 1$ and $m′_X(0) = E(X)$ to show that if $E(X) < 0$, then there exist $0<c<1$ with $P(S_n >a)\le c^n$, using the mean value theorem.
I have no idea for this one.
4.If one scales the random variable $Y_n$ using appropriate sequences of real numbers, $(a_n)_{n∈\Bbb{N^+}}$ and $(b_n)_{n∈\Bbb{N^+}}$, then $\frac{Y_n−a_n}{b_n}$ converges to a non-trivial random variable with a non-trivial distribution.
Let $(X_i)_{i∈\Bbb{N^+}}$ be a sequence of i.i.d. $exp(λ)$ random variables, show that for $n∈\Bbb{N^+}$, if we set $a_n =\lambda log(n)$ and $b_n =λ$ then
$$\lim_{n\to \infty} P(b_n^{−1}(Y_n−a_n)\le x)=e^{−e^{−x}}$$
I don't have much of a clue for this one as well.
 A: *

*Take Expectation on both sides. Then $E(I_{(X\leq c)})\leq E(e^{t(c-X)})=e^{tc}E(e^{-tX})=e^{tc}m_X(-t)$. But also note that $E(I_{(X\leq c)})=P(X\leq c)$, which immediately gives you the result.

*Taking expectation on both sides again, $P(S_n\geq a)\leq e^{-at}E(e^{tS_n})=e^{-at}E(e^{tX_1}e^{tX_2}...e^{tX_n})=e^{-at}(Ee^{tX_1})(Ee^{tX_2})...(Ee^{tX_n})=e^{-at}(m_X(t))^n$ due to i.i.d. property. 

*I think the estimate is extremely crude. $P(S_n\geq a)=s_n$ (say) then $0\leq s_n\leq 1$. Isn't it trivial that we can always find $c$ such that $s_n\leq cn$ i.e. $c>s_n/n$. Why not take the following path: Find $s_1=P(S_1\geq a)=P(X\geq a)$ and then put $c=\max\{0.99,\dfrac{s_1+0.99}{2}\}$ then you will see that this $c$ works. No need to use any Mean Value Theorem.

*You want $P((Y_n-a_n)/b_n\leq x)$. It is equivalent to find $P(Y_n\leq a_n + xb_n)$. Given that $X_n$ are i.i.d $Exp(\lambda)$, what is the cumulative distribution function $F(x)$? Can you realize that $P(Y_n\leq a_n + xb_n)=(F(a_n+xb_n))^n$? Now you can just take the limit as $n\to\infty$ which is a routine calculus task.
EDIT: As for 3. Did has mentioned a nice alternative, which can be found in the comments succeeding the question. We obtain that $P(S_n\geq na)\leq (e^{-at}m_X(t))^n$. The LHS is independent of $t$ while the RHS is dependent on $t$, hence we have $P(S_n\geq a)\leq \inf_{t}(e^{-at}m_X(t))^n=(\inf_{t}e^{-at}m_X(t))^n$. 
We want to show that $\inf_{t}e^{-at}m_X(t)=c\in(0,1)$. 
Define $f(t)=e^{-at}m_X(t)$ then $f(0)=1$ and $f'(0)<0$ as can be easily verified. Also $f'(t)$ is continuous for all $t\in(-h,h)$. So in particular, this means that there exists a neighbourhood around $0$, call it $(-\delta,\delta)$ for some $\delta>0$ so that for all $t\in(0,\delta)$, $f'(t)<0\implies f(t)<f(0)=1$ which means that we have obtained some $t_0$ such that $f(t_0)<1$, as desired. Hence $\inf_tf(t)\leq f(t_0)<1$ and the result is thus proved.
