[math] \newcommand{\smallfrac}[2]{\frac{#1}{#2}} \newcommand{\medfrac}[2]{\frac{#1}{#2}} \newcommand{\textfrac}[2]{\frac{#1}{#2}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\e}{\operatorname{e}} \newcommand{\B}{\operatorname{B}} \newcommand{\Bbar}{\overline{\operatorname{B}}} \newcommand{\pr}{\operatorname{pr}} \newcommand{\dd}{\operatorname{d}\hspace{-1pt}} \newcommand{\E}{\operatorname{E}} \newcommand{\V}{\operatorname{V}} \newcommand{\Cov}{\operatorname{Cov}} \newcommand{\Bigsum}[2]{\mathop{\textstyle\sum}_{#1}^{#2}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\card}{\#} \renewcommand{\P}{\operatorname{P}} \renewcommand{\L}{\operatorname{L}} \newcommand{\mathds}{\mathbb}[/math]

(‘naive exponential estimate’) Let [math]Y_1,\dots,Y_d[/math] be independent random variables and assume that [math]|\E(Y_i^k)|\leqslant k![/math] holds for all [math]k\geqslant0[/math] and [math]i=1,\dots,d[/math].

• Use the series expansion of [math]\exp(\cdot)[/math] and the assumption to get
  [[math]] \E(\exp(tY_i))\leqslant \Bigsum{k=0}{\infty}t^k=\left\{\begin{array}{cl}\textfrac{1}{1-t} &\text{ for } t\in(-1,1),\\\infty & \text{otherwise.}\end{array}\right. [[/math]]
• Show by means of calculus that
  [[math]] \inf_{t\in(0,1)}\exp(-ta)\prod_{i=1}^d\smallfrac{1}{1-t}=\left\{\begin{array}{cl}(\textfrac{a}{d})^d\exp(d-a) &\text{ if } a \gt d,\\ 1 & \text{otherwise.}\end{array}\right. [[/math]]
• Derive an estimate for [math]\P\bigl[|Y_1+\cdots+Y_d|\geqslant a\bigr][/math] from the above.
• Compare the bound in (iii) with the bound of Theorem.