[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}{\#} \newcommand{\mathds}{\mathbb} \renewcommand{\P}{\operatorname{P}} \renewcommand{\L}{\operatorname{L}} [/math]

Let [math]X_i\sim\mathcal{N}(0,1)[/math] and [math]X=X_1+\dots+X_d[/math].

• Use the formula [math]\E(f(X_i))=(2\pi)^{-1/2}\int_{\mathbb{R}}f(x)\exp(-x^2/2)\dd x[/math] to show that [math]\E(\exp(tX_i))=(1-2t)^{-d/2}[/math] holds for [math]t\in(0,1/2)[/math].
• Derive the estimate [math]\P\bigl[X\geqslant a\bigr] \leqslant\inf_{t\in(0,1/2)}\frac{\exp(-ta)}{(1-2t)^{d/2}}[/math] for [math]a \gt 0[/math].