exercise:B1e57f7b91: Difference between revisions

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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Let <math>\{X_k\}</math>, <math>1 \leq k \leq n</math>, be a sequence of independent random variables, all with mean 0 and variance 1, and let <math>S_n</math>, <math>S_n^*</math>, and <math>A_n</math> be their sum, standardized sum, and average,...")
 
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<div class="d-none"><math>
Let <math>\{X_k\}</math>, <math>1 \leq k \leq n</math>, be a sequence of independent random variables, all with mean 0 and variance 1, and let <math>S_n</math>, <math>S_n^*</math>, and <math>A_n</math> be
\newcommand{\NA}{{\rm NA}}
\newcommand{\mat}[1]{{\bf#1}}
\newcommand{\exref}[1]{\ref{##1}}
\newcommand{\secstoprocess}{\all}
\newcommand{\NA}{{\rm NA}}
\newcommand{\mathds}{\mathbb}</math></div>  Let <math>\{X_k\}</math>, <math>1 \leq k \leq n</math>, be a sequence of independent random
variables, all with mean 0 and variance 1, and let <math>S_n</math>, <math>S_n^*</math>, and <math>A_n</math> be
their sum, standardized sum, and average, respectively.  Verify directly that
their sum, standardized sum, and average, respectively.  Verify directly that
<math>S_n^* = S_n/\sqrt{n} = \sqrt{n} A_n</math>.
<math>S_n^* = S_n/\sqrt{n} = \sqrt{n} A_n</math>.

Latest revision as of 23:17, 14 June 2024

Let [math]\{X_k\}[/math], [math]1 \leq k \leq n[/math], be a sequence of independent random variables, all with mean 0 and variance 1, and let [math]S_n[/math], [math]S_n^*[/math], and [math]A_n[/math] be their sum, standardized sum, and average, respectively. Verify directly that [math]S_n^* = S_n/\sqrt{n} = \sqrt{n} A_n[/math].