exercise:1b8dc9cfa8: 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 random variables, all with mean <math>\mu</math> and variance <math>\sigma^2</math>, and <math>Y_k = X_k^*</math> be their standardized versions. Let <math>S_n</math> and <...") |
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Let <math>\{X_k\}</math>, <math>1 \leq k \leq n</math>, be a sequence of random variables, all with | |||
mean <math>\mu</math> and variance <math>\sigma^2</math>, and <math>Y_k = X_k^*</math> be their standardized | mean <math>\mu</math> and variance <math>\sigma^2</math>, and <math>Y_k = X_k^*</math> be their standardized | ||
versions. Let <math>S_n</math> and <math>T_n</math> be the sum of the <math>X_k</math> and <math>Y_k</math>, and <math>S_n^*</math> | versions. Let <math>S_n</math> and <math>T_n</math> be the sum of the <math>X_k</math> and <math>Y_k</math>, and <math>S_n^*</math> | ||
and <math>T_n^*</math> their standardized version. Show that <math>S_n^* = T_n^* = | and <math>T_n^*</math> their standardized version. Show that <math>S_n^* = T_n^* = | ||
T_n/\sqrt{n}</math>. | T_n/\sqrt{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 random variables, all with mean [math]\mu[/math] and variance [math]\sigma^2[/math], and [math]Y_k = X_k^*[/math] be their standardized versions. Let [math]S_n[/math] and [math]T_n[/math] be the sum of the [math]X_k[/math] and [math]Y_k[/math], and [math]S_n^*[/math] and [math]T_n^*[/math] their standardized version. Show that [math]S_n^* = T_n^* = T_n/\sqrt{n}[/math].