exercise:2b92b07290: 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> Show that, if <math>X</math> and <math>Y</math> are random variables taking on only two values each, and if <math>E(XY) = E(X)E(Y)</math>, then <math>X</math> and <math>Y</math> are independent.")
 
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<div class="d-none"><math>
Show that, if <math>X</math> and <math>Y</math> are random variables taking on only two values each, and if <math>E(XY) = E(X)E(Y)</math>, then <math>X</math> and <math>Y</math> are independent.
\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> Show that, if <math>X</math> and <math>Y</math> are random variables taking on
only two values each, and if <math>E(XY) = E(X)E(Y)</math>, then <math>X</math> and <math>Y</math> are independent.

Latest revision as of 10:39, 14 June 2024

Show that, if [math]X[/math] and [math]Y[/math] are random variables taking on only two values each, and if [math]E(XY) = E(X)E(Y)[/math], then [math]X[/math] and [math]Y[/math] are independent.