exercise:13b48eb117: 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</math> be a continuous random variable with mean <math>\mu(X)</math> and variance <math>\sigma^2(X)</math>, and let <math>X^* = (X - \mu)/\sigma</math> be its standardized version. Verify directly that <math>\mu(X^*) = 0</math> and <...")
 
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<div class="d-none"><math>
Let <math>X</math> be a continuous random variable with mean <math>\mu(X)</math> and variance
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\newcommand{\mathds}{\mathbb}</math></div>  Let <math>X</math> be a continuous random variable with mean <math>\mu(X)</math> and variance
<math>\sigma^2(X)</math>, and let <math>X^* = (X - \mu)/\sigma</math> be its standardized version.  
<math>\sigma^2(X)</math>, and let <math>X^* = (X - \mu)/\sigma</math> be its standardized version.  
Verify directly that <math>\mu(X^*) = 0</math> and <math>\sigma^2(X^*) = 1</math>.
Verify directly that <math>\mu(X^*) = 0</math> and <math>\sigma^2(X^*) = 1</math>.

Latest revision as of 00:17, 15 June 2024

Let [math]X[/math] be a continuous random variable with mean [math]\mu(X)[/math] and variance [math]\sigma^2(X)[/math], and let [math]X^* = (X - \mu)/\sigma[/math] be its standardized version. Verify directly that [math]\mu(X^*) = 0[/math] and [math]\sigma^2(X^*) = 1[/math].