exercise:C5e148cb87: 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> is a random variable with mean <math>\mu</math> and variance <math>\sigma^2</math>, and if <math>X^* = (X - \mu)/\sigma</math> is the standardized version of <math>X</math>, then <math display="block"> g_{X^*}(t) = e^{...")
 
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<div class="d-none"><math>
Show that if <math>X</math> is a random variable with mean <math>\mu</math> and variance <math>\sigma^2</math>, and if <math>X^* = (X - \mu)/\sigma</math> is the standardized version of <math>X</math>, then
\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> is a random variable with mean <math>\mu</math>
and variance <math>\sigma^2</math>, and if <math>X^* = (X - \mu)/\sigma</math> is the standardized
version of <math>X</math>, then


<math display="block">
<math display="block">
g_{X^*}(t) = e^{-\mu t/\sigma} g_X\left( \frac t\sigma \right)\ .
g_{X^*}(t) = e^{-\mu t/\sigma} g_X\left( \frac t\sigma \right)\ .
</math>
</math>

Latest revision as of 23:46, 14 June 2024

Show that if [math]X[/math] is a random variable with mean [math]\mu[/math] and variance [math]\sigma^2[/math], and if [math]X^* = (X - \mu)/\sigma[/math] is the standardized version of [math]X[/math], then

[[math]] g_{X^*}(t) = e^{-\mu t/\sigma} g_X\left( \frac t\sigma \right)\ . [[/math]]