exercise:7ddf221305: 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_1</math> and <math>X_2</math> be independent random variables and let <math>Y_1 = \phi_1(X_1)</math> and <math>Y_2 = \phi_2(X_2)</math>. <ul><li> Show that <math display="block"> P(Y_1 = r, Y_2 = s) = \sum_{\phi_1(a) = r \atop \phi_2(...") |
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Let <math>X_1</math> and <math>X_2</math> be independent random variables and let <math>Y_1 = | |||
\phi_1(X_1)</math> and <math>Y_2 = \phi_2(X_2)</math>. | \phi_1(X_1)</math> and <math>Y_2 = \phi_2(X_2)</math>. | ||
<ul><li> Show that | <ul style="list-style-type:lower-alpha"><li> Show that | ||
<math display="block"> | <math display="block"> | ||
Latest revision as of 01:13, 13 June 2024
Let [math]X_1[/math] and [math]X_2[/math] be independent random variables and let [math]Y_1 = \phi_1(X_1)[/math] and [math]Y_2 = \phi_2(X_2)[/math].
- Show that
[[math]] P(Y_1 = r, Y_2 = s) = \sum_{\phi_1(a) = r \atop \phi_2(b) = s} P(X_1 = a, X_2 = b)\ . [[/math]]
- Using (a), show that [math]P(Y_1 = r, Y_2 = s) = P(Y_1 = r)P(Y_2 = s)[/math] so that [math]Y_1[/math] and [math]Y_2[/math] are independent.