exercise:0876775b21: 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> and <math>Y</math> be random variables. The ''covariance'' <math>\rm {Cov}(X,Y)</math> is defined by (see Exercise \ref{sec 6.2}.) <math display="block"> \rm {cov}(X,Y) = E ((X - \mu(X))(Y -...") |
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Let <math>X</math> and <math>Y</math> be random variables. The ''covariance'' <math>\rm {Cov}(X,Y)</math> is defined by (see [[exercise:Afee895e05|Exercise]]) | |||
<math>\rm {Cov}(X,Y)</math> is defined by (see | |||
<math display="block"> | <math display="block"> | ||
\rm {cov}(X,Y) = E ((X - \mu(X))(Y - \mu(Y)) )\ . | \rm {cov}(X,Y) = E ((X - \mu(X))(Y - \mu(Y)) )\ . | ||
</math> | </math> | ||
<ul><li> Show that <math>\rm {cov}(X,Y) = E(XY) - E(X)E(Y)</math>. | <ul style="list-style-type:lower-alpha"><li> Show that <math>\rm {cov}(X,Y) = E(XY) - E(X)E(Y)</math>. | ||
</li> | </li> | ||
<li> Using (a), show that <math>{\rm cov}(X,Y) = 0</math>, if | <li> Using (a), show that <math>{\rm cov}(X,Y) = 0</math>, if | ||
Latest revision as of 22:43, 14 June 2024
Let [math]X[/math] and [math]Y[/math] be random variables. The covariance [math]\rm {Cov}(X,Y)[/math] is defined by (see Exercise)
[[math]]
\rm {cov}(X,Y) = E ((X - \mu(X))(Y - \mu(Y)) )\ .
[[/math]]
- Show that [math]\rm {cov}(X,Y) = E(XY) - E(X)E(Y)[/math].
- Using (a), show that [math]{\rm cov}(X,Y) = 0[/math], if [math]X[/math] and [math]Y[/math] are independent. (Caution: the converse is not always true.)
- Show that [math]V(X + Y) = V(X) + V(Y) + 2{\rm cov}(X,Y)[/math].