Revision as of 02:25, 9 June 2024 by Bot (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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Jun 09'24

Exercise

[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]

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]] \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].