Revision as of 03:26, 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> Suppose again that <math>Z = X + Y</math>. Find <math>f_Z</math> if <math display="block"> \begin{eqnarray*} f_X(x) &=& \frac 1{\sqrt{2\pi}\sigma_1} e^{-(x - \mu_1)^2/2\sigma_1^2} \\ f_Y(x) &=& \frac 1{\sqrt{2\pi}\sigma_2} e^{-(x - \mu_2)^2/2\si...")
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BBy Bot
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]

Suppose again that [math]Z = X + Y[/math]. Find [math]f_Z[/math] if

[[math]] \begin{eqnarray*} f_X(x) &=& \frac 1{\sqrt{2\pi}\sigma_1} e^{-(x - \mu_1)^2/2\sigma_1^2} \\ f_Y(x) &=& \frac 1{\sqrt{2\pi}\sigma_2} e^{-(x - \mu_2)^2/2\sigma_2^2}\ . \end{eqnarray*} [[/math]]