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A density function sometimes used by engineers to model
Chapter 6, Problem 34E(choose chapter or problem)
A density function sometimes used by engineers to model lengths of life of electronic components is the Rayleigh density, given by
\(f(y)=\left\{\begin{array}{ll} \left(\frac{2 y}{\theta}\right) e^{-y^{2} / \theta}, & y>0, \\ 0, & \text { elsewhere. } \end{array}\right.\)
a. If Y has the Rayleigh density, find the probability density function for \(U=Y^{2}\).
b. Use the result of part (a) to find E(Y) and V(Y).
Questions & Answers
QUESTION:
A density function sometimes used by engineers to model lengths of life of electronic components is the Rayleigh density, given by
\(f(y)=\left\{\begin{array}{ll} \left(\frac{2 y}{\theta}\right) e^{-y^{2} / \theta}, & y>0, \\ 0, & \text { elsewhere. } \end{array}\right.\)
a. If Y has the Rayleigh density, find the probability density function for \(U=Y^{2}\).
b. Use the result of part (a) to find E(Y) and V(Y).
ANSWER:Step 1 of 2
a) We have to find the probability density function of \(U=Y^{2}\)
The rayleigh function is \(\begin{array}{cc} f(y)=\left(\frac{2 y}{\theta}\right) e^{-y^{2 / \theta}}, y>0 \\ =0, & \text { Otherwise } \end{array}\)
Now \(\begin{aligned} U & =Y^{2} \\ Y & =\sqrt{U} \end{aligned}\)
Then \(\frac{d h}{d u}=\frac{d}{d u} \sqrt{u}=\frac{1}{2 \sqrt{u}}\)
Now the probability density function of \(U\) is
\(\begin{aligned} f_{U}(u) & =f_{Y}[h(u)]\left|\frac{d h}{d u}\right| \\ & =\left(\frac{2 \sqrt{u}}{\theta}\right) e^{-u / \theta}\left|\frac{1}{2 \sqrt{u}}\right|[\text { since } y=\sqrt{u}] \\ & =\frac{e^{-u / \theta}}{\theta} \end{aligned}\)
Hence the probability density function of \(U\) is \(f_{U}(u)=\frac{e^{-u / \theta}}{\theta}\)