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UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[\langle x\rangle=\int\limits_0^{\pi/2}\theta\cdot\rho(\theta)\cdot\text d\theta \]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.4whats \(\rho(\theta)\) ?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0probability density

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[1=\int\limits_0^{\pi/2}\rho(\theta)\cdot\text d\theta\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0assuming a unifirm probability distribution \[\rho(\theta)=A\] \[\frac 1A=\int\limits_0^{\pi/2}\text d\theta\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac 1A=\theta_0^{\pi/2}=\frac\pi2\] \[A=\frac 2\pi\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0so, \[\langle x\rangle=\frac 2\pi\int\limits_0^{\pi/2}\theta\cdot\text d\theta\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[=\frac 2\pi \frac{\theta^2}{2} _0^{\pi/2}\] \[=\frac 2\pi \frac{\left(\pi/2\right)^2}{2}\] \[=\frac\pi 4\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.4nice, pdf was assumed or given? assuming it as uniform simplified it a lot.....

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i probably should have stated uniform probability distribution in the question

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0so for \(0≤\theta≤2\pi\)\[\langle \theta \rangle =\pi\] the expected angle in a circle is 180°

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.4uniform pdf will always give you midpoint as expectation....
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