UnkleRhaukus
find \[\langle \theta \rangle \]
for \(0≤\theta≤\pi/2\)



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UnkleRhaukus
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\[\langle x\rangle=\int\limits_0^{\pi/2}\theta\cdot\rho(\theta)\cdot\text d\theta \]

hartnn
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whats \(\rho(\theta)\) ?

UnkleRhaukus
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probability density

UnkleRhaukus
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\[1=\int\limits_0^{\pi/2}\rho(\theta)\cdot\text d\theta\]

UnkleRhaukus
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assuming a unifirm probability distribution \[\rho(\theta)=A\]
\[\frac 1A=\int\limits_0^{\pi/2}\text d\theta\]

hartnn
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A=2/\(\pi\) then

UnkleRhaukus
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\[\frac 1A=\theta_0^{\pi/2}=\frac\pi2\]
\[A=\frac 2\pi\]

UnkleRhaukus
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so,
\[\langle x\rangle=\frac 2\pi\int\limits_0^{\pi/2}\theta\cdot\text d\theta\]

hartnn
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pi/4 ?

UnkleRhaukus
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\[=\frac 2\pi \frac{\theta^2}{2} _0^{\pi/2}\]
\[=\frac 2\pi \frac{\left(\pi/2\right)^2}{2}\]
\[=\frac\pi 4\]

hartnn
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nice, pdf was assumed or given?
assuming it as uniform simplified it a lot.....

UnkleRhaukus
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i probably should have stated uniform probability distribution in the question

UnkleRhaukus
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so
for \(0≤\theta≤2\pi\)\[\langle \theta \rangle =\pi\]
the expected angle in a circle is 180°

UnkleRhaukus
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makes sense

hartnn
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uniform pdf will always give you midpoint as expectation....

hartnn
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right ?

UnkleRhaukus
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right