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UnkleRhaukus

  • 3 years ago

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

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  1. UnkleRhaukus
    • 3 years ago
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    \[\langle x\rangle=\int\limits_0^{\pi/2}\theta\cdot\rho(\theta)\cdot\text d\theta \]

  2. hartnn
    • 3 years ago
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    whats \(\rho(\theta)\) ?

  3. UnkleRhaukus
    • 3 years ago
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    probability density

  4. UnkleRhaukus
    • 3 years ago
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    \[1=\int\limits_0^{\pi/2}\rho(\theta)\cdot\text d\theta\]

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

  6. hartnn
    • 3 years ago
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    A=2/\(\pi\) then

  7. UnkleRhaukus
    • 3 years ago
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    \[\frac 1A=\theta|_0^{\pi/2}=\frac\pi2\] \[A=\frac 2\pi\]

  8. UnkleRhaukus
    • 3 years ago
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    so, \[\langle x\rangle=\frac 2\pi\int\limits_0^{\pi/2}\theta\cdot\text d\theta\]

  9. hartnn
    • 3 years ago
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    pi/4 ?

  10. UnkleRhaukus
    • 3 years ago
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    \[=\frac 2\pi \frac{\theta^2}{2} |_0^{\pi/2}\] \[=\frac 2\pi \frac{\left(\pi/2\right)^2}{2}\] \[=\frac\pi 4\]

  11. hartnn
    • 3 years ago
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    nice, pdf was assumed or given? assuming it as uniform simplified it a lot.....

  12. UnkleRhaukus
    • 3 years ago
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    i probably should have stated uniform probability distribution in the question

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

  14. UnkleRhaukus
    • 3 years ago
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    makes sense

  15. hartnn
    • 3 years ago
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    uniform pdf will always give you mid-point as expectation....

  16. hartnn
    • 3 years ago
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    right ?

  17. UnkleRhaukus
    • 3 years ago
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    right

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