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UnkleRhaukus Group Title

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

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

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

    • 2 years ago
  3. UnkleRhaukus Group Title
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    probability density

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

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

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  6. hartnn Group Title
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    A=2/\(\pi\) then

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  7. UnkleRhaukus Group Title
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    \[\frac 1A=\theta|_0^{\pi/2}=\frac\pi2\] \[A=\frac 2\pi\]

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

    • 2 years ago
  9. hartnn Group Title
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    pi/4 ?

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  10. UnkleRhaukus Group Title
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    \[=\frac 2\pi \frac{\theta^2}{2} |_0^{\pi/2}\] \[=\frac 2\pi \frac{\left(\pi/2\right)^2}{2}\] \[=\frac\pi 4\]

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

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

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

    • 2 years ago
  14. UnkleRhaukus Group Title
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    makes sense

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

    • 2 years ago
  16. hartnn Group Title
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    right ?

    • 2 years ago
  17. UnkleRhaukus Group Title
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    right

    • 2 years ago
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