## UnkleRhaukus 2 years ago find $\langle \theta \rangle$ for $$0≤\theta≤\pi/2$$

1. UnkleRhaukus

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

2. hartnn

whats $$\rho(\theta)$$ ?

3. UnkleRhaukus

probability density

4. UnkleRhaukus

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

5. UnkleRhaukus

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

6. hartnn

A=2/$$\pi$$ then

7. UnkleRhaukus

$\frac 1A=\theta|_0^{\pi/2}=\frac\pi2$ $A=\frac 2\pi$

8. UnkleRhaukus

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

9. hartnn

pi/4 ?

10. UnkleRhaukus

$=\frac 2\pi \frac{\theta^2}{2} |_0^{\pi/2}$ $=\frac 2\pi \frac{\left(\pi/2\right)^2}{2}$ $=\frac\pi 4$

11. hartnn

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

12. UnkleRhaukus

i probably should have stated uniform probability distribution in the question

13. UnkleRhaukus

so for $$0≤\theta≤2\pi$$$\langle \theta \rangle =\pi$ the expected angle in a circle is 180°

14. UnkleRhaukus

makes sense

15. hartnn

uniform pdf will always give you mid-point as expectation....

16. hartnn

right ?

17. UnkleRhaukus

right