Question

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

Answer 1

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

Answer 2

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

Answer 3

probability density

Answer 4

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

Answer 5

assuming a unifirm probability distribution \[\rho(\theta)=A\]

\[\frac 1A=\int\limits_0^{\pi/2}\text d\theta\]

Answer 6

A=2/\(\pi\) then

Answer 7

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

Answer 8

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

Answer 9

pi/4 ?

Answer 10

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

Answer 11

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

Answer 12

i probably should have stated uniform probability distribution in the question

Answer 13

so


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

the expected angle in a circle is 180°

Answer 14

makes sense

Answer 15

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

Answer 16

right ?

Answer 17

right