Find the probability density function (pdf) for a uniform distribution on the unit circle.

To find the probability density function (pdf) for a uniform distribution on the unit circle, we start by noting that the unit circle can be expressed in terms of its angular coordinate $\theta$, where

txt1

txt1

txt1

\leq \theta < 2\pi$.

Since the distribution is uniform, the probability density function (pdf) must be constant. The integral of the pdf over the entire circle must be 1:

$ \int_0^{2\pi} f(\theta) \, d\theta = 1 $

Let $f(\theta) = c$ be the constant pdf. Then:

$ c \int_0^{2\pi} \, d\theta = 1 $

Evaluating the integral gives:

$ c \cdot 2\pi = 1 $

Solving for $c$, we get:

$ c = \frac{1}{2\pi} $

Therefore, the pdf for a uniform distribution on the unit circle is:

$ f(\theta) = \frac{1}{2\pi}, \quad 0 \leq \theta < 2\pi $

The pdf for a uniform distribution on the unit circle can be found by considering the angular coordinate $ heta$ with

txt2

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txt2

leq heta < 2pi$. The pdf must be constant and integrate to 1:

$ int_0^{2pi} f( heta) , d heta = 1 $

Assuming $f( heta) = c$:

$ c int_0^{2pi} , d heta = 1 Rightarrow c cdot 2pi = 1 Rightarrow c = frac{1}{2pi} $

Thus, the pdf is:

$ f( heta) = frac{1}{2pi} $

On the unit circle, the pdf for a uniform distribution is constant and must integrate to 1:

$ int_0^{2pi} f( heta) , d heta = 1 $

If $f( heta) = c$:

$ c cdot 2pi = 1 Rightarrow c = frac{1}{2pi} $

The pdf is:

$ f( heta) = frac{1}{2pi} $

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