spin arrow of random variables

Question

Spin an arrow attached to the center of a circular board, let theta be the final angle of the arrow, theta<= 2pi. The probability that theta falls in a subinterval (0, 2pi] is proportional to the length of the subinterval. The random variable is dened by X(theta) = 2theta. Find the CDF and pdf for X.

Answer 1

From the tag, you are aware that you are dealing with a (continuous) uniform distribution.

If the random variable $X$ has distribution that is uniform in the interval $(a,b]$, then the density function $f_X(x)$ is $\frac{1}{b-a}$ for $x$ in the interval, and $0$ elsewhere.

The cumulative distribution function $F_X(x)$ gives $\Pr(X\le x)$.

First the easy parts: If $x\le a$, then $F_X(z)=0$. For $X\gt b$, $F_X(x)=1$.

For $a\lt x\le b$, the probability that $X\le x$ is the length of the interval $[a,x]$, divided by $b-a$. So $F_X(x)=\frac{x-a}{b-a}$.

Remark: Please note that for example the specification that $f_X(x)$ is $0$ when $x$ is not in $(a,b]$ is an essential part of the specification of $f_X(x)$. This is important for two reasons: (i) If you do not so specify, you may lose marks on a test and (ii) Forgetting about the fact that the density is given by different formulas in different parts of the world is a fairly common cause of computational errors.