A problem with uniform distribution?

Question

In a game Alis and Daniel shoot arrows on a circular target with a radius of R.

Alis throws arrows such that their distance from the center have Continuous uniform distribution (0,R) - o,R are the parameters for uniform distribution.

While Daniel throws arrows such that their distance from the center have random Continuous uniform distribution (From the target)

I was asked to calculate the expected value for the distance of the arrows for both participant.

For Alis it's R/2 but what about Daniel? I didn't understand what's given about him, what are the parameters for him...

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I claimed: "For Alis it's R/2" Proof:

The expected value for random variable with uniform distribution (a,b) (with parameters a and b) is (a+b)/2

Answer 1

HINT: Presumably it means that the probability that Daniel’s arrow lands in a given region of the target is proportional to the area of the region. The area of an annulus of mean radius $r$ and width $dr$ is $dA=2\pi r\,dr$, and the area of the target is $\pi R^2$, so the probability that it lands in that annulus is

$$\frac{dA}{\pi R^2}=\frac{2\pi r\,dr}{\pi R^2}=\frac2{R^2}r\,dr\,.$$

Alternatively, the cumulative distribution function $F(r)$ for $0\le r\le R$ is given by

$$F(r)=\frac{\pi r^2}{\pi R^2}=\frac{r^2}{R^2}\,.$$

Use whichever approach you like to get the probability density function, and then use that to calculated the expected value from the definition of expected value.

Answer 2

Assuming the question is:

1. A's arrows hit the target so that the distribution of the distance from the center is uniform, i.e., $f_A(r,\theta )= \begin{cases} \frac{1}{2\pi R} & 0<r<R, 0<\theta<2\pi \\ 0, &\text{otherwise} \end{cases} $,

2. D's arrows hit the target uniformly distributed over the target, i.e., $f_D(r,\theta )= \begin{cases} \frac{r}{\pi R^2} & 0<r<R, 0<\theta<2\pi \\ 0, &\text{otherwise} \end{cases} $.

Further, if the question asks for the expected distance of the arrow from the center for each of them, then:

$E_A\{r\} = \int\limits_{r=0}^{R} r\frac{1}{R}dr = \frac{1}{2}R$, and

$E_D\{r\} = \int\limits_{r=0}^{R} r\frac{2r}{R^2}dr=\frac{2}{3}R$.