Question

1. A point P is chosen with a uniform probability distribution around a circle of radius r Let Z be a random variable that measures the absolute value of the distance of P from the y-axis (a) What is the mean and the variance of Z? (Hint, define an appropriately normalized uniform probability density function for the angle 0 describing the polar angle of the position P on the circle.) (b) Does your answer for the mean make sense? (c) What is the probability of Z being greater than its mean?

Answer 1

The joint PDF is, $f_{XY}\left ( x,y \right )=\frac{1}{\pi r^2};0<x^2+y^2\leqslant r^2$ . are the X-Y coordinates of the point P.  The region is a circle with radius .

Define the random variable $Z=\left | X \right |$ , the absolute value of the distance of from Y-axis.

a) The marginal PDF is, dependent on , varies from $\sqrt{r^2-X^2} \textup{ to } \sqrt{r^2-X^2}.$

$f_X\left ( x \right )=\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}f_{XY}\left ( x,y \right )dy\\ f_X\left ( x \right )=\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\left ( \frac{1}{\pi r^2} \right )dy\\ f_X\left ( x \right )=\frac{2\sqrt{r^2-x^2}}{\pi r^2} ;-r\leqslant x\leqslant r$

So the PDF of $Z=\left | X \right |$ is

$f_Z\left ( z \right )=\frac{4\sqrt{r^2-z^2}}{\pi r^2} ;0\leqslant z\leqslant r$

The mean is

$E\left ( Z \right )=\int_{0}^{r}zf_Z\left ( z \right )\\ E\left ( Z \right )=\int_{0}^{r}z\frac{4\sqrt{r^2-z^2}}{\pi r^2}dz\\ E\left ( Z \right )=\left [-\frac{4\left (r^2-z^2 \right )^{3/2}}{3\pi r^2} \right ]_{z=0}^{r}\\ {\color{Blue} E\left ( Z \right )=\frac{4r}{3\pi }}$

$E\left ( Z ^2\right )=\int_{0}^{r}z^2f_Z\left ( z \right )dz\\ E\left ( Z^2 \right )=\int_{0}^{r}z^2\frac{4\sqrt{r^2-z^2}}{\pi r^2}dz\\ E\left ( Z^2 \right )=\frac{r^2}{4}$

The variance is

$Var\left ( Z \right )=E\left ( Z ^2\right )-E\left ( Z \right )^2\\ Var\left ( Z \right )=\frac{r^2}{4}-\left (\frac{4r}{3\pi } \right )^2\\ {\color{Blue} Var\left ( Z \right )=\left ( \frac{9\pi ^2-64 }{36\pi ^2}\right )r^2}$

b) The mean is $E\left ( Z \right )=\frac{4r}{3\pi }<r$ so it makes sense.

c) The required probability is

$P\left ( Z>\frac{4r}{3\pi } \right )=\int_{\frac{4r}{3\pi }}^{r}f_Z\left ( z \right )dz\\ P\left ( Z>\frac{4r}{3\pi } \right )=\int_{\frac{4r}{3\pi }}^{r}\frac{4\sqrt{r^2-z^2}}{\pi r^2} dz\\ {\color{Blue} P\left ( Z>\frac{4r}{3\pi } \right )=0.4763124186}$