Question

William Tell is a very bad shot. In practice, he places a small green apple on top of a straight wall which stretches to infinity in both directions. He then takes up position at a distance of one perch from the apple, so that his line of sight to the target is perpendicular to the wall. He now selects an angle uniformly at random from his entire field of view and shoots his arrow in this direction. Assuming that his arrow hits the wall somewhere, what is the distribution function of the horizontal distance (measured in perches) between the apple and the point which the arrow strikes? There is no wind.

Answer 1

Answer:

Step-by-step explanation:

Let d be the perpendicular distance of the target from the person. His angle of view is a random variable $\Theta$ which uniformly distributed (given). $\Theta \sim U\left [-\pi /2,\pi /2 \right ]$ ., since his field view is from $-\pi /2$ to $\pi /2$ . That is

$f_{\Theta }\left ( \theta \right )=\frac{1}{\pi };-\pi /2$

Now the distance X where the arrow strikes from the target is

$\tan \theta =\frac{x}{d}$ Or $x=d\tan \theta .$

Now we haver to find the distribution of $X=d\tan\left | \Theta \right |$ which is a function of known $RV \Theta$ .

We know the distribution of the transformation $X=g\left (\Theta \right )$ is

$f_X\left (x \right )=\sum f_{\Theta }\left ( g^{-1}\left (y\right ) \right )\left | \frac{\mathrm{d} g^{-1}\left (y\right ) }{\mathrm{d} x} \right |$.

The sum is because if we have more than one inverse functions of $\Theta$ .

Now, the inverse function is $\left |\Theta \right |=\tan^{-1}\left ( \frac{X}{d} \right )$ . So the PDF of  $X=d\tan\left | \Theta \right |$ is

$f_X\left (x \right )=\sum f_{\Theta }\left ( \tan^{-1}\left ( \frac{x}{d} \right )\right )\left | \frac{\mathrm{d} \tan^{-1}\left ( \frac{x}{d} \right ) }{\mathrm{d} x} \right |\\ f_X\left (x \right )=2\frac{1}{\pi }\frac{d}{d^2+x^2};0$

I do not understand "perches". If you want you can set d=1 and your distribution becomes,

${\color{Blue} f_X\left (x \right )=\frac{2}{\pi }\frac{1}{1+x^2};0$

Then X measured in so called perches.