Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 16

The slope of the particle's straight trajectory is $\tan {\theta }_i=y_i/x_0$, soThe endpoint directions $\pm \pi /2$ are parallel to the screen and do not give finite collisions, but each has zero probability in the continuous angular model. The relation therefore holds almost surely.

The angular interval has width $\pi$, so its continuous uniform distribution has probability density functionIntegrating this probability density function gives the cumulative distribution functionValues of the probability density function at the two individual endpoints may be assigned arbitrarily without changing any probability.

The tangent function is strictly increasing on the angular interval. Thus $Y\le y$ is equivalent, up to the zero-probability endpoints, to $\Theta \le \arctan (y/x_0)$. Substituting into the angular cumulative distribution function givesIts limits are zero and one at the two infinite endpoints. This is the construction in which the tangent of a uniform angle has a Cauchy distribution.

Differentiate the cumulative distribution function using the chain rule:This is a Cauchy distribution with location zero and scale $x_0$. The probability density function is nonnegative and integrates to $F(\infty )-F(-\infty )=1$.

The expected value is not defined: the integral of $yf(y)$ over the positive half-line diverges logarithmically to $+\infty$, and the negative half-line gives $-\infty$. In particular, symmetry alone does not supply a mean. More explicitly,The second moment also diverges because $y^{2}f(y)\to x_0/\pi >0$ as $|y|\to \infty$. ThusThe symmetric Cauchy principal value of the first-moment integral is zero, but it is not an expected value and does not define a variance.

The new angular interval has width $\pi /2$, so its probability density function is $2/\pi$ there. Its endpoints stay away from the poles of the tangent function, making $Y$ bounded and its expected value finite. Integrate over the continuous uniform distribution of angles:The two cosines are $\sqrt{3}/2$ and $1/2$. HenceIt has units of length and is positive because the permitted positive angles extend farther than the negative ones.

The $i$th screen distance is $x_0/2^i$, starting at $i=1$. The previous expected value therefore scales to $E[Y_i]=x_0\ln 3/(\pi 2^i)$. To justify the infinite linearity of expectation, note that $|\tan {\theta }_i|\le \sqrt{3}$ throughout the restricted angular interval. ThusThis deterministic integrable bound justifies the expectation of an absolutely summable random series, even without assuming independent firing angles. Summing the geometric series gives