= Solution
The <probability density> is an <even function>. Its <derivative> simplifies to
$$
h'(z)=\frac1\pi\left[\frac1{1+(z+1/2)^2}-\frac1{1+(z-1/2)^2}\right]
=\frac{-2z}{\pi[1+(z+1/2)^2][1+(z-1/2)^2]}.
$$
It increases for $z<0$ and decreases for $z>0$, with maximum height $h(0)=2\arctan(1/2)/\pi$. Thus the most probable value, or <mode>, is $\boxed{z=0}$. The original sketch shows the symmetric, slightly flattened peak and long tails.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2-q14-cauchy-smoothing.png]
{title=Density of a standard Cauchy variable after an independent uniform displacement}
There is a genuine qualification to the request for a mean: \b[the mean does not exist]. The integral representation shows that $h(z)\sim1/(\pi z^2)$ as $|z|\to\infty$, so
$$
\int_0^\infty z h(z)\,dz=\infty,\qquad
\int_{-\infty}^0 z h(z)\,dz=-\infty.
$$
The two divergent integrals cannot be subtracted to define an <expected value>. Symmetry does give the <Cauchy principal value> $\lim_{A\to\infty}\int_{-A}^A zh(z)\,dz=0$, but that is not the mean. Equivalently, adding the bounded variable $Y$ cannot make the first absolute moment of a <Cauchy distribution> finite.
Back to article page
