Put $x=\sqrt{2\nu t}+ct$ and $y=c^2t/\nu$. Then
$$
\frac{cx}{\nu}=\sqrt{2y}+y,
\qquad
\sqrt{1+2(\sqrt{2y}+y)}=1+\sqrt{2y}.
$$
Consequently
$$
h\left(\frac{cx}{\nu}\right)
=1+\sqrt{2y}+y-(1+\sqrt{2y})=y.
$$
The tail bound from part c therefore has exponent $-(\nu/c^2)y=-t$, and
$$
\mathbb P\left(X\geq\sqrt{2\nu t}+ct\right)\leq e^{-t}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 208 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-208/1/d/solution