Because a strictly monotone map fixing both endpoints is increasing, the <change of variables formula> $t=h(s)$ gives
$$
\lVert Y\rVert^2
=\int_0^1X(h^{-1}(t))^2\,dt
=\int_0^1X(s)^2h'(s)\,ds.
$$
Assuming $\mathbb E\lVert X\rVert^2>0$, define
$$
A=\frac{\mathbb E\int_0^1X(s)^2h'(s)\,ds}
{\mathbb E\int_0^1X(s)^2\,ds}.
$$
The Hilbert-space variance identity then gives
$$
\begin{aligned}
\mathbb E\lVert Y-\mu\rVert^2
&=\mathbb E\lVert Y\rVert^2-\lVert\mu\rVert^2\\
&=A\mathbb E\lVert X\rVert^2-\lVert\mu\rVert^2\\
&=A\mathbb E\lVert X-\nu\rVert^2+A\lVert\nu\rVert^2-\lVert\mu\rVert^2.
\end{aligned}
$$

Under the regularity needed to interchange expectation and differentiation, $\mathbb Eh'(s)=1$ because $\mathbb Eh(s)=s$. Hence
$$
A=1+
\frac{\int_0^1\operatorname{Cov}(X(s)^2,h'(s))\,ds}
{\mathbb E\lVert X\rVert^2}.
$$
Thus $A=1$ exactly when the integrated covariance in the numerator vanishes. In particular, this holds when the amplitude $X$ and the time warp $h$ are <independent random variables>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 225 / 3 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-225/3/a/i/solution