If (Eleft(|X|_{p}^{alpha}+|Y|_{q}^{alpha} ight)

If \(E\left(|X|_{p}^{\alpha}+|Y|_{q}^{\alpha}\right)<\infty\) we define \(\mathcal{V}^{(\alpha)}\) by its square \(\mathcal{V}^{2}(X, Y ; w)(15.2)\) with weight function \(w(t, s ; \alpha)(15.3)\). That is, if \(X \in \mathbb{R}^{p}\) and \(Y \in \mathbb{R}^{q}\),

\[
\begin{aligned}
\mathcal{V}^{2(\alpha)}(X, Y) & =\left\|\hat{f}_{X, Y}(t, s)-\hat{f}_{X}(t) \hat{f}_{Y}(s)\right\|_{\alpha}^{2} \\
& =\frac{1}{C(p, \alpha) C(q, \alpha)} \int_{\mathbb{R}^{p+q}} \frac{\left|\hat{f}_{X, Y}(t, s)-\hat{f}_{X}(t) \hat{f}_{Y}(s)\right|^{2}}{|t|_{p}^{\alpha+p}|s|_{q}^{\alpha+q}} d t d s,
\end{aligned}
\]

where \(C(p, \alpha)\) is defined in Lemma 15.1.

Lemma 15.1 holds for all exponents \(0<\alpha<2\). Thus the Remark that follows Lemma 15.1 can be generalized for \(|X|^{\alpha},|Y|^{\alpha}\). Without using Brownian covariance, prove that

\[
\begin{aligned}
\mathcal{V}^{2(\alpha)}(X, Y)=E\left|X-X^{\prime}\right|^{\alpha}\left|Y-Y^{\prime}\right|^{\alpha}+E\left|X-X^{\prime}\right|^{\alpha} E\left|Y-Y^{\prime}\right|^{\alpha} \\
-E\left|X-X^{\prime}\right|\left|Y-Y^{\prime \prime}\right|^{\alpha}-E\left|X-X^{\prime \prime}\right|^{\alpha}\left|Y-Y^{\prime}\right|^{\alpha} .
\end{aligned}
\]