LeJEPA: Provable and Scalable Self-Supervised Learning Without the Heuristics — part 7/10

Cleaned-text code points [90244, 103101). Bibliography omitted here; source.md includes it.

Let the spectral radius be constrained by ρ ⁡ ( Σ ) = max i ⁡ λ i ≤ r \rho(\Sigma)=\max_{i}\lambda_{i}\leq r for some r > 0 r>0 . Since x ↦ 1 / x x\mapsto 1/x is strictly decreasing on ( 0 , ∞ ) (0,\infty) , ∑ i = 1 d 1 λ i ≥ ∑ i = 1 d 1 r = d r , \sum_{i=1}^{d}\frac{1}{\lambda_{i}}\;\geq\;\sum_{i=1}^{d}\frac{1}{r}\;=\;\frac{d}{r}, with equality if and only if λ i = r \lambda_{i}=r for all i i . Therefore min Σ ≻ 0 : ρ ⁡ ( Σ ) ≤ r tr ( Σ − 1 ) = d r , attained at Σ = r I d . \min_{\Sigma\succ 0:\ \rho(\Sigma)\leq r}\ \mathrm{tr}(\Sigma^{-1})\;=\;\frac{d}{r},\quad\text{attained at}\quad\Sigma=rI_{d}. (The same conclusion holds if the constraint is ρ ⁡ ( Σ ) = r \rho(\Sigma)=r , since one may take all eigenvalues equal to r r .)

### Conclusion: Isotropic Gaussian is optimal

Combining Lemma B.4 with the solutions (a)–(d), we obtain:

###### Proof.

For any admissible p p with covariance Σ \Sigma , Lemma B.4 gives J ⁡ ( p ) ≥ tr ⁡ ( Σ − 1 ) J(p)\geq\mathrm{tr}(\Sigma^{-1}) . Minimizing the right-hand side under the stated scalar constraint yields Σ = s ​ I d \Sigma=sI_{d} by the calculations in (a)–(d). Equality in Lemma B.4 holds if and only if p p is Gaussian with that covariance, hence p G p_{G} uniquely attains the bound. ∎

∎

### B.5 Proof of Section A.2

###### Proof.

Write the numerator and denominator of m ^ ​ ( x ) \widehat{m}(x) as B n ​ ( x ) := ∑ i = 1 n K h ​ ( x − X i ) ​ Y i , A n ​ ( x ) := ∑ i = 1 n K h ​ ( x − X i ) , B_{n}(x):=\sum_{i=1}^{n}K_{h}(x-X_{i})Y_{i},\qquad A_{n}(x):=\sum_{i=1}^{n}K_{h}(x-X_{i}), so that m ^ ​ ( x ) = B n ​ ( x ) A n ​ ( x ) \widehat{m}(x)=\frac{B_{n}(x)}{A_{n}(x)} . Bias. Compute expectations using independence and change of variables. For the denominator, 𝔼 ​ [ A n ​ ( x ) ] \displaystyle\mathbb{E}[A_{n}(x)] = n ​ 𝔼 ​ [ K h ​ ( x − X ) ] \displaystyle=n\mathbb{E}\big[K_{h}(x-X)\big] = n ​ ∫ ℝ d h − d ​ K ​ ( x − u h ) ​ p ​ ( u ) ​ 𝑑 u \displaystyle=n\int_{\mathbb{R}^{d}}h^{-d}K\Big(\frac{x-u}{h}\Big)p(u)du = n ∫ ℝ d K ( t ) p ( x − h t ) d t ( t := ( x − u ) / h ) \displaystyle=n\int_{\mathbb{R}^{d}}K(t)p(x-ht)dt\qquad(t:=(x-u)/h) = n ∫ ℝ d K ( t ) ( p ( x ) − h t ⊤ ∇ p ( x ) + h 2 2 t ⊤ ∇ 2 p ( x ) t + o ( h 2 ) ) d t \displaystyle=n\int_{\mathbb{R}^{d}}K(t)\Big(p(x)-ht^{\top}\nabla p(x)+\frac{h^{2}}{2}t^{\top}\nabla^{2}p(x)t+o(h^{2})\Big)dt = n ⁡ ( p ⁡ ( x ) + h 2 2 ​ ∫ t ⊤ ​ ∇ 2 p ​ ( x ) ​ t ​ K ​ ( t ) ​ d t ⏟ = μ 2 ​ ( K ) ​ Δ ​ p ​ ( x ) + o ⁡ ( h 2 ) ) , \displaystyle=n\Big(p(x)+\frac{h^{2}}{2}\underbrace{\int t^{\top}\nabla^{2}p(x)tK(t)dt}_{=\mu_{2}(K)\Delta p(x)}+o(h^{2})\Big), where we used symmetry ∫ t ​ K ​ ( t ) ​ 𝑑 t = 0 \int tK(t)dt=0 and isotropy ∫ t ​ t ⊤ ​ K ​ ( t ) ​ 𝑑 t = μ 2 ​ ( K ) ​ I d \int tt^{\top}K(t)dt=\mu_{2}(K)I_{d} , which implies ∫ t ⊤ ​ ∇ 2 p ​ ( x ) ​ t ​ K ​ ( t ) ​ 𝑑 t = μ 2 ​ ( K ) ​ tr ​ ( ∇ 2 p ​ ( x ) ) = μ 2 ​ ( K ) ​ Δ ​ p ​ ( x ) \int t^{\top}\nabla^{2}p(x)tK(t)dt=\mu_{2}(K)\mathrm{tr}(\nabla^{2}p(x))=\mu_{2}(K)\Delta p(x) . Similarly, for the numerator, 𝔼 ​ [ B n ​ ( x ) ] \displaystyle\mathbb{E}[B_{n}(x)] = n ​ 𝔼 ​ [ K h ​ ( x − X ) ​ Y ] = n ​ ∫ K ⁡ ( t ) ​ ( m ​ p ) ​ ( x − h ​ t ) ​ 𝑑 t \displaystyle=n\mathbb{E}\big[K_{h}(x-X)Y\big]=n\int K(t)(mp)(x-ht)dt = n ∫ K ( t ) ( ( m p ) ( x ) − h t ⊤ ∇ ( m p ) ( x ) + h 2 2 t ⊤ ∇ 2 ( m p ) ( x ) t + o ( h 2 ) ) d t \displaystyle=n\int K(t)\Big((mp)(x)-ht^{\top}\nabla(mp)(x)+\frac{h^{2}}{2}t^{\top}\nabla^{2}(mp)(x)t+o(h^{2})\Big)dt = n ⁡ ( m ⁡ ( x ) ​ p ​ ( x ) + h 2 2 ​ μ 2 ​ ( K ) ​ tr ​ ( ∇ 2 ( m ​ p ) ​ ( x ) ) + o ⁡ ( h 2 ) ) \displaystyle=n\Big(m(x)p(x)+\frac{h^{2}}{2}\mu_{2}(K)\mathrm{tr}\big(\nabla^{2}(mp)(x)\big)+o(h^{2})\Big) = n ( m ( x ) p ( x ) + h 2 ​ μ 2 ​ ( K ) 2 ( p Δ m + m Δ p + 2 ∇ m ⊤ ∇ p ) ( x ) + o ( h 2 ) ) , \displaystyle=n\Big(m(x)p(x)+\frac{h^{2}\mu_{2}(K)}{2}\big(p\Delta m+m\Delta p+2\nabla m^{\top}\nabla p\big)(x)+o(h^{2})\Big), where the last step uses the fact that tr ( ∇ 2 ( m p ) ) = p Δ m + m Δ p + 2 ∇ m ⊤ ∇ p \mathrm{tr}\big(\nabla^{2}(mp)\big)=p\Delta m+m\Delta p+2\nabla m^{\top}\nabla p by the product rule and symmetry of mixed derivatives.

Now expand the ratio 𝔼 ​ [ B n ​ ( x ) ] 𝔼 ​ [ A n ​ ( x ) ] \frac{\mathbb{E}[B_{n}(x)]}{\mathbb{E}[A_{n}(x)]} using the identity a 0 + h 2 ​ a 2 + o ⁡ ( h 2 ) b 0 + h 2 ​ b 2 + o ⁡ ( h 2 ) = a 0 b 0 + h 2 ​ a 2 ​ b 0 − a 0 ​ b 2 b 0 2 + o ⁡ ( h 2 ) , \frac{a_{0}+h^{2}a_{2}+o(h^{2})}{b_{0}+h^{2}b_{2}+o(h^{2})}=\frac{a_{0}}{b_{0}}+h^{2}\frac{a_{2}b_{0}-a_{0}b_{2}}{b_{0}^{2}}+o(h^{2}), with a 0 = m ⁡ ( x ) ​ p ​ ( x ) a_{0}=m(x)p(x) , a 2 = μ 2 ​ ( K ) 2 ( p Δ m + m Δ p + 2 ∇ m ⊤ ∇ p ) ( x ) a_{2}=\frac{\mu_{2}(K)}{2}\big(p\Delta m+m\Delta p+2\nabla m^{\top}\nabla p\big)(x) , b 0 = p ⁡ ( x ) b_{0}=p(x) , and b 2 = μ 2 ​ ( K ) 2 ​ Δ ​ p ​ ( x ) b_{2}=\frac{\mu_{2}(K)}{2}\Delta p(x) . This yields 𝔼 ​ [ B n ​ ( x ) ] 𝔼 ​ [ A n ​ ( x ) ] \displaystyle\frac{\mathbb{E}[B_{n}(x)]}{\mathbb{E}[A_{n}(x)]} = m ⁡ ( x ) + h 2 ​ μ 2 ​ ( K ) 2 ​ ( p Δ m + m Δ p + 2 ∇ m ⊤ ∇ p ) p − m p Δ p p 2 | x + o ⁡ ( h 2 ) \displaystyle=m(x)+\frac{h^{2}\mu_{2}(K)}{2}\frac{\big(p\Delta m+m\Delta p+2\nabla m^{\top}\nabla p\big)p-mp\Delta p}{p^{2}}\Big|_{x}+o(h^{2}) = m ( x ) + h 2 ​ μ 2 ​ ( K ) 2 ( Δ m ( x ) + 2 ∇ m ( x ) ⊤ ∇ p ​ ( x ) p ⁡ ( x ) ) + o ( h 2 ) , \displaystyle=m(x)+\frac{h^{2}\mu_{2}(K)}{2}\Big(\Delta m(x)+2\nabla m(x)^{\top}\frac{\nabla p(x)}{p(x)}\Big)+o(h^{2}), which recovers our statement. Variance. Linearize m ^ ​ ( x ) = B n ​ ( x ) / A n ​ ( x ) \widehat{m}(x)=B_{n}(x)/A_{n}(x) around ( 𝔼 ⁡ [ B n ​ ( x ) ] , 𝔼 ⁡ [ A n ​ ( x ) ] ) (\mathbb{E}[B_{n}(x)],\mathbb{E}[A_{n}(x)]) and use independence. To leading order, Var ​ [ m ^ ​ ( x ) ] ≈ Var ​ [ B n ​ ( x ) ] ( 𝔼 ⁡ [ A n ​ ( x ) ] ) 2 . \mathrm{Var}[\widehat{m}(x)]\approx\frac{\mathrm{Var}[B_{n}(x)]}{(\mathbb{E}[A_{n}(x)])^{2}}. Compute Var ​ [ B n ​ ( x ) ] \displaystyle\mathrm{Var}[B_{n}(x)] = ∑ i = 1 n Var ( K h ( x − X i ) Y i ) (independence) \displaystyle=\sum_{i=1}^{n}\mathrm{Var}\big(K_{h}(x-X_{i})Y_{i}\big)\quad\text{(independence)} = n ​ 𝔼 ​ [ K h ​ ( x − X ) 2 ​ Var ​ ( Y ∣ X ) ] = n ​ 𝔼 ​ [ K h ​ ( x − X ) 2 ​ v ​ ( X ) ] \displaystyle=n\mathbb{E}\big[K_{h}(x-X)^{2}\mathrm{Var}(Y\mid X)\big]=n\mathbb{E}\big[K_{h}(x-X)^{2}v(X)\big] = n ​ ∫ h − 2 ​ d ​ K ​ ( x − u h ) 2 ​ v ​ ( u ) ​ p ​ ( u ) ​ 𝑑 u \displaystyle=n\int h^{-2d}K\Big(\frac{x-u}{h}\Big)^{2}v(u)p(u)du = n ​ h − d ​ ∫ K ​ ( t ) 2 ​ v ​ ( x − h ​ t ) ​ p ​ ( x − h ​ t ) ​ 𝑑 t = n ​ h − d ​ ( R ⁡ ( K ) ​ v ​ ( x ) ​ p ​ ( x ) + o ⁡ ( 1 ) ) , \displaystyle=nh^{-d}\int K(t)^{2}v(x-ht)p(x-ht)dt=nh^{-d}\Big(R(K)v(x)p(x)+o(1)\Big), while 𝔼 ⁡ [ A n ​ ( x ) ] = n ⁡ ( p ⁡ ( x ) + o ⁡ ( 1 ) ) . \mathbb{E}[A_{n}(x)]=n\big(p(x)+o(1)\big). Therefore, Var ⁡ [ m ^ ​ ( x ) ] ≈ n ​ h − d ​ R ​ ( K ) ​ v ​ ( x ) ​ p ​ ( x ) n 2 ​ p ​ ( x ) 2 = R ⁡ ( K ) n ​ h d ​ v ⁡ ( x ) p ⁡ ( x ) + o ⁡ ( ( n ​ h d ) − 1 ) , \mathrm{Var}[\widehat{m}(x)]\approx\frac{nh^{-d}R(K)v(x)p(x)}{n^{2}p(x)^{2}}=\frac{R(K)}{nh^{d}}\frac{v(x)}{p(x)}+o\big((nh^{d})^{-1}\big), completing the proof. ∎

### B.6 Proof of Equation 6 to Equation 7

###### Proof.

Let 𝐳 ¯ = 1 V g ​ ∑ v = 1 V g 𝐳 n , v \bar{\mathbf{z}}=\frac{1}{V_{g}}\sum_{v=1}^{V_{g}}\mathbf{z}_{n,v} denote the mean of the first V g V_{g} vectors.

We prove that: 1 V g ​ ∑ v = 1 V g 1 V ​ ∑ v ′ = 1 V ‖ 𝐳 n , v − 𝐳 n , v ′ ‖ 2 2 = 1 V ​ ∑ v ′ = 1 V ‖ 𝐳 ¯ − 𝐳 n , v ′ ‖ 2 2 \frac{1}{V_{g}}\sum_{v=1}^{V_{g}}\frac{1}{V}\sum_{v^{\prime}=1}^{V}\|\mathbf{z}_{n,v}-\mathbf{z}_{n,v^{\prime}}\|_{2}^{2}=\frac{1}{V}\sum_{v^{\prime}=1}^{V}\left\|\bar{\mathbf{z}}-\mathbf{z}_{n,v^{\prime}}\right\|_{2}^{2} (16)

Expanding the left-hand side: LHS = 1 V g ​ V ​ ∑ v = 1 V g ∑ v ′ = 1 V ‖ 𝐳 n , v − 𝐳 n , v ′ ‖ 2 2 \displaystyle=\frac{1}{V_{g}V}\sum_{v=1}^{V_{g}}\sum_{v^{\prime}=1}^{V}\|\mathbf{z}_{n,v}-\mathbf{z}_{n,v^{\prime}}\|_{2}^{2} (17) = 1 V g ​ V ​ ∑ v = 1 V g ∑ v ′ = 1 V ( ‖ 𝐳 n , v ‖ 2 2 − 2 ​ 𝐳 n , v T ​ 𝐳 n , v ′ + ‖ 𝐳 n , v ′ ‖ 2 2 ) \displaystyle=\frac{1}{V_{g}V}\sum_{v=1}^{V_{g}}\sum_{v^{\prime}=1}^{V}\left(\|\mathbf{z}_{n,v}\|_{2}^{2}-2\mathbf{z}_{n,v}^{T}\mathbf{z}_{n,v^{\prime}}+\|\mathbf{z}_{n,v^{\prime}}\|_{2}^{2}\right) (18) = 1 V g ​ ∑ v = 1 V g ‖ 𝐳 n , v ‖ 2 2 − 2 V g ​ V ​ ∑ v = 1 V g ∑ v ′ = 1 V 𝐳 n , v T ​ 𝐳 n , v ′ + 1 V ​ ∑ v ′ = 1 V ‖ 𝐳 n , v ′ ‖ 2 2 \displaystyle=\frac{1}{V_{g}}\sum_{v=1}^{V_{g}}\|\mathbf{z}_{n,v}\|_{2}^{2}-\frac{2}{V_{g}V}\sum_{v=1}^{V_{g}}\sum_{v^{\prime}=1}^{V}\mathbf{z}_{n,v}^{T}\mathbf{z}_{n,v^{\prime}}+\frac{1}{V}\sum_{v^{\prime}=1}^{V}\|\mathbf{z}_{n,v^{\prime}}\|_{2}^{2} (19) = 1 V g ​ ∑ v = 1 V g ‖ 𝐳 n , v ‖ 2 2 − 2 V ​ 𝐳 ¯ T ​ ∑ v ′ = 1 V 𝐳 n , v ′ + 1 V ​ ∑ v ′ = 1 V ‖ 𝐳 n , v ′ ‖ 2 2 \displaystyle=\frac{1}{V_{g}}\sum_{v=1}^{V_{g}}\|\mathbf{z}_{n,v}\|_{2}^{2}-\frac{2}{V}\bar{\mathbf{z}}^{T}\sum_{v^{\prime}=1}^{V}\mathbf{z}_{n,v^{\prime}}+\frac{1}{V}\sum_{v^{\prime}=1}^{V}\|\mathbf{z}_{n,v^{\prime}}\|_{2}^{2} (20)

Expanding the right-hand side: RHS = 1 V ​ ∑ v ′ = 1 V ( ‖ 𝐳 ¯ ‖ 2 2 − 2 ​ 𝐳 ¯ T ​ 𝐳 n , v ′ + ‖ 𝐳 n , v ′ ‖ 2 2 ) \displaystyle=\frac{1}{V}\sum_{v^{\prime}=1}^{V}\left(\|\bar{\mathbf{z}}\|_{2}^{2}-2\bar{\mathbf{z}}^{T}\mathbf{z}_{n,v^{\prime}}+\|\mathbf{z}_{n,v^{\prime}}\|_{2}^{2}\right) (21) = ‖ 𝐳 ¯ ‖ 2 2 − 2 V ​ 𝐳 ¯ T ​ ∑ v ′ = 1 V 𝐳 n , v ′ + 1 V ​ ∑ v ′ = 1 V ‖ 𝐳 n , v ′ ‖ 2 2 \displaystyle=\|\bar{\mathbf{z}}\|_{2}^{2}-\frac{2}{V}\bar{\mathbf{z}}^{T}\sum_{v^{\prime}=1}^{V}\mathbf{z}_{n,v^{\prime}}+\frac{1}{V}\sum_{v^{\prime}=1}^{V}\|\mathbf{z}_{n,v^{\prime}}\|_{2}^{2} (22)

To complete the proof, we verify that: 1 V g ​ ∑ v = 1 V g ‖ 𝐳 n , v ‖ 2 2 = ‖ 𝐳 ¯ ‖ 2 2 \frac{1}{V_{g}}\sum_{v=1}^{V_{g}}\|\mathbf{z}_{n,v}\|_{2}^{2}=\|\bar{\mathbf{z}}\|_{2}^{2} (23)

Expanding the right-hand side: ‖ 𝐳 ¯ ‖ 2 2 \displaystyle\|\bar{\mathbf{z}}\|_{2}^{2} = ‖ 1 V g ​ ∑ v = 1 V g 𝐳 n , v ‖ 2 2 \displaystyle=\left\|\frac{1}{V_{g}}\sum_{v=1}^{V_{g}}\mathbf{z}_{n,v}\right\|_{2}^{2} (24) = 1 V g 2 ​ ∑ v = 1 V g ∑ v ′′ = 1 V g 𝐳 n , v T ​ 𝐳 n , v ′′ \displaystyle=\frac{1}{V_{g}^{2}}\sum_{v=1}^{V_{g}}\sum_{v^{\prime\prime}=1}^{V_{g}}\mathbf{z}_{n,v}^{T}\mathbf{z}_{n,v^{\prime\prime}} (25) = 1 V g ​ ∑ v = 1 V g ‖ 𝐳 n , v ‖ 2 2 \displaystyle=\frac{1}{V_{g}}\sum_{v=1}^{V_{g}}\|\mathbf{z}_{n,v}\|_{2}^{2} (26)

Therefore, LHS = RHS, completing the proof. ∎

### B.7 Proof of Section A.2

###### Proof.

For each x x , Bias [ m ^ ( x ) ] = h 2 ​ μ 2 ​ ( K ) 2 ( Δ m ( x ) + 2 ∇ m ( x ) ⊤ ∇ log p ( x ) ) + o ( h 2 ) . \mathrm{Bias}[\widehat{m}(x)]=\frac{h^{2}\mu_{2}(K)}{2}\Big(\Delta m(x)+2\nabla m(x)^{\top}\nabla\log p(x)\Big)+o(h^{2}). Square and integrate against p ⁡ ( x ) p(x) : ℬ 2 ​ ( h , p , m ) \displaystyle\mathcal{B}^{2}(h;p,m) = ( h 2 ​ μ 2 ​ ( K ) 2 ) 2 ∫ ( Δ m ( x ) + 2 ∇ m ( x ) ⊤ ∇ log p ( x ) ) 2 p ( x ) d x + o ( h 4 ) \displaystyle=\Big(\frac{h^{2}\mu_{2}(K)}{2}\Big)^{2}\int\Big(\Delta m(x)+2\nabla m(x)^{\top}\nabla\log p(x)\Big)^{2}p(x)dx+o(h^{4}) ≤ ( h 2 ​ μ 2 ​ ( K ) 2 ) 2 ∫ ( 2 ( Δ m ( x ) ) 2 + 2 ( 2 ∇ m ( x ) ⊤ ∇ log p ( x ) ) 2 ) p ( x ) d x + o ( h 4 ) \displaystyle\leq\Big(\frac{h^{2}\mu_{2}(K)}{2}\Big)^{2}\int\Big(2(\Delta m(x))^{2}+2(2\nabla m(x)^{\top}\nabla\log p(x))^{2}\Big)p(x)dx+o(h^{4}) = ( h 2 ​ μ 2 ​ ( K ) 2 ) 2 ( 2 ∫ ( Δ m ( x ) ) 2 p ( x ) d x + 8 ∫ ( ∇ m ( x ) ⊤ ∇ log p ( x ) ) 2 p ( x ) d x ) + o ( h 4 ) , \displaystyle=\Big(\frac{h^{2}\mu_{2}(K)}{2}\Big)^{2}\Big(2\int(\Delta m(x))^{2}p(x)dx+8\int(\nabla m(x)^{\top}\nabla\log p(x))^{2}p(x)dx\Big)+o(h^{4}), where we used ( a + b ) 2 ≤ 2 ​ a 2 + 2 ​ b 2 (a+b)^{2}\leq 2a^{2}+2b^{2} pointwise. Since | Δ ​ m ​ ( x ) | ≤ B |\Delta m(x)|\leq B for all x x , we have ∫ ( Δ ​ m ) 2 ​ p ≤ ∫ B 2 ​ p = B 2 . \int(\Delta m)^{2}p\leq\int B^{2}p=B^{2}. For the second term, first use Cauchy–Schwarz and then integrate against p ⁡ ( x ) p(x) to obtain ( ∇ m ( x ) ⊤ ∇ log p ( x ) ) 2 ≤ ∥ ∇ m ( x ) ∥ 2 ∥ ∇ log p ( x ) ∥ 2 ≤ L 2 ∥ ∇ log p ( x ) ∥ 2 \displaystyle(\nabla m(x)^{\top}\nabla\log p(x))^{2}\leq\|\nabla m(x)\|^{2}\|\nabla\log p(x)\|^{2}\leq L^{2}\|\nabla\log p(x)\|^{2} ⟹ ∫ ( ∇ m ( x ) ⊤ ∇ log p ( x ) ) 2 p ( x ) d x ≤ L 2 ∫ ∥ ∇ log p ( x ) ∥ 2 p ( x ) d x = L 2 J ( p ) . \displaystyle\implies\int(\nabla m(x)^{\top}\nabla\log p(x))^{2}p(x)dx\leq L^{2}\int\|\nabla\log p(x)\|^{2}p(x)dx=L^{2}J(p). which can be combined with the bounds above to obtain the desired result. We similarly have for the integrated variance 𝒱 ⁡ ( h , p ) \displaystyle\mathcal{V}(h;p) = ∫ ( R ⁡ ( K ) n ​ h d ​ v ⁡ ( x ) p ⁡ ( x ) + o ⁡ ( ( n ​ h d ) − 1 ) ) ​ p ​ ( x ) ​ 𝑑 x = R ⁡ ( K ) n ​ h d ​ ∫ v ⁡ ( x ) ​ 𝑑 x + o ⁡ ( ( n ​ h d ) − 1 ) , \displaystyle=\int\Big(\frac{R(K)}{nh^{d}}\frac{v(x)}{p(x)}+o\big((nh^{d})^{-1}\big)\Big)p(x)dx=\frac{R(K)}{nh^{d}}\int v(x)dx+o\big((nh^{d})^{-1}\big), which is independent of p p . ∎

### B.8 Proof of Section 4.1

###### Proof.

We first start by reminding the reader about the original Cramér-Wold theorem that is a function of all possible directions (not unit-norm ones).