Proof of Rotation Invariance of a Pair of Independent Standard Normal Random Variables

Throughout, $\mathcal{B}_2=\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$ is the product $\sigma$-algebra on $\mathbb{R}^{2}$, and a function on $\mathbb{R}^{2}$ is called jointly Borel if it is measurable from $(\mathbb{R}^{2},\mathcal{B}_2)$ to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, as in Joint Distribution, Expectations, and Block Independence for Independent Random Variables. Let $g(x)=\exp(-x^{2}/2)$ and $c=\int_{\mathbb{R}}g\,d\lambda\in(0,\infty)$ be as in Standard Normal Distribution, with $\lambda$ Lebesgue measure, which is $\sigma$-finite, so the product measure $\lambda\otimes\lambda$ on $\mathcal{B}_2$ is defined. Set $G(x,y)=g(x)\,g(y)$.

Step 0: Measurability inventory. By Claim 4 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables, the coordinate projections $\pi_1,\pi_2$ are jointly Borel, and $t\circ\varphi$ is jointly Borel whenever $\varphi$ is jointly Borel and $t$ is Borel; since $g$ is Borel (recorded in Standard Normal Distribution), $g\circ\pi_1$ and $g\circ\pi_2$ are jointly Borel, and their product $G$ is jointly Borel by the closure of measurable functions under products recorded in the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product, applied on the measurable space $(\mathbb{R}^{2},\mathcal{B}_2)$. Next, a map $T:\mathbb{R}^{2}\to\mathbb{R}^{2}$ whose two coordinate functions are jointly Borel is measurable from $(\mathbb{R}^{2},\mathcal{B}_2)$ to itself: the preimage of a measurable rectangle $B\times B'$ is the intersection of the coordinate preimages, which lie in $\mathcal{B}_2$, and rectangles generate $\mathcal{B}_2$ by Product Sigma-Algebra, so the generator criterion applies. Finally, for a real $\alpha$ the function $h_\alpha(x,y)=x+\alpha y$ is jointly Borel: the map $\varphi_\alpha(x,y)=(x,\alpha y)$ has jointly Borel coordinates ($\pi_1$, and $t\circ\pi_2$ with the Borel map $t(u)=\alpha u$), hence is measurable by the rectangle argument, and $h_\alpha$ is the composition of the jointly Borel addition map of Claim 4 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables with $\varphi_\alpha$, and compositions of measurable maps are measurable (preimages compose).

Step 1: The joint law of $(Z_1,Z_2)$ has density $G$. By Claim 1 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables, the map $V=(Z_1,Z_2):\Omega\to\mathbb{R}^{2}$ is measurable and its distribution is the product measure $N\otimes N$, where $N$ is the standard normal distribution. Define $\rho:\mathcal{B}_2\to[0,\infty]$ by

$$\rho(C)=c^{-2}\int_{\mathbb{R}^{2}}\mathbf{1}_{C}\,G\,d(\lambda\otimes\lambda).$$

Then $\rho$ is a measure: $\rho(\emptyset)=0$, and for pairwise disjoint $(C_k)_{k\in\mathbb{N}}$ the partial sums of $\sum_k\mathbf{1}_{C_k}G$ increase pointwise to $\mathbf{1}_{\cup_kC_k}G$, so countable additivity follows from Monotone Convergence Theorem together with the additivity of the integral in Linearity and Monotonicity of the Lebesgue Integral. On a measurable rectangle $B\times B'$, the Tonelli theorem and the fact that constants factor out of the integral (by Linearity and Monotonicity of the Lebesgue Integral) give

$$\rho(B\times B')=c^{-2}\int_{\mathbb{R}}\mathbf{1}_{B'}(y)\,g(y)\Bigl(\int_{\mathbb{R}}\mathbf{1}_{B}(x)\,g(x)\,d\lambda(x)\Bigr)d\lambda(y)=N(B)\,N(B').$$

Since $\rho$ and $N\otimes N$ are two measures on $\mathcal{B}_2$ assigning $N(B)N(B')$ to every measurable rectangle, the uniqueness assertion of Existence and Uniqueness of the Product Measure yields $\rho=N\otimes N$. Hence

$$P(V\in C)=c^{-2}\int_{\mathbb{R}^{2}}\mathbf{1}_{C}\,G\,d(\lambda\otimes\lambda)\qquad(C\in\mathcal{B}_2). \tag{1}$$

Step 2: Shears preserve the integral. For a real $\alpha$ define the shears $S_\alpha(x,y)=(x+\alpha y,\;y)$ and $S'_\alpha(x,y)=(x,\;y+\alpha x)$; by Step 0 both are measurable from $(\mathbb{R}^{2},\mathcal{B}_2)$ to itself. We claim that for every $\mathcal{B}_2$-measurable $f:\mathbb{R}^{2}\to\mathbb{R}$ with $f\ge0$,

$$\int_{\mathbb{R}^{2}}f\circ S_\alpha\,d(\lambda\otimes\lambda)=\int_{\mathbb{R}^{2}}f\,d(\lambda\otimes\lambda),$$

and likewise for $S'_\alpha$. Indeed, $f\circ S_\alpha$ is $\mathcal{B}_2$-measurable as a composition, and by the Tonelli theorem, integrating first in $x$,

$$\int_{\mathbb{R}^{2}}f\circ S_\alpha\,d(\lambda\otimes\lambda)=\int_{\mathbb{R}}\Bigl(\int_{\mathbb{R}}f(x+\alpha y,\,y)\,d\lambda(x)\Bigr)d\lambda(y).$$

For each fixed $y$, the inner integrand is the translate by $t=\alpha y$ of the section $x\mapsto f(x,y)$, which is measurable by the section assertion of Tonelli and Fubini Theorems; Claim 2 of Translation Invariance of Lebesgue Measure and the Lebesgue Integral shows the inner integral equals $\int_{\mathbb{R}}f(x,y)\,d\lambda(x)$. Applying the Tonelli theorem once more gives the claim. The claim for $S'_\alpha$ is proved symmetrically, integrating first in $y$ and using the other iterated-integral identity of Tonelli and Fubini Theorems.

Step 3: Factorization of the rotation into shears. Let $R(x,y)=(a\,x+b\,y,\;-b\,x+a\,y)$; its coordinates are jointly Borel by Step 0 (sums of the jointly Borel functions $t\circ\pi_i$ are jointly Borel, again by the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product on $(\mathbb{R}^{2},\mathcal{B}_2)$), so $R$ is measurable. Suppose first $b\ne0$, and set

$$\alpha=\gamma=\frac{1-a}{b},\qquad \beta=-b .$$

Then $R=S_\alpha\circ S'_\beta\circ S_\gamma$. Indeed, applying $S_\gamma$, then $S'_\beta$, then $S_\alpha$ to $(x,y)$ yields the pair with second coordinate $\beta x+(1+\beta\gamma)y$ and first coordinate $(1+\alpha\beta)x+\bigl(\gamma+\alpha(1+\beta\gamma)\bigr)y$; here $1+\alpha\beta=1+\beta\gamma=1-(1-a)=a$, and

$$\gamma+\alpha\,a=\frac{(1-a)(1+a)}{b}=\frac{1-a^{2}}{b}=\frac{b^{2}}{b}=b$$

using $a^{2}+b^{2}=1$, so the pair is exactly $(a\,x+b\,y,\;-b\,x+a\,y)$. Consequently, for $\mathcal{B}_2$-measurable $f\ge0$, writing $f\circ R=\bigl((f\circ S_\alpha)\circ S'_\beta\bigr)\circ S_\gamma$ and applying Step 2 three times,

$$\int_{\mathbb{R}^{2}}f\circ R\,d(\lambda\otimes\lambda)=\int_{\mathbb{R}^{2}}f\,d(\lambda\otimes\lambda). \tag{2}$$

Step 4: The density is rotation invariant. By the multiplicative property of the exponential function (Basic Properties of the Exponential Function), $G(x,y)=\exp\bigl(-(x^{2}+y^{2})/2\bigr)$, and

$$(a\,x+b\,y)^{2}+(-b\,x+a\,y)^{2}=(a^{2}+b^{2})(x^{2}+y^{2})=x^{2}+y^{2},$$

since the cross terms $2ab\,xy$ and $-2ab\,xy$ cancel. Hence $G\circ R=G$ pointwise on $\mathbb{R}^{2}$.

Step 5: Conclusion for $b\ne0$. The functions $W_1,W_2$ are random variables by the closure preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product, and $(W_1,W_2)=R\circ V$ pointwise on $\Omega$. Fix Borel sets $B,B'$ and put $C=R^{-1}(B\times B')\in\mathcal{B}_2$. Then $\{W_1\in B\}\cap\{W_2\in B'\}=V^{-1}(C)$, and $\mathbf{1}_{C}=\mathbf{1}_{B\times B'}\circ R$, so by Step 4, $\mathbf{1}_{C}\,G=(\mathbf{1}_{B\times B'}\,G)\circ R$. Combining the identity (1) of Step 1, the invariance (2) of Step 3 applied to $f=\mathbf{1}_{B\times B'}\,G$, and Step 1 again,

$$P(W_1\in B,\;W_2\in B')=c^{-2}\int_{\mathbb{R}^{2}}(\mathbf{1}_{B\times B'}\,G)\circ R\,d(\lambda\otimes\lambda)=c^{-2}\int_{\mathbb{R}^{2}}\mathbf{1}_{B\times B'}\,G\,d(\lambda\otimes\lambda)=N(B)\,N(B').$$

Taking $B'=\mathbb{R}$ gives $P(W_1\in B)=N(B)$ for every Borel $B$, so $W_1$ is standard normal; taking $B=\mathbb{R}$ gives the same for $W_2$. The displayed identity then reads $P(W_1\in B,\;W_2\in B')=P(W_1\in B)\,P(W_2\in B')$ for all Borel $B,B'$, which is independence of $W_1$ and $W_2$.

Step 6: The case $b=0$. Then $a^{2}=1$. If $a=1$, then $(W_1,W_2)=(Z_1,Z_2)$ and there is nothing to prove. If $a=-1$, then $(W_1,W_2)=(-Z_1,-Z_2)$: each is standard normal by Claim 3 of Reflection Invariance of Lebesgue Measure and Symmetry of the Standard Normal Distribution, and for Borel $B,B'$ the events $\{-Z_1\in B\}=\{Z_1\in-B\}$ and $\{-Z_2\in B'\}=\{Z_2\in-B'\}$ involve the Borel sets $-B,-B'$ by Claim 1 of Reflection Invariance of Lebesgue Measure and Symmetry of the Standard Normal Distribution, so the product identity follows from the independence of $Z_1$ and $Z_2$. $\blacksquare$