[proofplan] We first reduce both rank correlations to probabilities involving signs of centered normal variables. The common analytic input is the bivariate normal orthant formula: if two centered standard normal variables have correlation $\rho$, then the probability that both are positive is $\frac14+\frac{1}{2\pi}\arcsin(\rho)$. Kendall's $\tau$ follows by applying this formula to the difference of two independent copies. Spearman's $\rho_S$ follows by standardising the margins, writing $F_X(X)$ and $F_Y(Y)$ as standard normal distribution functions, and introducing independent threshold variables to convert the expectation into another bivariate normal orthant probability. Throughout the proof, $\mathcal{L}^1$ denotes Lebesgue measure on $\mathbb{R}$ and $\mathcal{L}^2$ denotes Lebesgue measure on $\mathbb{R}^2$. [/proofplan]

[step:Establish the bivariate normal orthant formula]We prove the auxiliary formula used twice below. Let $(U,V)$ be a centered bivariate normal random vector with $\mathbb{E}[U]=\mathbb{E}[V]=0$, $\operatorname{Var}(U)=\operatorname{Var}(V)=1$, and correlation $\rho \in [-1,1]$. Then \begin{align*} \mathbb{P}(U>0,\ V>0)=\frac14+\frac{1}{2\pi}\arcsin(\rho). \end{align*} For $|\rho|<1$, the joint density $f_\rho:\mathbb{R}^2\to[0,\infty)$ of $(U,V)$ with respect to $\mathcal{L}^2$ is \begin{align*} f_\rho(u,v) = \frac{1}{2\pi\sqrt{1-\rho^2}} \exp\left( -\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)} \right). \end{align*} Define $Q:(-1,1)\to[0,1]$ by \begin{align*} Q(\rho):=\int_0^\infty\int_0^\infty f_\rho(u,v)\,d\mathcal{L}^1(v)\,d\mathcal{L}^1(u). \end{align*} A direct differentiation of $f_\rho$ gives \begin{align*} \frac{\partial f_\rho}{\partial \rho}(u,v) = \frac{\partial^2 f_\rho}{\partial u\,\partial v}(u,v). \end{align*} Since $f_\rho$ and its first derivatives decay exponentially on every compact subinterval of $\rho\in(-1,1)$, differentiating under the integral sign and integrating by parts on $(0,\infty)^2$ are justified. Hence \begin{align*} Q'(\rho) &= \int_0^\infty\int_0^\infty \frac{\partial^2 f_\rho}{\partial u\,\partial v}(u,v) \,d\mathcal{L}^1(v)\,d\mathcal{L}^1(u)\\ &= f_\rho(0,0)\\ &= \frac{1}{2\pi\sqrt{1-\rho^2}}. \end{align*} At $\rho=0$, the variables $U$ and $V$ are independent standard normal variables, so \begin{align*} Q(0)=\mathbb{P}(U>0)\mathbb{P}(V>0)=\frac12\cdot\frac12=\frac14. \end{align*} Therefore, for $|\rho|<1$, \begin{align*} Q(\rho) &= Q(0)+\int_0^\rho Q'(s)\,d\mathcal{L}^1(s)\\ &= \frac14+\int_0^\rho \frac{1}{2\pi\sqrt{1-s^2}}\,d\mathcal{L}^1(s)\\ &= \frac14+\frac{1}{2\pi}\arcsin(\rho). \end{align*} For $\rho=1$, we have $V=U$ almost surely, hence $\mathbb{P}(U>0,V>0)=\mathbb{P}(U>0)=\frac12$. For $\rho=-1$, we have $V=-U$ almost surely, hence $\mathbb{P}(U>0,V>0)=0$. These agree with the same formula at $\rho=\pm1$.[/step]

[guided]The orthant formula is the only analytic computation in the proof. We want to compute the probability that a centered bivariate normal vector falls in the first quadrant. Assume first that $|\rho|<1$, so the joint distribution has a density. The density is the map \begin{align*} f_\rho:\mathbb{R}^2&\to[0,\infty)\\ (u,v)&\mapsto \frac{1}{2\pi\sqrt{1-\rho^2}} \exp\left( -\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)} \right). \end{align*} Define the quadrant probability as the function \begin{align*} Q:(-1,1)&\to[0,1]\\ \rho&\mapsto \int_0^\infty\int_0^\infty f_\rho(u,v)\,d\mathcal{L}^1(v)\,d\mathcal{L}^1(u). \end{align*} The key identity is \begin{align*} \frac{\partial f_\rho}{\partial \rho}(u,v) = \frac{\partial^2 f_\rho}{\partial u\,\partial v}(u,v). \end{align*} This identity is obtained by differentiating the displayed density with respect to $\rho$, $u$, and $v$ and comparing the resulting expressions. Because the Gaussian exponential gives uniform exponential decay on compact subintervals of $\rho\in(-1,1)$, we may differentiate under the integral sign. Thus \begin{align*} Q'(\rho) = \int_0^\infty\int_0^\infty \frac{\partial f_\rho}{\partial \rho}(u,v) \,d\mathcal{L}^1(v)\,d\mathcal{L}^1(u). \end{align*} Substituting the derivative identity gives \begin{align*} Q'(\rho) = \int_0^\infty\int_0^\infty \frac{\partial^2 f_\rho}{\partial u\,\partial v}(u,v) \,d\mathcal{L}^1(v)\,d\mathcal{L}^1(u). \end{align*} Integrating first in $v$ and then in $u$, the boundary terms at infinity vanish because the density and its derivatives decay exponentially. The only remaining boundary value is the value at $(0,0)$: \begin{align*} Q'(\rho) = f_\rho(0,0) = \frac{1}{2\pi\sqrt{1-\rho^2}}. \end{align*} At $\rho=0$, the covariance matrix is the identity matrix, so $U$ and $V$ are independent standard normal variables. Hence \begin{align*} Q(0)=\mathbb{P}(U>0)\mathbb{P}(V>0)=\frac12\cdot\frac12=\frac14. \end{align*} Integrating the derivative from $0$ to $\rho$ gives \begin{align*} Q(\rho) &= Q(0)+\int_0^\rho Q'(s)\,d\mathcal{L}^1(s)\\ &= \frac14+\int_0^\rho \frac{1}{2\pi\sqrt{1-s^2}}\,d\mathcal{L}^1(s)\\ &= \frac14+\frac{1}{2\pi}\arcsin(\rho). \end{align*} The endpoint cases are singular but immediate. If $\rho=1$, then $V=U$ almost surely, so the first-quadrant event is just $\{U>0\}$ and has probability $\frac12$. If $\rho=-1$, then $V=-U$ almost surely, so the event $\{U>0,V>0\}$ has probability $0$. These are exactly the values of $\frac14+\frac{1}{2\pi}\arcsin(\rho)$ at $\rho=1$ and $\rho=-1$.[/guided]

[step:Reduce Kendall's $\tau$ to a centered normal orthant probability]Let $(X_1,Y_1)$ and $(X_2,Y_2)$ be independent copies of $(X,Y)$. Define the difference variables \begin{align*} D_X&:=X_1-X_2,\\ D_Y&:=Y_1-Y_2. \end{align*} Since independent linear combinations of jointly normal random variables are jointly normal, $(D_X,D_Y)$ is a centered bivariate normal random vector. Moreover, \begin{align*} \operatorname{Var}(D_X)&=\operatorname{Var}(X_1)+\operatorname{Var}(X_2)=2\operatorname{Var}(X),\\ \operatorname{Var}(D_Y)&=\operatorname{Var}(Y_1)+\operatorname{Var}(Y_2)=2\operatorname{Var}(Y),\\ \operatorname{Cov}(D_X,D_Y) &= \operatorname{Cov}(X_1,Y_1)+\operatorname{Cov}(X_2,Y_2)\\ &= 2\operatorname{Cov}(X,Y). \end{align*} Therefore the correlation of $D_X$ and $D_Y$ is \begin{align*} \frac{\operatorname{Cov}(D_X,D_Y)} {\sqrt{\operatorname{Var}(D_X)\operatorname{Var}(D_Y)}} = \frac{2\operatorname{Cov}(X,Y)} {2\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}} = r. \end{align*} Because $(D_X,D_Y)$ has continuous marginals unless $r=\pm1$, ties have probability zero for $|r|<1$. In the endpoint cases $r=\pm1$, the conclusion follows directly from $D_Y=\pm cD_X$ almost surely for some $c>0$. Thus it remains to consider $|r|<1$. Let \begin{align*} U:=\frac{D_X}{\sqrt{\operatorname{Var}(D_X)}}, \qquad V:=\frac{D_Y}{\sqrt{\operatorname{Var}(D_Y)}}. \end{align*} Then $(U,V)$ is centered standard bivariate normal with correlation $r$. Since multiplication by positive constants preserves signs, \begin{align*} \mathbb{P}(D_X>0,D_Y>0)=\mathbb{P}(U>0,V>0). \end{align*} By symmetry of the centered bivariate normal distribution under $(U,V)\mapsto(-U,-V)$, \begin{align*} \mathbb{P}(D_X<0,D_Y<0)=\mathbb{P}(D_X>0,D_Y>0). \end{align*} Thus \begin{align*} \mathbb{P}(D_XD_Y>0)=2\mathbb{P}(U>0,V>0). \end{align*} Since $\mathbb{P}(D_XD_Y=0)=0$ for $|r|<1$, \begin{align*} \tau &= \mathbb{E}[\operatorname{sgn}(D_XD_Y)]\\ &= \mathbb{P}(D_XD_Y>0)-\mathbb{P}(D_XD_Y<0)\\ &= 2\mathbb{P}(D_XD_Y>0)-1\\ &= 4\mathbb{P}(U>0,V>0)-1. \end{align*} Applying the orthant formula with $\rho=r$ gives \begin{align*} \tau = 4\left(\frac14+\frac{1}{2\pi}\arcsin(r)\right)-1 = \frac{2}{\pi}\arcsin(r). \end{align*}[/step]

[guided]Kendall's $\tau$ compares the signs of the two coordinate differences in two independent observations. We therefore introduce the random variables \begin{align*} D_X&:=X_1-X_2,\\ D_Y&:=Y_1-Y_2. \end{align*} The event $D_XD_Y>0$ means the two observations are concordant, and the event $D_XD_Y<0$ means they are discordant. Because $(X_1,Y_1)$ and $(X_2,Y_2)$ are independent copies of a bivariate normal vector, every linear combination of their coordinates is normal. Hence $(D_X,D_Y)$ is bivariate normal. Its mean is zero, and its covariance data are \begin{align*} \operatorname{Var}(D_X)&=\operatorname{Var}(X_1)+\operatorname{Var}(X_2)=2\operatorname{Var}(X),\\ \operatorname{Var}(D_Y)&=\operatorname{Var}(Y_1)+\operatorname{Var}(Y_2)=2\operatorname{Var}(Y),\\ \operatorname{Cov}(D_X,D_Y) &= \operatorname{Cov}(X_1-X_2,Y_1-Y_2)\\ &= \operatorname{Cov}(X_1,Y_1)+\operatorname{Cov}(X_2,Y_2)\\ &= 2\operatorname{Cov}(X,Y). \end{align*} The mixed covariance terms vanish because the two copies are independent. Therefore the correlation of $D_X$ and $D_Y$ is \begin{align*} \frac{\operatorname{Cov}(D_X,D_Y)} {\sqrt{\operatorname{Var}(D_X)\operatorname{Var}(D_Y)}} = \frac{2\operatorname{Cov}(X,Y)} {2\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}} = r. \end{align*} Now standardise the two difference variables by defining \begin{align*} U:=\frac{D_X}{\sqrt{\operatorname{Var}(D_X)}}, \qquad V:=\frac{D_Y}{\sqrt{\operatorname{Var}(D_Y)}}. \end{align*} The constants in the denominators are positive, so $U$ and $D_X$ have the same sign, and $V$ and $D_Y$ have the same sign. Thus \begin{align*} \mathbb{P}(D_X>0,D_Y>0)=\mathbb{P}(U>0,V>0). \end{align*} For $|r|<1$, the bivariate normal law is nondegenerate, so $\mathbb{P}(D_XD_Y=0)=0$. The centered normal law is invariant under the map $(U,V)\mapsto(-U,-V)$, so the two concordant quadrants have the same probability: \begin{align*} \mathbb{P}(D_X<0,D_Y<0)=\mathbb{P}(D_X>0,D_Y>0). \end{align*} Therefore \begin{align*} \mathbb{P}(D_XD_Y>0)=2\mathbb{P}(U>0,V>0). \end{align*} Using the definition of Kendall's $\tau$, \begin{align*} \tau &= \mathbb{E}[\operatorname{sgn}(D_XD_Y)]\\ &= \mathbb{P}(D_XD_Y>0)-\mathbb{P}(D_XD_Y<0)\\ &= 2\mathbb{P}(D_XD_Y>0)-1\\ &= 4\mathbb{P}(U>0,V>0)-1. \end{align*} The orthant formula applies to $(U,V)$ because it is centered standard bivariate normal with correlation $r$. Hence \begin{align*} \tau = 4\left(\frac14+\frac{1}{2\pi}\arcsin(r)\right)-1 = \frac{2}{\pi}\arcsin(r). \end{align*} If $r=1$, then $Y$ is an increasing affine function of $X$ almost surely, so $D_Y=cD_X$ almost surely for some $c>0$ and $\tau=1$. If $r=-1$, then $D_Y=-cD_X$ almost surely for some $c>0$ and $\tau=-1$. These values agree with $\frac{2}{\pi}\arcsin(r)$ at $r=\pm1$.[/guided]

[step:Represent Spearman's $\rho_S$ using standard normal distribution functions] Let $\mu_X:=\mathbb{E}[X]$, $\mu_Y:=\mathbb{E}[Y]$, $\sigma_X:=\sqrt{\operatorname{Var}(X)}$, and $\sigma_Y:=\sqrt{\operatorname{Var}(Y)}$. Define the standardised variables \begin{align*} \widetilde{X}:=\frac{X-\mu_X}{\sigma_X}, \qquad \widetilde{Y}:=\frac{Y-\mu_Y}{\sigma_Y}. \end{align*} Then $(\widetilde{X},\widetilde{Y})$ is centered standard bivariate normal with correlation $r$. Let $\Phi:\mathbb{R}\to[0,1]$ denote the standard normal distribution function. Since the marginal laws of $X$ and $Y$ are normal, \begin{align*} F_X(X)=\Phi(\widetilde{X}), \qquad F_Y(Y)=\Phi(\widetilde{Y}) \end{align*} almost surely. The random variables $\Phi(\widetilde{X})$ and $\Phi(\widetilde{Y})$ are uniform on $[0,1]$, hence each has expectation $\frac12$ and variance $\frac{1}{12}$. Therefore \begin{align*} \rho_S &= \operatorname{Corr}(\Phi(\widetilde{X}),\Phi(\widetilde{Y}))\\ &= \frac{\mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})]-\frac14}{\frac{1}{12}}\\ &= 12\mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})]-3. \end{align*} [/step]

[step:Convert the Spearman expectation into an orthant probability]Let $Z_1$ and $Z_2$ be independent standard normal random variables, independent of $(\widetilde{X},\widetilde{Y})$. Define \begin{align*} A:=\widetilde{X}-Z_1, \qquad B:=\widetilde{Y}-Z_2. \end{align*} Conditioning on $(\widetilde{X},\widetilde{Y})$ gives \begin{align*} \mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})] &= \mathbb{P}(Z_1\leq \widetilde{X},\ Z_2\leq \widetilde{Y})\\ &= \mathbb{P}(A\geq 0,\ B\geq 0). \end{align*} The vector $(A,B)$ is centered bivariate normal. Its variances and covariance are \begin{align*} \operatorname{Var}(A)&=\operatorname{Var}(\widetilde{X})+\operatorname{Var}(Z_1)=2,\\ \operatorname{Var}(B)&=\operatorname{Var}(\widetilde{Y})+\operatorname{Var}(Z_2)=2,\\ \operatorname{Cov}(A,B)&=\operatorname{Cov}(\widetilde{X},\widetilde{Y})=r. \end{align*} Hence the correlation of $A$ and $B$ is \begin{align*} \frac{\operatorname{Cov}(A,B)} {\sqrt{\operatorname{Var}(A)\operatorname{Var}(B)}} = \frac{r}{2}. \end{align*} Since $(A,B)$ has continuous marginals, the events $\{A\geq0,B\geq0\}$ and $\{A>0,B>0\}$ have the same probability. Applying the orthant formula with $\rho=r/2$ gives \begin{align*} \mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})] = \frac14+\frac{1}{2\pi}\arcsin\left(\frac{r}{2}\right). \end{align*}[/step]

[guided]The expectation $\mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})]$ is converted into a probability by adding independent threshold variables. Let $Z_1$ and $Z_2$ be independent standard normal random variables, independent of $(\widetilde{X},\widetilde{Y})$, and define \begin{align*} A:=\widetilde{X}-Z_1, \qquad B:=\widetilde{Y}-Z_2. \end{align*} Condition on the values of $\widetilde{X}$ and $\widetilde{Y}$. Since $Z_1$ and $Z_2$ are independent standard normal variables, \begin{align*} \mathbb{P}(Z_1\leq \widetilde{X},\ Z_2\leq \widetilde{Y}\mid \widetilde{X},\widetilde{Y}) = \Phi(\widetilde{X})\Phi(\widetilde{Y}). \end{align*} Taking expectations gives \begin{align*} \mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})] = \mathbb{P}(Z_1\leq \widetilde{X},\ Z_2\leq \widetilde{Y}) = \mathbb{P}(A\geq0,\ B\geq0). \end{align*} Now compute the distribution of $(A,B)$. Since $A$ and $B$ are linear combinations of jointly normal variables, $(A,B)$ is bivariate normal. It is centered because each of $\widetilde{X},\widetilde{Y},Z_1,Z_2$ has mean zero. Its variances are \begin{align*} \operatorname{Var}(A)&=\operatorname{Var}(\widetilde{X})+\operatorname{Var}(Z_1)=1+1=2,\\ \operatorname{Var}(B)&=\operatorname{Var}(\widetilde{Y})+\operatorname{Var}(Z_2)=1+1=2. \end{align*} The cross terms vanish because $Z_1$ and $Z_2$ are independent of $(\widetilde{X},\widetilde{Y})$ and of each other. The covariance is \begin{align*} \operatorname{Cov}(A,B) &= \operatorname{Cov}(\widetilde{X}-Z_1,\widetilde{Y}-Z_2)\\ &= \operatorname{Cov}(\widetilde{X},\widetilde{Y})\\ &= r. \end{align*} Therefore the correlation coefficient of $A$ and $B$ is \begin{align*} \frac{\operatorname{Cov}(A,B)} {\sqrt{\operatorname{Var}(A)\operatorname{Var}(B)}} = \frac{r}{2}. \end{align*} Because $(A,B)$ has continuous marginals, replacing $\geq0$ by $>0$ does not change the probability. The orthant formula applies to the standardised version of $(A,B)$, whose correlation is still $r/2$. Hence \begin{align*} \mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})] = \mathbb{P}(A\geq0,\ B\geq0) = \frac14+\frac{1}{2\pi}\arcsin\left(\frac{r}{2}\right). \end{align*}[/guided]

[step:Substitute the orthant probability into Spearman's formula] From the previous two steps, \begin{align*} \rho_S &= 12\mathbb{E}[\Phi(\widetilde{X})\Phi(\widetilde{Y})]-3\\ &= 12\left( \frac14+\frac{1}{2\pi}\arcsin\left(\frac{r}{2}\right) \right)-3\\ &= \frac{6}{\pi}\arcsin\left(\frac{r}{2}\right). \end{align*} Together with the Kendall computation, this proves both displayed identities. [/step]