[proofplan] We first rewrite tail dependence in terms of exceedances of the Student $t$ margins. The key computation is the conditional distribution of $T_2$ given $T_1=x$, which is again Student $t$, now with $\nu+1$ degrees of freedom, location $rx$, and a scale depending on $x$. We then pass to the upper-tail limit by conditioning on $T_1>x$ and using the regular variation of the univariate Student $t$ tail; the resulting one-dimensional limit evaluates to the stated Student $t_{\nu+1}$ distribution value. Finally, central symmetry of the bivariate Student $t$ law identifies the lower tail coefficient with the upper one. [/proofplan] [step:Rewrite the copula tail coefficients as Student $t$ exceedance limits] Let $F_\nu:\mathbb{R}\to(0,1)$ denote the continuous distribution function of a univariate Student $t$ random variable with $\nu$ degrees of freedom. Since each marginal of $(T_1,T_2)$ has distribution function $F_\nu$, the random variables \begin{align*} U_1 &:= F_\nu(T_1),& U_2 &:= F_\nu(T_2) \end{align*} are uniformly distributed on $(0,1)$, and $C_{\nu,r}$ is the distribution function of $(U_1,U_2)$. For $u\in(0,1)$, define the quantile \begin{align*} x_u := F_\nu^{-1}(u). \end{align*} As $u\uparrow 1$, $x_u\to\infty$, and therefore \begin{align*} \mathbb{P}(U_2>u\mid U_1>u) = \mathbb{P}(T_2>x_u\mid T_1>x_u). \end{align*} Hence \begin{align*} \lambda_U = \lim_{x\to\infty}\mathbb{P}(T_2>x\mid T_1>x), \end{align*} provided the limit exists. Similarly, as $u\downarrow0$, $x_u\to-\infty$, and \begin{align*} \lambda_L = \lim_{x\to-\infty}\mathbb{P}(T_2\leq x\mid T_1\leq x), \end{align*} provided the limit exists. [/step] [step:Compute the conditional Student distribution of $T_2$ given $T_1=x$] Let $f_{\nu,r}:\mathbb{R}^2\to(0,\infty)$ denote the joint density of $(T_1,T_2)$, and let $f_\nu:\mathbb{R}\to(0,\infty)$ denote the univariate Student $t_\nu$ density. They are \begin{align*} f_{\nu,r}(x,y) &= \frac{\Gamma\left(\frac{\nu+2}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\nu\pi\sqrt{1-r^2}} \left( 1+ \frac{x^2-2rxy+y^2}{\nu(1-r^2)} \right)^{-(\nu+2)/2},\\ f_\nu(x) &= \frac{\Gamma\left(\frac{\nu+1}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\sqrt{\nu\pi}} \left(1+\frac{x^2}{\nu}\right)^{-(\nu+1)/2}. \end{align*} For each fixed $x\in\mathbb{R}$, the conditional density \begin{align*} f_{2\mid1}(\,\cdot\,\mid x):\mathbb{R}\to(0,\infty) \end{align*} of $T_2$ given $T_1=x$ is defined by \begin{align*} f_{2\mid1}(y\mid x) := \frac{f_{\nu,r}(x,y)}{f_\nu(x)}. \end{align*} Complete the square in the quadratic form: \begin{align*} x^2-2rxy+y^2 = (1-r^2)x^2+(y-rx)^2. \end{align*} Substituting this identity into the ratio $f_{\nu,r}(x,y)/f_\nu(x)$ gives \begin{align*} f_{2\mid1}(y\mid x) = \frac{1}{s_x} \frac{\Gamma\left(\frac{\nu+2}{2}\right)} {\Gamma\left(\frac{\nu+1}{2}\right)\sqrt{(\nu+1)\pi}} \left( 1+ \frac{1}{\nu+1} \left(\frac{y-rx}{s_x}\right)^2 \right)^{-(\nu+2)/2}, \end{align*} where the scale parameter $s_x>0$ is \begin{align*} s_x := \sqrt{\frac{(\nu+x^2)(1-r^2)}{\nu+1}}. \end{align*} Thus, for every $x\in\mathbb{R}$, \begin{align*} \frac{T_2-rx}{s_x}\,\Big|\,\{T_1=x\} \end{align*} has the univariate Student $t$ distribution with $\nu+1$ degrees of freedom. Consequently, for every $a\in\mathbb{R}$ and every $x\in\mathbb{R}$, \begin{align*} \mathbb{P}(T_2>a\mid T_1=x) = 1- F_{\nu+1}\left(\frac{a-rx}{s_x}\right). \end{align*} [guided] We need an explicit conditional law because the event $T_2>x$ is easiest to analyse after $T_1$ has been fixed. Let $f_{\nu,r}:\mathbb{R}^2\to(0,\infty)$ be the joint density of the centred bivariate Student $t$ vector with correlation parameter $r$, and let $f_\nu:\mathbb{R}\to(0,\infty)$ be the marginal Student $t_\nu$ density: \begin{align*} f_{\nu,r}(x,y) &= \frac{\Gamma\left(\frac{\nu+2}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\nu\pi\sqrt{1-r^2}} \left( 1+ \frac{x^2-2rxy+y^2}{\nu(1-r^2)} \right)^{-(\nu+2)/2},\\ f_\nu(x) &= \frac{\Gamma\left(\frac{\nu+1}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\sqrt{\nu\pi}} \left(1+\frac{x^2}{\nu}\right)^{-(\nu+1)/2}. \end{align*} Because $f_\nu(x)>0$ for every $x\in\mathbb{R}$, the conditional density of $T_2$ given $T_1=x$ is \begin{align*} f_{2\mid1}(y\mid x) := \frac{f_{\nu,r}(x,y)}{f_\nu(x)}. \end{align*} The algebraic point is that the dependence on $y$ appears through a completed square: \begin{align*} x^2-2rxy+y^2 = (1-r^2)x^2+(y-rx)^2. \end{align*} Substituting this into the joint density separates the part depending only on $x$ from the part depending on $y-rx$. After cancelling the marginal density $f_\nu(x)$, the conditional density becomes \begin{align*} f_{2\mid1}(y\mid x) = \frac{1}{s_x} \frac{\Gamma\left(\frac{\nu+2}{2}\right)} {\Gamma\left(\frac{\nu+1}{2}\right)\sqrt{(\nu+1)\pi}} \left( 1+ \frac{1}{\nu+1} \left(\frac{y-rx}{s_x}\right)^2 \right)^{-(\nu+2)/2}, \end{align*} where \begin{align*} s_x := \sqrt{\frac{(\nu+x^2)(1-r^2)}{\nu+1}}. \end{align*} This is exactly the density of a Student $t$ random variable with $\nu+1$ degrees of freedom, shifted by $rx$ and scaled by $s_x$. Therefore \begin{align*} \frac{T_2-rx}{s_x}\,\Big|\,\{T_1=x\} \end{align*} has distribution function $F_{\nu+1}$. Hence, for every threshold $a\in\mathbb{R}$, \begin{align*} \mathbb{P}(T_2>a\mid T_1=x) = 1- F_{\nu+1}\left(\frac{a-rx}{s_x}\right). \end{align*} This formula is the mechanism behind tail dependence: when $x$ is large, the conditional location $rx$ is large and the conditional scale $s_x$ is also proportional to $x$. [/guided] [/step] [step:Pass from conditional exceedances to the upper tail limit] Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. For $x>0$, define the conditional exceedance ratio \begin{align*} R_x := \frac{T_1}{x} \end{align*} as a random variable on the conditional probability space with probability measure $\mathbb{P}(\,\cdot\,\mid T_1>x)$. The Student $t_\nu$ density satisfies \begin{align*} f_\nu(t) &= \frac{\Gamma\left(\frac{\nu+1}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\sqrt{\nu\pi}} \left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2} \sim A_\nu t^{-\nu-1} \end{align*} as $t\to\infty$, where \begin{align*} A_\nu := \frac{\Gamma\left(\frac{\nu+1}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\sqrt{\nu\pi}} \nu^{(\nu+1)/2}. \end{align*} We record the tail integration argument explicitly. Since \begin{align*} \frac{f_\nu(t)}{A_\nu t^{-\nu-1}}\to1 \end{align*} as $t\to\infty$, for every $\varepsilon\in(0,1)$ there is $M_\varepsilon>0$ such that, for all $t\geq M_\varepsilon$, \begin{align*} (1-\varepsilon)A_\nu t^{-\nu-1} \leq f_\nu(t) \leq (1+\varepsilon)A_\nu t^{-\nu-1}. \end{align*} Integrating this two-sided estimate over $[x,\infty)$ with respect to $\mathcal{L}^1$ gives, for $x\geq M_\varepsilon$, \begin{align*} (1-\varepsilon)\frac{A_\nu}{\nu}x^{-\nu} \leq \mathbb{P}(T_1>x) \leq (1+\varepsilon)\frac{A_\nu}{\nu}x^{-\nu}. \end{align*} Letting $\varepsilon\downarrow0$ implies \begin{align*} \mathbb{P}(T_1>x) \sim \frac{A_\nu}{\nu}x^{-\nu}. \end{align*} Consequently, for every $z\geq1$, \begin{align*} \lim_{x\to\infty} \mathbb{P}(R_x>z) &= \lim_{x\to\infty} \frac{\mathbb{P}(T_1>xz)}{\mathbb{P}(T_1>x)} = z^{-\nu}. \end{align*} Thus $R_x$ converges in distribution on $[1,\infty)$ to a Pareto random variable $R$ with density \begin{align*} p_R:(1,\infty)&\to(0,\infty)\\ z&\mapsto \nu z^{-\nu-1}. \end{align*} Define, for $x>0$, the bounded measurable function \begin{align*} g_x:[1,\infty)&\to[0,1]\\ z&\mapsto 1- F_{\nu+1}\left( \frac{x-rxz} {\sqrt{\frac{(\nu+x^2z^2)(1-r^2)}{\nu+1}}} \right), \end{align*} and define \begin{align*} g:[1,\infty]&\to[0,1]\\ z&\mapsto 1- F_{\nu+1}\left( \sqrt{\nu+1}\, \frac{1-rz}{z\sqrt{1-r^2}} \right),\qquad z<\infty,\\ \infty&\mapsto 1-F_{\nu+1}\left(-r\sqrt{\frac{\nu+1}{1-r^2}}\right). \end{align*} The function $g$ is continuous on the compactified half-line $[1,\infty]$, and $g_x\to g$ uniformly there because \begin{align*} \frac{x-rxz} {\sqrt{\frac{(\nu+x^2z^2)(1-r^2)}{\nu+1}}} = \frac{1-rz} {\sqrt{\left(\frac{\nu}{x^2}+z^2\right)\frac{1-r^2}{\nu+1}}} \end{align*} converges uniformly on $[1,\infty]$ to \begin{align*} \sqrt{\nu+1}\, \frac{1-rz}{z\sqrt{1-r^2}}, \end{align*} with the displayed finite value at $z=\infty$. Using the conditional law from the previous step with threshold $a=x$ and conditioning on $T_1$, we obtain \begin{align*} \mathbb{P}(T_2>x\mid T_1>x) = \mathbb{E}\left[g_x(R_x)\mid T_1>x\right]. \end{align*} The [weak convergence](/page/Weak%20Convergence) $R_x\Rightarrow R$ and the [uniform convergence](/page/Uniform%20Convergence) $g_x\to g$ give the limit as follows: the uniform error satisfies \begin{align*} \left| \mathbb{E}\left[g_x(R_x)\mid T_1>x\right] - \mathbb{E}\left[g(R_x)\mid T_1>x\right] \right| \leq \sup_{z\in[1,\infty]}|g_x(z)-g(z)|\to0, \end{align*} and, because $g$ is bounded and continuous on the compactified half-line, weak convergence gives $\mathbb{E}[g(R_x)\mid T_1>x]\to\mathbb{E}[g(R)]$. Hence \begin{align*} \lambda_U &= \int_1^\infty \left[ 1- F_{\nu+1}\left( \sqrt{\nu+1}\, \frac{1-rz}{z\sqrt{1-r^2}} \right) \right] \nu z^{-\nu-1}\,d\mathcal{L}^1(z). \end{align*} It remains to evaluate this integral. Put \begin{align*} m&:=\nu+1, & q&:=\sqrt{\frac{m(1-r)}{1+r}}, & a&:=\frac{\sqrt{m}}{\sqrt{1-r^2}}, \end{align*} let $F_m:\mathbb{R}\to(0,1)$ denote the Student $t_m$ distribution function, and let $f_m:\mathbb{R}\to(0,\infty)$ denote the Student $t_m$ density. With the substitution $y=1/z$, for which $z=1/y$ and $\nu z^{-\nu-1}\,d\mathcal{L}^1(z)=-\nu y^{\nu-1}\,d\mathcal{L}^1(y)$, the integral becomes \begin{align*} I &:= \nu\int_0^1 \left[1-F_m\left(a(y-r)\right)\right] y^{\nu-1}\,d\mathcal{L}^1(y). \end{align*} Integrating by parts with $u(y)=1-F_m(a(y-r))$ and $dv=\nu y^{\nu-1}\,d\mathcal{L}^1(y)$ gives \begin{align*} I &= 1-F_m(q) + a\int_0^1 y^\nu f_m(a(y-r))\,d\mathcal{L}^1(y). \end{align*} The remaining integral equals $1-F_m(q)$. Indeed, \begin{align*} a\int_0^1 y^\nu f_m(a(y-r))\,d\mathcal{L}^1(y) &= \sqrt{m}\,c_m(1-r^2)^{(\nu+1)/2} \int_0^1 \frac{y^\nu}{(y^2-2ry+1)^{(\nu+2)/2}}\,d\mathcal{L}^1(y), \end{align*} where \begin{align*} c_m := \frac{\Gamma\left(\frac{m+1}{2}\right)} {\Gamma\left(\frac{m}{2}\right)\sqrt{m\pi}}. \end{align*} On the other hand, using the substitution $s=a(t-r)$ in the Student tail integral and then $t=1/y$ gives \begin{align*} 1-F_m(q) &= \int_q^\infty f_m(s)\,d\mathcal{L}^1(s)\\ &= \sqrt{m}\,c_m(1-r^2)^{(\nu+1)/2} \int_1^\infty \frac{1}{(t^2-2rt+1)^{(\nu+2)/2}}\,d\mathcal{L}^1(t)\\ &= \sqrt{m}\,c_m(1-r^2)^{(\nu+1)/2} \int_0^1 \frac{y^\nu}{(y^2-2ry+1)^{(\nu+2)/2}}\,d\mathcal{L}^1(y). \end{align*} Therefore $I=2(1-F_m(q))$. Since the Student $t_m$ distribution is symmetric about $0$, $1-F_m(q)=F_m(-q)$, and hence \begin{align*} \lambda_U = 2F_{\nu+1}\left( -\sqrt{\frac{(\nu+1)(1-r)}{1+r}} \right). \end{align*} [guided] The upper tail coefficient is not the same as the pointwise conditional probability $\mathbb{P}(T_2>x\mid T_1=x)$; it is the conditional probability given the whole event $\{T_1>x\}$. We therefore introduce the scaled exceedance \begin{align*} R_x := \frac{T_1}{x} \end{align*} as a random variable under the conditional probability measure $\mathbb{P}(\,\cdot\,\mid T_1>x)$. First we justify the limiting distribution of $R_x$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. The Student $t_\nu$ density has the asymptotic form \begin{align*} f_\nu(t) &= \frac{\Gamma\left(\frac{\nu+1}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\sqrt{\nu\pi}} \left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2} \sim A_\nu t^{-\nu-1} \end{align*} as $t\to\infty$, where \begin{align*} A_\nu := \frac{\Gamma\left(\frac{\nu+1}{2}\right)} {\Gamma\left(\frac{\nu}{2}\right)\sqrt{\nu\pi}} \nu^{(\nu+1)/2}. \end{align*} We now integrate this tail asymptotic directly. Since \begin{align*} \frac{f_\nu(t)}{A_\nu t^{-\nu-1}}\to1, \end{align*} for every $\varepsilon\in(0,1)$ there exists $M_\varepsilon>0$ such that, whenever $t\geq M_\varepsilon$, \begin{align*} (1-\varepsilon)A_\nu t^{-\nu-1} \leq f_\nu(t) \leq (1+\varepsilon)A_\nu t^{-\nu-1}. \end{align*} For $x\geq M_\varepsilon$, integrating over $[x,\infty)$ with respect to $\mathcal{L}^1$ gives \begin{align*} (1-\varepsilon)\frac{A_\nu}{\nu}x^{-\nu} \leq \mathbb{P}(T_1>x) = \int_x^\infty f_\nu(t)\,d\mathcal{L}^1(t) \leq (1+\varepsilon)\frac{A_\nu}{\nu}x^{-\nu}. \end{align*} Letting $\varepsilon\downarrow0$ proves \begin{align*} \mathbb{P}(T_1>x) \sim \frac{A_\nu}{\nu}x^{-\nu}. \end{align*} Therefore, for each $z\geq1$, \begin{align*} \lim_{x\to\infty} \mathbb{P}(R_x>z) &= \lim_{x\to\infty} \frac{\mathbb{P}(T_1>xz)}{\mathbb{P}(T_1>x)} = z^{-\nu}. \end{align*} This is exactly the survival function of a Pareto random variable $R$ supported on $[1,\infty)$ with density \begin{align*} p_R:(1,\infty)&\to(0,\infty)\\ z&\mapsto \nu z^{-\nu-1}. \end{align*} Thus $R_x\Rightarrow R$. Now we average the conditional Student law over the exceedance $R_x$. Define \begin{align*} g_x:[1,\infty)&\to[0,1]\\ z&\mapsto 1- F_{\nu+1}\left( \frac{x-rxz} {\sqrt{\frac{(\nu+x^2z^2)(1-r^2)}{\nu+1}}} \right). \end{align*} The previous step shows that $g_x(z)$ is precisely $\mathbb{P}(T_2>x\mid T_1=xz)$. Hence conditioning on $T_1$ gives \begin{align*} \mathbb{P}(T_2>x\mid T_1>x) = \mathbb{E}\left[g_x(R_x)\mid T_1>x\right]. \end{align*} For fixed $z\geq1$, \begin{align*} \frac{x-rxz} {\sqrt{\frac{(\nu+x^2z^2)(1-r^2)}{\nu+1}}} &= \frac{1-rz} {\sqrt{\left(\frac{\nu}{x^2}+z^2\right)\frac{1-r^2}{\nu+1}}}\\ &\longrightarrow \sqrt{\nu+1}\, \frac{1-rz}{z\sqrt{1-r^2}}. \end{align*} To justify passing this convergence through the conditional expectation, compactify $[1,\infty)$ by adding the point $\infty$. The limiting function \begin{align*} g:[1,\infty]&\to[0,1]\\ z&\mapsto 1- F_{\nu+1}\left( \sqrt{\nu+1}\, \frac{1-rz}{z\sqrt{1-r^2}} \right),\qquad z<\infty,\\ \infty&\mapsto 1-F_{\nu+1}\left(-r\sqrt{\frac{\nu+1}{1-r^2}}\right) \end{align*} is continuous, and the displayed formula for the argument shows $g_x\to g$ uniformly on this compactified half-line. We now justify the limit of the conditional expectations. The uniform convergence gives \begin{align*} \left| \mathbb{E}\left[g_x(R_x)\mid T_1>x\right] - \mathbb{E}\left[g(R_x)\mid T_1>x\right] \right| \leq \sup_{z\in[1,\infty]}|g_x(z)-g(z)|\to0. \end{align*} Since $g$ is bounded and continuous on the compactified half-line, the weak convergence $R_x\Rightarrow R$ gives \begin{align*} \mathbb{E}\left[g(R_x)\mid T_1>x\right]\to\mathbb{E}[g(R)]. \end{align*} Combining these two limits, we obtain \begin{align*} \lambda_U &= \int_1^\infty \left[ 1- F_{\nu+1}\left( \sqrt{\nu+1}\, \frac{1-rz}{z\sqrt{1-r^2}} \right) \right] \nu z^{-\nu-1}\,d\mathcal{L}^1(z). \end{align*} We now evaluate the integral rather than quoting it. Put \begin{align*} m&:=\nu+1, & q&:=\sqrt{\frac{m(1-r)}{1+r}}, & a&:=\frac{\sqrt{m}}{\sqrt{1-r^2}}, \end{align*} let $F_m:\mathbb{R}\to(0,1)$ denote the Student $t_m$ distribution function, and let $f_m:\mathbb{R}\to(0,\infty)$ denote the Student $t_m$ density. Use the substitution $y=1/z$. Then $z=1/y$, the domain $z\in[1,\infty)$ becomes $y\in(0,1]$, and \begin{align*} \nu z^{-\nu-1}\,d\mathcal{L}^1(z) = -\nu y^{\nu-1}\,d\mathcal{L}^1(y). \end{align*} Thus the integral equals \begin{align*} I &:= \nu\int_0^1 \left[1-F_m\left(a(y-r)\right) \right] y^{\nu-1}\,d\mathcal{L}^1(y). \end{align*} Integrating by parts with $u(y)=1-F_m(a(y-r))$ and $dv=\nu y^{\nu-1}\,d\mathcal{L}^1(y)$ yields \begin{align*} I &= 1-F_m(q) + a\int_0^1 y^\nu f_m(a(y-r))\,d\mathcal{L}^1(y), \end{align*} because $a(1-r)=q$ and $y^\nu u(y)$ vanishes at $y=0$. It remains to identify the last integral as the same Student tail. Write \begin{align*} c_m := \frac{\Gamma\left(\frac{m+1}{2}\right)} {\Gamma\left(\frac{m}{2}\right)\sqrt{m\pi}}, \end{align*} so that \begin{align*} f_m(s)=c_m\left(1+\frac{s^2}{m}\right)^{-(m+1)/2}. \end{align*} Since $m=\nu+1$, \begin{align*} a\int_0^1 y^\nu f_m(a(y-r))\,d\mathcal{L}^1(y) &= \sqrt{m}\,c_m(1-r^2)^{(\nu+1)/2} \int_0^1 \frac{y^\nu}{(y^2-2ry+1)^{(\nu+2)/2}}\,d\mathcal{L}^1(y). \end{align*} Now compute the tail $1-F_m(q)$ directly. With $s=a(t-r)$, the lower endpoint $s=q$ corresponds to $t=1$, and \begin{align*} 1-F_m(q) &= \int_q^\infty f_m(s)\,d\mathcal{L}^1(s)\\ &= \sqrt{m}\,c_m(1-r^2)^{(\nu+1)/2} \int_1^\infty \frac{1}{(t^2-2rt+1)^{(\nu+2)/2}}\,d\mathcal{L}^1(t). \end{align*} Finally substitute $t=1/y$. Then $d\mathcal{L}^1(t)=-y^{-2}\,d\mathcal{L}^1(y)$, the domain $t\in[1,\infty)$ becomes $y\in(0,1]$, and \begin{align*} t^2-2rt+1 = \frac{y^2-2ry+1}{y^2}. \end{align*} Therefore \begin{align*} 1-F_m(q) &= \sqrt{m}\,c_m(1-r^2)^{(\nu+1)/2} \int_0^1 \frac{y^\nu}{(y^2-2ry+1)^{(\nu+2)/2}}\,d\mathcal{L}^1(y). \end{align*} This proves \begin{align*} a\int_0^1 y^\nu f_m(a(y-r))\,d\mathcal{L}^1(y) = 1-F_m(q). \end{align*} Hence \begin{align*} I=2(1-F_m(q)). \end{align*} The Student $t_m$ density is even, so its distribution function satisfies $1-F_m(q)=F_m(-q)$. Returning to $m=\nu+1$ gives \begin{align*} \lambda_U = 2F_{\nu+1}\left( -\sqrt{\frac{(\nu+1)(1-r)}{1+r}} \right). \end{align*} [/guided] [/step] [step:Use central symmetry to identify the lower tail coefficient] The centred bivariate Student $t$ distribution is centrally symmetric: \begin{align*} (T_1,T_2)\overset{d}{=}(-T_1,-T_2). \end{align*} Therefore, for $x<0$, \begin{align*} \mathbb{P}(T_2\leq x\mid T_1\leq x) &= \mathbb{P}(-T_2\geq -x\mid -T_1\geq -x)\\ &= \mathbb{P}(T_2\geq -x\mid T_1\geq -x). \end{align*} Since the marginal distributions are continuous, replacing $\geq$ by $>$ does not change these probabilities. Letting $x\to-\infty$ is therefore the same as letting $-x\to\infty$, and the upper-tail computation gives \begin{align*} \lambda_L = \lambda_U = 2F_{\nu+1}\left( -\sqrt{\frac{(\nu+1)(1-r)}{1+r}} \right). \end{align*} This proves the formula for both tail dependence coefficients. [/step]