Relation between bivariate survial function and cumulative density function I am trying to understand why
$Pr(T_1> t_1,T_2 > 2)=1-F_1(t_1)-F_2(t_2)+F_{12}(t_1,t_2)$
My derivation is as follows:
\begin{align} Pr(T_1> t_1,T_2 > t_2) &=Pr(T_1> t_1\mid T_2> t_2)Pr(T_2> 2)\\&=(1-Pr(T_1\leq t_1\mid T_2> t_2))(1-Pr(T_2\leq t_2))
\end{align}
and 
\begin{align} Pr(T_1> t_1,T_2> t_2) &=Pr(T_2> t_2\mid T_1> t_2)Pr(T_2> t_2)\\&=(1-Pr(T_2\leq t_2\mid T_1> t_1))(1-Pr(T_1\leq t_1))
\end{align}
By symmetry (or equality): $Pr(T\leq t_1)=Pr(T\leq t_1\mid T_2> t_2)$ (and likwise for $T_2$).
But, hey does not this imply that $T_1$ and $T_2$ are independent?
 A: The equality $Pr(T_1\leq t_1)=Pr(T_1\leq t_1\mid T_2> t_2)$ for every $t_1,t_2$ is indeed equivalent to independence of $T_1$ and $T_2$. And the initial equality does not need any independence assumptions. It is an identity. 
First, the events $T_1>t_1$ and $T_1\leq t_1$ are the opposite, so
$$
Pr(T_1>t_1, T_2> t_2)+Pr(T_1\leq t_1, T_2> t_2) = Pr(T_2> t_2) = 1-F_2(t_2).
$$
So, 
$$
Pr(T_1>t_1, T_2> t_2)= 1-F_2(t_2) - Pr(T_1\leq t_1, T_2> t_2). \tag{1}\label{1}
$$
Next, 
$$
Pr(T_1\leq t_1, T_2> t_2) +Pr(T_1\leq t_1, T_2\leq t_2) = Pr(T_1\leq t_1) = F_1(t_1),
$$
so
$$
Pr(T_1\leq t_1, T_2> t_2) = F_1(t_1) - Pr(T_1\leq t_1, T_2\leq t_2) = F_1(t_1)-F_{1,2}(t_1,t_2).
$$
Substitute the last value into \eqref{1} instead of $Pr(T_1\leq t_1, T_2> t_2)$:
$$
Pr(T_1>t_1, T_2> t_2)= 1-F_2(t_2) - \left(F_1(t_1)-F_{1,2}(t_1,t_2)\right) = 1-F_2(t_2) - F_1(t_1)+F_{1,2}(t_1,t_2).
$$
A: You can simply find it via De-Morgan's Law. Let $A=T_1>t_1,B=T_2>t_2$
$$P(A\cap B)=1-P(A'\cup B')=1-(P(A')+P(B')-P(A'\cap B'))$$
