Conditional independence and independence when CDF is given I have a problem in my probability exercises and I have the following question regarding it,
I am given two random variables, $X_1$ and $X_2$. 
$X_1$ has a CDF of $F_{X_1}(x,\theta)$ and $X_2$ has a CDF of $F_{X_2}(x,\theta)$. 
It is given that $T_1|\theta$ is independent from $T_2|\theta$.
Edit: It is given that $\color{red}{X_1}|\theta$ is independent from $\color{red}{X_2}|\theta$ 
Does this mean that $T_1$ is independent of $T_2$ ? 
Edit: Does this mean that $\color{red}{X_1}$ is independent of $\color{red}{X_2}$ ?
Can I write, say,
$$ Pr(X_1 >x_1, X_2 > x_2) = \bigg(1-\int_{-\infty}^{\infty}F_{X_1}(x_1,\theta)F_{\theta}(\theta) d\theta\bigg) \times  \bigg(1-\int_{-\infty}^{\infty}F_{X_2}(x_2,\theta)F_{\theta}(\theta) d\theta\bigg)$$
Is this correct? The distribution of $\theta$ is given as Gamma, if that helps. Any help would be greatly appreciated.  
 A: It looks like using $T_1, T_2$ is a typo, I think you mean $X_1$ and $X_2$ are conditionally independent given $\theta$. This does not mean $X_1$ and $X_2$ are independent (see counter-example in my comment above). So your proposed equation is incorrect.  
Fortunately, a correct equation can be obtained just as simply. If you are given $X_1, X_2$ are conditionally independent given $\theta$, you should directly use that information to compute $P[X_1>x_1, X_2>x_2]$ by conditioning on $\theta$ (via the law of total probability).  You know how to condition on all possible values of $\theta$ to compute the probability of an event involving one random variable.  You can use the same technique to condition on an event involving two random variables. Note that in general, if $A$ is an event, then the law of total probability says: 
$$ P[A] = \int_{-\infty}^{\infty} P[A|\theta=u] f_{\theta}(u)du$$
Can you give an answer now?  Possibly typing up your own answer to the question here, or giving it in a comment? 
