I don't understand how to find the probability $P(T_x>T_y)$, when we know that $f_{T_x|T_y}(x|y)$ is $exp\left(\frac{1}{y}\right)$ and $f_{T_y}$ is $exp(\beta)$. I would do this to start with: $$P(T_x>t|T_y=t),$$ but I cannot use condtitional probability, as $T_y=t$ is $0$. However, I could find $f_{T_x,T_y}(x,y)$, and then find $f_{T_x}$ and so on, however, when I asked my teacher he said: "You can easily see from the densities, that $P(T_x>T_y)$ must be $e^{-1}$", but I cannot. Anyone can tell me how to easily see it?
EDIT: After quite some time, I figured this out:
We know that $f_{T_x|T_y}(x|y)$ is $exp\left(\frac{1}{y}\right)$, so $$F_{T_x|T_y}(x|y)=P(X\leq x|T_y=y)=1-e^{-\frac{1}{y}x}$$ and then: $$\bar{F}_{T_x|T_y}(x|y)=P(X>x|T_y=y)=1-F(x)=1-\left(1-e^{-\frac{1}{y}x}\right)=e^{-\frac{x}{y}}$$ os if, when have that $T_x>T_y$, that $T_y=x$ we get: $$\bar{F}_{T_x|T_y}(x|y)=P(X>x|T_y=x)=e^{-1}$$ and we would just like to find $\bar{F_X}=P(T_x>x)$, as we said $T_y=x$, so $$\bar{F_{T_x}}=\int_{0}^\infty \bar{F}_{T_x|T_y}(x|y)f_{T_y}{y}=e^{-1}\int_{0}^\infty f_{T_y}{y}=e^{-1}.$$