Hazard crossing for survival curve I am very confused because people say the following pic is showing hazard crossing, but the y_aixs is actually not h(t), it is s(x) the survival probability, then how do you conclude they are showing the hazard crossing when they are not a function of hazard ?

 A: The hazard is an arbitrary function of time. The survival is related to the hazard by:
$$ S(t) = \exp(-\Lambda(t))$$
where $$\Lambda(t) = \int_{0}^t h(s) ds$$
is called the cumulative hazard.
So if you tell me, $S_a(t_1) < S_b(t_1)$ and $S_a(t_2) > S_b(t_2)$  then it follows that: $\Lambda_a(t_1) > \Lambda_b(t_1)$ and $\Lambda_a(t_2) < \Lambda_b(t_2)$ because the exponential is monotonic. An interesting corollary for which I'm sure of any short, easy proof is that this necessarily implies that for at least one point in time $h_a(s) > h_b(s)$ for some $0 < s < t_1$ and $h_a(r) < h_b(r)$ for another $t_1 < r$.
A: The hazard function $h(t)$ is related to the survival function $S(t)$ through 
$$S(t) = \exp\left(-\int_0^t h(x) \,\mathrm dx\right).$$
Now suppose $h_i(t)$, $i=1,2$, with the property that $h_1(t) \geq h_2(t)$ for all $t \geq 0$. Then it is true that 
$$\int_0^t h_1(x)) \,\mathrm dx \geq \int_0^t h_2(x)) \,\mathrm dx$$
and so
$S_2(t) \geq S_1(t)$ for all $t \geq 0.$
Note that the two survival functions might have the same value for some choices of $t$, that is, the two curves might meet at some point(s) (they do meet at $t=0$ where they both have value $1$) but they will not cross each other anywhere. Thus we have the result that,
$$h_1(t) \geq h_2(t) ~\text{for all}~ t\geq 0\implies S_2(t) \geq S_1(t)~\text{for all}~ t\geq 0.\tag{1}$$
Now, logically, a proposition of the form $A \implies B$ is equivalent to the proposition $\lnot B \implies \lnot A$ and so $(1)$ is the same as saying that if there is at least one nonnegative $t_0$ such that $S_2(t_0)$ is strictly less than $S_1(t_0)$, that is, the survival function curves cross each other, then it must the case that $h_1(t)$ is strictly smaller than $h_2(t)$ for some values of $t$, that is, the hazard rate curves must also cross somewhere.
