Hey everyone
I can’t really wrap my head around the correct option of this question:
1 is the wrong asnwer, but how would we mathematically start getting the results? If S1(T) = S2(T) then they are the same exact graph, no?. How would H1(T) be represented graphically? Would that be a y = x type of graph? And if H2(T)=1 (a constant) then that would be simply a horizontal line, no? How do we get to t = 2 ?
Thanks
Hey @Rafael_Oliveira
To determine the time T when S1(T) = S2(T), we need to use the definition of survival functions and the given hazard functions.
We know that the survival function S1(t) is related to the hazard function h1(t) by the equation:
S1(t) = exp(-Integral[0,t] h1(s) ds)
Since h1(t) = t, we can integrate it to obtain:
Integral[0,t] h1(s) ds = 1/2 * t^2
Substituting this back into the first equation, we get:
S1(t) = exp(-1/2 * t^2)
Similarly, for h2(t) = 1, we have:
S2(t) = exp(-Integral[0,t] h2(s) ds) = exp(-t)
To find the time T when S1(T) = S2(T), we can set the two expressions equal to each other:
exp(-1/2 * T^2) = exp(-T)
Taking the natural logarithm of both sides, we get:
-1/2 * T^2 = -T
Simplifying this equation, we get:
T * (T - 2) = 0
So, T = 0 or T = 2.
However, T = 0 corresponds to the initial time point and is not a valid solution for the survival function, since S(T) is defined as the probability of survival at time T given that the subject has survived up to time T. Therefore, the correct answer is T = 2.
Graphically, h1(t) = t corresponds to a linearly increasing hazard function, which can be represented as a diagonal line with a slope of 1 in a hazard rate plot. The corresponding survival function S1(t) decreases smoothly from 1 to 0 as time increases. On the other hand, h2(t) = 1 corresponds to a constant hazard function, which can be represented as a horizontal line in a hazard rate plot. The corresponding survival function S2(t) decreases exponentially from 1 to 0 as time increases.
