# Calculations behind this quiz question | Week 4, AI for Prognosis, question 3

Hey everyone

I recently got a question on the last week of AI for Prognosis (week 4, question 3) and the question is this one:

I can’t wrap my head around this conceptually. It makes sense that if the survival probablity of a patient is greater than the other, the hazard should be lesser than the other (hence if S1(T) > S2(T), then H1(T)<H2(T), if they’re proportional), but I don’t understand how that is translated mathematically on the exponential function

Thanks!

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Hey @Rafael_Oliveira
If the hazards for two patients are proportional to each other, then the hazard ratio between them is constant over time. Mathematically, we can write:

h1(t) / h2(t) = C

where C is a constant hazard ratio. This means that the ratio of the hazards of the two patients remains constant at all time points. Therefore, if we compare the hazards of the two patients at a specific time T, we have:

h1(T) / h2(T) = C

Solving for h1(T), we get:

h1(T) = C * h2(T)

So, if we know that S1(T) > S2(T) for some T > 0, then we can use the relationship between survival and hazard to say:

S1(T) = exp(-Integral[0,T] h1(s) ds) > exp(-Integral[0,T] h2(s) ds) = S2(T)

Taking the natural logarithm of both sides, we get:

-Integral[0,T] h1(s) ds < -Integral[0,T] h2(s) ds

Multiplying both sides by -1 and reversing the inequality, we get:

Integral[0,T] h1(s) ds > Integral[0,T] h2(s) ds

Using the relationship between hazard and survival, we can write:

Integral[0,T] h1(s) ds = -log(S1(T))

Integral[0,T] h2(s) ds = -log(S2(T))

Substituting these back into the inequality, we get:

-log(S1(T)) > -log(S2(T))

Taking the exponential of both sides, we get:

S1(T) < S2(T)

Since we know that S1(T) > S2(T) for some T > 0, this inequality cannot hold. Therefore, the assumption that the hazards for the two patients are proportional to each other and S1(T) > S2(T) for some T > 0 cannot be true.

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