Survival Function Issues: f= qpt^(p-1)/(((q+t^(p))^2) where q>0 and p>0

mathman45

New member
Joined
Mar 26, 2018
Messages
1
Good Evening,

I am struggling with a survival function problem, I always get a value over 1, just wondering if someone could have a look at my maths and give me a hand.

so given the function

f= qpt^(p-1)/(((q+t^(p))^2) where q>0 and p>0

I get

F(t) as = -q/(q+(t^p))

S(t) = 1-F(t) as 1 - (-q/(q+(t^p)))

if I hypothetically say q is 2 and p is 2 and t is 1
then trying to plug numbers in survival function we get 1 - (-2/(2+(1^2))) = 1 - (-2/3) = 1.6666666 which cannot be a survival function as it is above 1...........
I'm lost........
 
My first suspicion is that p and q are not well-defined. Do they have more definition that you are not sharing? It is odd that the formula would use 'p' and 'q', which are normally defined as friends, \(\displaystyle 0 \le p \le 1\;and\;q = 1-p\) in the probability world.

Let's make sure we know what we're dealing with, first.
 
Good Evening,

I am struggling with a survival function problem, I always get a value over 1, just wondering if someone could have a look at my maths and give me a hand.

so given the function

f= qpt^(p-1)/(((q+t^(p))^2) where q>0 and p>0

I get

F(t) as = -q/(q+(t^p))

S(t) = 1-F(t) as 1 - (-q/(q+(t^p)))

if I hypothetically say q is 2 and p is 2 and t is 1
then trying to plug numbers in survival function we get 1 - (-2/(2+(1^2))) = 1 - (-2/3) = 1.6666666 which cannot be a survival function as it is above 1...........
I'm lost........
You opened three parenthesis in a row but only closed two of them. Please fix this.
 
Top