1

This is a really basic question I know but for some reason I'm failing to convince myself of the right answer here.

Given a survival model that has CDF $F(t) = \mathbb{P}(\text{failure before}\ t)$

I would like to calculate $\mathbb{P}(t < T\ |\ t > s)$

Is it

1) $\mathbb{P}(t < T\ |\ t > s) = \frac{F(T)}{1-F(s)}$

or

2) $\mathbb{P}(t < T\ |\ t > s) = \frac{F(T) - F(s)}{1-F(s)}$?

I believe it is #2 because $\mathbb{P}(t < T\ |\ t > s) = \frac{\mathbb{P}(t < T \ \cap\ t > s)}{\mathbb{P}(t > s)}$ and $\mathbb{P}(t < T \cap t > s) = F(T) - F(s)$. Is this correct?

sp59b2
  • 340
LoLa
  • 215

1 Answers1

2

Assuming the variable is continuous, i.e. no jumps in the CDF due to probability mass, the second option is the correct one, as you've also said (assuming $T>s$, otherwise the asked probability is $0$), because: $$P(t<T|t>s)=\frac{P(s<t<T)}{P(t>s)}=\frac{F(T)-F(s)}{1-F(s)}$$ Note we're implicitly assuming that $P(t<T)=P(t\leq T)=F(T)\ \ , \ \forall\ T$ because the CDF is assumed to be continuous, therefore has no jumps and no single point has non-zero probability.

gunes
  • 57,205
  • Thanks for the explanation. I guess the part I was really somehow missing was that P(t < T & t > s) = P(s < t < T)! – LoLa Jul 05 '19 at 21:45