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This is a basic question, I am however not able to reduce my solution to the simple form proposed in a book and thus seeking help.

A patient can go via two stages in a disease. first stage is followed up by second stage and that is followed up by death.

Notation:

$p({t^{a}_{1})}$ be the pdf of time spent in stage A/stage 1

$p({t^{b}_{2})}$ be the pdf of time spent in stage B / stage 2 before death and this is independent of time spent in stage A

let $p(t^{d})$ be the convolution of $p(t^{a})$ and $p(t^{b})$ that gives probability distribution of months from start of disease till death i.e $p(t^{a}_{1}+t^{b}_{2})$.

From here on notation P refers to probability (& not pdf):

I further calculated probability of atleast total 'T' months (a constant) till death given stage 1 already happened should be $P(\frac{t>T}{t_{1}<T})= \frac{P(t^{d}>T) {\cap}P(t^{a}_{1} <= T)}{P(t^{a}_{1} <= T)}$.

I am however not able to tie this out to solution that's given out for

$P(\frac{t>T}{t_{1}<T}) = P(t^{a}_{1}<=T)-P(t^{d}<=T)$

Please can you suggest what I am doing incorrect here and help me derive?

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    $\begingroup$ At any given time, you can be in one of four states - no disease, stage 1, stage 2, or dead. In order to be in stage 2, you have to have contracted the disease, completed stage 1, and not died. So the probability of being in stage 2 is the probability of having completed stage 1 minus the probability of having completed both stages (i.e. died). Those are the two terms in the book answer. $\endgroup$
    – MikeP
    Commented Apr 11, 2019 at 19:01
  • $\begingroup$ @MikeP - you may be correct here. I am having difficulty in internalizing this. If that's true, isn't basic assumption that pdf of t1 and t2 are independent, incorrect? In any case, will it be possible for you to expand on what you have written or help me to derive it from the last step I have left things at? $\endgroup$
    – toing
    Commented Apr 11, 2019 at 21:10
  • $\begingroup$ Will add it would be great if someone can help derive it rigorously instead of using intuition / constraint $\endgroup$
    – toing
    Commented Apr 12, 2019 at 22:41
  • $\begingroup$ @ArtemMavrin: I have simplied the notation. Is it better? let me know what other detail you need? $\endgroup$
    – toing
    Commented Apr 12, 2019 at 23:08
  • $\begingroup$ @ArtemMavrin - Thanks for being patient and trying. I am trying to figure out probability of time spent in stage 2 conditioned on fact that stage 1 is completed. May be my notation is off the mark. $\endgroup$
    – toing
    Commented Apr 12, 2019 at 23:12

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