1
$\begingroup$

I'm a doctor so I'm sorry for my basic knowledge of statisctics. I have a prior probability of the disease D $$P(D) = 0.12$$

I know that risk factor RF can cause the disease. Total probability of RF among population is $$P(RF) = 0.05$$

Also I know the probability of disease in the case when the person has the risk factor is

$$P(D|RF = +) = 0.1$$

Now I have to calculate conditional probability of

  1. The person having the risk factor and disease
  2. The person not having the risk factor but has the disease

My variant $$P(D) = P(D|RF = +) \times P(RF = +) + P(D|RF = -) \times P(RF = -)$$

So I have

$$ 0.12 = 0.1 \times 0.05 + P(D|RF = -) \times (1 - P(RF = +))$$

So $$ P(D|RF = -) = 0.121 $$

It seems strange to me, because prior probability is 0.12 and probability of disease in case when a person doesn't have a risk factor is more then when he has a risk factor.

Am I right? Thank you for your time and any clarifications/explanations/advice.

$\endgroup$
  • $\begingroup$ "Risk factor" is a misnomer here - it's a protective factor, since you are less likely to have the disease if you exhibit this factor! $\endgroup$ – Nuclear Wang Mar 6 at 15:52
0
$\begingroup$

Is your risk factor really a risk factor if the probability of having a disease with it being positive is less than the probability of disease straight out?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, now I think also about it. But due to medical article I have next info: Risk factor progresses to clinical disease in 10% of females. $\endgroup$ – Aliru Mar 6 at 10:57
0
$\begingroup$

Your working so far looks correct. The weird result is because you have a disease with a 12% background rate, but after you know about the risk factor the risk drops to 10%. Normally you would expect a "risk factor" to increase your chance of getting the disease. In this case the risk factor appears to be protective. If this is a real world situation, is it possible that the 0.1 that you have is not in fact a conditional probability? Maybe it is an increase in odd ratio or the increase in risk?

Edit In Light of Comments

Ok, so what is going wrong here is combining incidence with prevalence with risk. We also have some problems with repeat infection it seems.

  • The annual incidence of UTI is 12%. In a given year, 12% of women will have at least one UTI at some point in the year (sometime incidence includes repeates, but here it appears not to).
  • The risk of getting a UTI within 1 week, given that you have asymptomatic bacteriuria (AB) is 10%.
  • The prevalence of AB is 5%. At any point in time 5% of women have AB.

Combining these quantities together is much more subtle that a simply probability exercise, and will require a few assumptions. In particular we could do with knowing how long a UTI lasts as well as the incidence of AB (or how long it lasts).

If we have 100 women, then we expect 12 women to have UTI in a year. We have 5200 woman-weeks in the year, and 5% = 260 women-weeks are AB positive. Of those weeks 10%=26 are expected to progress to UTI, which suggests quite a lot of women are having multiple UTIs in a single year - perhaps because BA is persisting?

Perhaps another thing that is going on here, is that all of these numbers are very approximate, and so by mixing them together, we've ended up with an unrealistic combination. It is also not totally clear that whether the 8-15% (10% in our example) that progress to UTI within a week is from the onset of AB or for every week that you have AB. If it is per AB infection then that totally changes things.

However the answers to your questions are also documented elsewhere in the papers (sort of). They also state that 24-57% of cases of UTI also test positive for AB. Taking that number as 40% then that means that $40\% \times 12\% = 4.8\%$ of the population have the risk factor (AB) and the disease (UTI). The complement to that is then clearly 7.2% have the disease without the risk factor.

Working backwards from these numbers then 4.8 cases come from 5% of women with AB and 7.2 cases come from the rest, meaning that having AB makes you $\frac{4.8}{7.2}\times\frac{95}{5} = 12.6$ times more likely to have a UTI if you have AB

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ From the journal: Risk factor progresses to clinical disease in 8–12% of females. $\endgroup$ – Aliru Mar 6 at 10:54
  • $\begingroup$ @Aliru and where do you get the base rate of 12% from? Can you link to the article? $\endgroup$ – Korone Mar 6 at 11:46
  • $\begingroup$ ncbi.nlm.nih.gov/books/NBK482435 The incidence of UTI is 12% in women based on self-reported annual incidence. $\endgroup$ – Aliru Mar 6 at 12:10
  • $\begingroup$ @Aliru ah... I think the problem here is the difference between incidence and probability. The 12% incidence is the proportion that have it at some point in the year. This is going to be a rather more complex matter to combine - do you have a link to the "Risk factor progresses to clinical disease in 8-12% of females"? We would need to know more about that context that risk factor to combine it with an incidence. $\endgroup$ – Korone Mar 6 at 13:45
  • $\begingroup$ sciencedirect.com/science/article/pii/B9780323040730100251 Asymptomatic bacteriuria progresses to clinical UTI within 1 week in 8–15% of females. Thank you a lot @Corone $\endgroup$ – Aliru Mar 6 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.