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Let's see the 7-day average rate to test positive is $4\%$, and I'll be in school for one month. What's the possibility I'll catch a virus this month? $1-0.96^{30}=0.7$, so the possibility is $70\%$ I will catch the virus in one month. This result cannot be right, what's wrong with my model? Thanks.

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  • $\begingroup$ If you assume that there's 4% to catch a virus in one day, then your number is right. It's a power of compounding! If the propensity was so high, I'd skip the activity. $\endgroup$ – Aksakal Apr 22 at 14:08
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    $\begingroup$ @Aksakal Would you believe the events to be independent? // There's a "related" question on the right that gives me an idea for an analogy. Say I want to go to one of the top 30 colleges in the world, each of which has a 4% admission rate. I apply to all, so my chance of getting in is $1-0.96^{30} = 70\%$. I don't believe that number. $\endgroup$ – Dave Apr 22 at 14:12
  • $\begingroup$ @Dave, assumption of independence is implicit here. If OP didnt think the hazard rate is independent then the hazard rate itself is obviously not enough to make conclusions. $\endgroup$ – Aksakal Apr 22 at 14:39
  • $\begingroup$ independence assumption is actually quite common in epidemiology models, e.g. take a look at SIR. the problem in these calculations is truly not the independence, but conditionality. you need to personalize the probability to your situation, e.g. risk factors such as health (diabetes etc.) and behavior. $\endgroup$ – Aksakal Apr 22 at 14:45
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    $\begingroup$ @Dave, schools are probably independent in their decisions, they don't collude. they use common factors, of course, but that's conditional part. think of probability as $y_i=X_i\beta+\varepsilon_i$, then independence is about $\varepsilon_i$ not $X_i\beta$, which can be conditional part $\endgroup$ – Aksakal Apr 22 at 15:19
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Your computation is right, but the assumption of independence is not met. Your risk of getting an infection on the secand day is not the same as on the first day.

Let's say we are talking chickenpox, which is a "catchy" disease. If one child in your class has chickenpox, you have a high risk of catching it the very first day. If you did not get it on the first day, then there is probably no child with chickenpox in your class and you are quite safe of getting it the next few days.

Let's say we are talking CoVid19. Let's say there is some kind of background immunity gained from other Corona viruses in the years gone by. If the infection rate is as low as 4% on one day because many pupils already have background immunity, then their risk is not going to grow much from repeated exposition.

All these and likely many more are reasons for the assumptions behind your formula not being met.

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  • $\begingroup$ that's not how diseases are spread or reported. if you're given 4% unconditional probability of contracting a disease, then this number probably includes the overall environment. in the chickenpox example it would include prevalence of chickenpox in the country. of course, the conditional probability would be drastically different if you knew who's sick. that doesn't invalidate unconditional calculations. you're confusing these things $\endgroup$ – Aksakal Apr 22 at 14:43
  • $\begingroup$ Thank you for the discussion. I feel it all depends on the assumption of a 4% positive test rate. If it's based on a large sample and 4% is the infection rate(which is not necessarily equal to the positive test rate), then my model is right, the odds of catching the virus in 30 days is very high. On the other hand, if 4% is based on a small sample, the independence btw. each day doesn't hold, then the odds will be much lower. Thank you, guys! $\endgroup$ – Lynette Apr 23 at 15:04
  • $\begingroup$ forty thousand people have been tested each day on average and the 4% positive test rate is drawn out of them. The population in the city is 80 million. The sample size is only 0.5% of the population, so it's counted as a small sample. The independence doesn't hold here. I never took the case-design class. I hope my reasoning is correct. $\endgroup$ – Lynette Apr 23 at 16:31

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