I am aware that the efficacy of a vaccine is calculated with $${\displaystyle VE={\frac {ARU-ARV}{ARU}}\times 100\%,} \text{with}\\ \text{VE=Vaccine efficacy}\\ \text {ARU= Attack rate of unvaccinated people}\\ \text {ARV= Attack rate of vaccinated people}$$ but besides echoing the definition of efficacy, I cannot explain even to myself what does it model. Initially I thought that the efficacy is the probability of you not getting the disease in question. However a close look at the formula reveals that efficacy cannot be a mathematical probability, since it can be a negative value: to my understanding during the pandemic negative efficacies have sometimes been reported regarding proposed vaccines. So what does efficacy model at an individual level, and what, if any, connection efficacy has to probabilities and probability theory?

  • $\begingroup$ @DonThousand The formula is $VE = \frac{ARU-ARV}{ARU} \times 100\%$, where $ARU$ is the attack rate of unvaccinated people and $ARV$ the attack rate of vaccinated people. So if vaccinated people had a higher attack rate than unvaccinated, $VE$ would be negative. $\endgroup$ May 11, 2021 at 14:13
  • $\begingroup$ @Don Thousand To my understanding the ARV can be in practise higher than ARU, causing the numerator to be negative. If efficacy were to model probabilities, shouldn't this be taken into account? $\endgroup$ May 11, 2021 at 14:16
  • $\begingroup$ Check out my explanation here: What does vaccine efficacy mean? An excerpt: “Technically, vaccine efficacy is the proportionate reduction in disease occurrence conferred by the vaccine during the clinical trial. Think of vaccine efficacy as a measure of the increase in protection conferred by the vaccine during the clinical trial—not as an indication of how frequently the vaccine works (either across time, or among the vaccinated population).” $\endgroup$
    – ryang
    May 11, 2021 at 15:02
  • $\begingroup$ Theoretically, statistical significance aside, a positive vaccine efficacy (say, 5%) means that the vaccine did confer production to the clinical-trial participants. $\endgroup$
    – ryang
    May 11, 2021 at 15:08
  • $\begingroup$ Isn't it just measuring the decrease in probability of catching the disease if vaccinated? If the probability of catching the vaccine goes down on getting vaccinated, that's a positive VE, whereas if the probability stays flat, that's a zero VE, and if the probability increases on getting vaccinated, that's a negative VE. $\endgroup$
    – mweiss
    Aug 12, 2022 at 22:39

1 Answer 1


Suppose you have two near-identical populations of $N$ people. The only difference is that one of those populations gets vaccinated and the other population does not. Also suppose that $n_U$ people from the unvaccinated population get the disease while $n_V$ people from the vaccinated population get the disease. Then

$$ n_U=ARU \cdot N\\ n_V=ARV \cdot N $$

and so

$$ n_V=\frac{ARV}{ARU} n_U=(1-VE) \cdot n_U \, . $$

That is, you would expect that only $(1-VE) \cdot n_U$ people from the vaccinated population would get the disease, meaning that on average the vaccine reduces the number of people from the vaccinated population who got the disease by $VE \cdot n_U$.

Equivalently, suppose you either get vaccinated or don't. In either case, you then go about your life for a while in such a way that the probability of you getting the disease would have been $p$ had you not been vaccinated. Imagine a large collection of $N$ alternate universes where you got vaccinated, and $N$ where you didn't, and construct populations like the above out of these universes.

You expect to get the disease in $n_U=N \cdot p$ of the unvaccinated universes, by assumption. So, as above, you'd expect to get the disease in $(1-VE) \cdot N \cdot p$ vaccinated universes, meaning that the probability of vaccinated-you getting the disease is $(1-VE) \cdot p$.

In other words, getting vaccinated reduces the absolute probability that you get the disease by $VE \cdot p$: roughly speaking, the probability of you not getting the disease as the result of being vaccinated is $VE \cdot p$. More carefully, this is the probability that you don't get the disease as a result of being vaccinated, minus the probability that you do get the disease when you otherwise wouldn't have as a result of being vaccinated.

In particular, if the "vaccine" somehow makes you more susceptible to the disease, then $VE$ will be negative.

  • $\begingroup$ Could you walk me through why the probability of me not getting the disease is $VE\cdot p$? Perhaps mathematics has nulled my mind, but I find it easier to follow this sort of story symbolically, where all information is presented explicitly. $\endgroup$ May 11, 2021 at 15:01
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    $\begingroup$ @EpsilonAway: I added a bunch more detail. Hope this helps (and isn't too painfully boring!) $\endgroup$
    – Micah
    May 11, 2021 at 15:19

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