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Canada at the present time:

  • $80\%$ of the population fully vaccinated,
  • $160$ people newly hospitalized (Feb. 5, 2022) with Covid-19,
  • $118$ of them fully vaccinated (i.e., $73.75\%$ fully vaccinated).

Source: Making sense of the numbers: Greater proportion of unvaccinated are being hospitalized (Feb. 7, 2022).

Using the notation

  • $p$ = the percentage of the total population (of Canada) fully vaccinated,
  • $h$ = the percentage of the people in hospital fully vaccinated,

I postulated that $\frac{h}{p}=1$ for $η=0$, where $η$ is the efficiency of vaccines, and got $$η=1-\frac{h}{p}=1-\frac{73.75\%}{80\%}=7.81\%.$$ (The above formula is empirical, and I do not have a proof for it.)

My formula can not be so wrong because it is self-evident that, had the vaccines had no effect (had they been just placebos), $80\%$ of the above newly hospitalized people would have been fully vaccinated. Since the actual percentage $73.75\%$ is quite close to this $80\%,$ the efficiency of the vaccines ought to be close to zero; you do not need a formula to see that.

UPDATE

I have rewritten the above more clearly by adapting the notation (and translation of my formula) introduced by @ryang in his answer below, in which he argued to the contrary that $$η=1-\frac{\frac{h}{1-h}}{\frac{p}{1-p}}=\frac{p-h}{p(1-h)}=30\%.$$ For clarity, using Mathcad, for $h\leq p$ (for which the efficiencies (or effectiveness) are nonnegative), I have represented $η(h,p)$ for both formulae:

enter image description here

ryang
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Robert Werner
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2 Answers2

1

Your math is incorrect.

We know that 42 people who were unvaccinated got the disease.

Therefore, if vaccines were totally ineffective, we would expect a total of $42 / 20\% = 210$ people to get the disease. In particular, we would have expected $210 - 52 = 168$ vaccinated infections. Thus, the vaccine prevented $(168 - 118)/168 = 50 / 168 \approx 30\%$ of infections.

Of course, this assumes that the vaccines were randomly distributed and did not affect people’s behaviour. Neither of these factors holds. People who are more vulnerable to COVID-19 are more likely to get vaccinated and also more likely to be hospitalised. People who get vaccinated may be less reluctant to get exposed fo COVID-19 after getting vaccinated. And people who don’t get vaccinated because they are anti-science may be less likely to go to the hospital once they contract COVID, preventing their infection from getting counted.

Ultimately, the only way to definitively establish vaccine effectiveness is through a controlled experiment.

Mark Saving
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  • Yes, 42 unvaccinated (26.25%) of the newly admitted to hospitals got a serious form of Covid-19. Had the vaccine had zero effect only 20% (32 people) of those admitted to the hospital should have been unvaccinated. However, 26.25% and 20% (which means zero efficiency!) are quite close to each other and in consequence that 26.25% can not lead to a high efficiency like say over 90%. – Robert Werner Feb 08 '22 at 03:34
  • @RobertWerner You do not understand what the percent effectiveness means. According to your definition, if everyone is vaccinated, then the vaccine will always have zero effectiveness, since 100% of the infections are of vaccinated people and 100% of the population is vaccinated. – Mark Saving Feb 08 '22 at 03:45
  • If everyone is vaccinated then it is impossible to determine the efficiency of the vaccines and in consequence that zero you obtained has a meaning (the efficiency does not make sense!). – Robert Werner Feb 08 '22 at 03:52
  • "if vaccines were totally ineffective, we would expect a total of 42/20 people to get the disease"?!! 42/20 = 2.1 people. How did you reach such a conclusion? Let me explain my reasoning like this: If 80% of Canadians have a stamp on the hand with "Vaccinated" (but they are not vaccinated) and the rest of 20% a stamp with "Unvaccinated" than statistically 80% of the people infected with Covid-19 being admitted to hospitals would have the word "Vaccinated" written on their hand and the rest of 20% the word "Unvaccinated" and the efficiency of the vaccine must be 0 because nobody received it. – Robert Werner Feb 08 '22 at 04:00
  • @RobertWerner That was an issue with formatting; it was supposed to be 42 / 20%. – Mark Saving Feb 08 '22 at 04:02
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    @Robert Effectiveness is defined as (probability of getting infected given you’re vaccinated) / (probability of getting infected given you’re unvaccinated). Equivalently, it equals (number of vaccinated people who actually got infected) / (number of vaccinated people you’d expect to get infected if the vaccine were ineffective). You are using a completely made up formula that you have provided zero justification for to calculate this quantity. Your formula got the wrong result, as I demonstrated. – Mark Saving Feb 08 '22 at 04:08
  • I know but I can not apply that formula you mentioned because there is not enough data. "Probability of getting infected given you’re vaccinated" and the "Probability of getting infected given you’re unvaccinated" are not available in newspaper articles. You can not calculate them based on raw data like the percentage of people admitted to hospitals and the total percentage of vaccinated people in the country. – Robert Werner Feb 08 '22 at 04:12
  • If vaccines were totally ineffective we would expect, at each 100 people admitted to hospitals, 80 vaccinated and 20 unvaccinated. My formula is based on this idea which I have already explained a few times. – Robert Werner Feb 08 '22 at 04:45
  • @ryang , I thought about your formula months ago but I abandoned it when I noticed that it does not behave well when the number of vaccinated people is close to 100%. For example, for 99.99% vaccinated in hospital and 99.999% vaccinated out of the entire population the efficiency of the vaccines will be 1-((99.99/0.01)/(99.999/0.001)) = 90%. If 99% in hospitals are vaccinated then the same efficiency will be 1-((99/1)/(99.999/0.001)) = 99.9%. Using my formula the efficiencies will be below 1% in both cases. – Robert Werner Feb 08 '22 at 08:13
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    @RobertWerner Yes, when the population has been almost entirely vaccinated, of course the COVID-hospitalised patients correspondingly have been almost entirely vaccinated too. (I.e., the latter being due to the former is not a poor reflection of vaccine effectiveness.) My Answer's formula $(✔)$ does behave well in describing the vaccine effectiveness of this near-edge case as over 90%, whereas your speculation $(✗)$ wrongly asserts that value as under 1%. – ryang Feb 08 '22 at 20:50
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  1. Let $p\%\:\:(0<p<100)$ of the population be vaccinated,

    and $h\%\:\:(0\leq h<100)$ of the COVID-hospitalised patients be vaccinated.

    Then, ignoring immunity duration and confounding factors like patient behaviour, \begin{gather}\text{vaccine effectiveness against hospitalisation}\\=1-\left(\frac{h}{100-h}\right)\div\left(\frac{p}{100-p}\right)\\=\frac{100(p-h)}{p(100-h)}.\tag✔\end{gather}

    I have explained here and here my derivation of this formula from the originating definition.

    Note that the technical term is vaccine effectiveness (not vaccine efficiency), which is a measure of the increase in protection conferred by the vaccine.

    Regarding the restrictions on $p$ and $h:$

    • it makes sense to define vaccine effectiveness only for a partially-vaccinated population, since no inter-group comparison is possible otherwise

    • and to exclude $h=100,$ because in this case any valuation of the (negative) hospitalisation vaccine effectiveness would be necessarily arbitrary, and we don't need a formula for that!

  2. Your made-up formula

    efficiency of vaccine = (percentage of the hospitalized who is unvaccinated - percentage of the country who is unvaccinated)/(percentage of the country who is fully vaccinated)

    can be rewritten as $$\text{vaccine effectiveness against hospitalisation}=\frac{p-h}p;\tag✗$$ it is consistent with the correct formula above if and only if $h=0$ or $p$ among a partially vaccinated population.

    Since the actual percentage 73.75% is quite close to this 80%, the efficiency of the vaccines ought to be close to zero; you do not need a formula to see that.

    Your formula $(✗)$ appears to be motivated by a muddling of percentage change and change in percentage points.

    Let's compare a change from 1% to 5% with a change from 95% to 99%. While both are an increase of $4$ percentage points, the former is a whopping $400\%$ increase, whereas the latter is merely a $0.42\%$ increase.

  3. In your given example, the hospitalisation vaccine effectiveness is given by formula $(✔)$—and Mark Saving's Answer, which I agree with—as $\displaystyle\frac{25}{84}\approx30\%.$ Formula $(✗)$ on the other hand gives the incorrect value $\displaystyle\frac{25}{84}\left(1-\frac h{100}\right)=7.8\%.$

ryang
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  • You simply multiply my efficiency (effectiveness) with a compensation factor ( 1/(1-(h/100%)) ) that increases your effectiveness strongly when the percentage of vaccinated people in hospital goes to large percentages. For h=90% your effectives is 10 times mine, for h=50% only twice and for h=5% just 1.05 times mine. You do not really have a demonstration for your formula (like me) because in your justifications you started from the idea (which looks intuitive) that effectiveness = 1-(vaccinated/unvaccinated)_in hospital/(vaccinated/unvaccinated)_in total, in other words from the final result. – Robert Werner Feb 08 '22 at 18:11
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    @RobertWerner 1. No, I proved formula $(✔)$ from definition, fully explained why its conditions make sense, and showed via your 'compensation factor' that your speculation $(✗)$ is wrong. (The fact that both formulae share the property that $p=h⟹$effectiveness$=0%$ is not pertinent.) 2. I've just expanded my Answer to shed light on the fact that your motivation for $(✗)$ is misguided. 3. I likely shan't have further comments. – ryang Feb 08 '22 at 19:27
  • Basically, I postulated that $\frac{h}{p}=1$ for ${\eta}=0$, where ${\eta}$ is the efficiency, and got ${\eta}=1-\frac{h}{p}$. You postulated that $\frac{\frac{h}{1-h}}{\frac{p}{1-p}}=1$ for ${\eta}=0$ and you got ${\eta} =1-\frac{\frac{h}{1-h}}{\frac{p}{1-p}}$. That is all regarding our justifications. Say now that $p=80%$ and $h=40%$. My efficiency will be $50%$ while yours $83.3%$ which appears to be way too big. – Robert Werner Feb 09 '22 at 20:12
  • Are you sure that your formula is correct? $h\to 100$ becomes singular and for fixed $p$ you can even make the expression smaller than $0$. The metric should be between $0$ and $1$. The singularity should also be removable. – MachineLearner Feb 10 '22 at 07:33
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    My formula is correct, and the metric (vaccine effectiveness) indeed ought to lie on $(−∞,1]:$ in my second link I explained that a vaccine with negative effectiveness is one that makes the subject more susceptible to the disease, and gave the standard definition of relative risk, which can be seen to approach infinity—and the vaccine effectiveness correspondingly approach negative infinity—as $h$ approaches $100.$ – ryang Feb 10 '22 at 11:35
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    Thank you for explaining. But I have the impression that this metric is not very useful from an intuitive perspective. – MachineLearner Feb 10 '22 at 19:22
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    Oh yes, it is extremely easy for laypersons to misunderstand vaccine efficacy & vaccine effectiveness (the former is regarding the clinical-trial data). Most wrongly believe that it measures one of the following: 1) the recipient's chance of non-infection, 2) how frequently the vaccine successfully blocks infection, 3) the proportion of recipients to whom the vaccine confers protection. – ryang Feb 10 '22 at 20:50
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    To be clear: 1. relative risk and vaccine effectiveness are theoretical measures, 2. the latter is not a probability (nor a measure of chance of protection). @MachineLearner – ryang Feb 11 '22 at 06:00