1
$\begingroup$

I found a study that compared minor physical anomalies(MPA) between certain group of patients with the control group to determine if MPAs occur more frequently among these patients compared to the control group. One of the MPA item is epicanthus which is an extremely common trait among Asians(esp. Chinese, Japanese, Korean, ect.). In fact major studies I found show that among the Han Chinese population(which represents 92% of all people in China), the prevalence rate of epicanthus among these people is between 31.79% to 41.75%(https://pubmed.ncbi.nlm.nih.gov/31492581/). I believe that this may still be an underestimation because other sources describe higher. According to my search, it appears that 31.79% to 41.75% is about the minimum. Going back to the patient-control comparison study on MPAs, this MPA study describes that 29.1% of the patients(in this MPA study) had epicanthus while only 3.3% of the Chinese control group had epicanthus. As a result, they concluded that epicanthus is mucher higher among the patients which therefore, shows that manifestation of epicanthus among the Chinese population is a risk factor for the patients' illness.
However, I believe that this study included a selection bias(for the control group) which dramatically distorted the results. Considering that epicanthus is known to be very common among the general Chinese population(which seems to be at least about 31% to 41%), I believe that their control group's epicanthus prevalence rate of 3.3% would represent selection bias(since 3.3% would not represent the Chinese general population's reasonable prevalence rate for epicanthus). Rather, if we assume that the Chinese population's epicanthus prevalence rate is abut 31% to 41%, we can see that the patients' 29.1% for epicanthus(in the MPA study) is very similar to the Chinese general population's prevalence rate(which I would think is most likely the correct conclusion).
Since the Chinese general population's prevalence rate for epicanthus is at least about 31% to 41%(while the 3.3% epicanthus prevalence rate seems like a selection bias), can I use the general population's 31% to 41% epicanthus prevalence rate to compare it with the patients' 29.1% epicanthus prevalence rate and conclude that this MPA study does not show(i.e., does not provide evidence) that epicanthus is a risk actor for the patients' illness(since 29.1% is actually slightly lower than 31% to 41% among the general population)?

$\endgroup$

1 Answer 1

0
$\begingroup$

Let $E$ be the random variable describing whether the patient has the anomaly epicanthus ($E=0$ means this anomaly is not present, $E=1$ means it is present), and let $I$ be the random variable describing whether the patient is having the illness that is considered here ($I=0$ means the patient is not sick, $I=1$ means the patient is sick).

Now, IIUC, your question is, whether it is possible to have $P(E=1|I=0) = 0.03$ while $P(E=1) \approx 0.3$ and $P(E=1|I=1) \approx 0.3$. In this case, the numbers you gave would result from proper sampling without bias.

First, imagine that the people with epicanthus are a subset of the people with the illness.

enter image description here

Then it is clear that we even have $P(E=1|I=0)=0$. You didn't say what the frequency of the illness is, but if it is larger than that of epicanthus, this possibility cannot be ruled out. Of course, if the illness is rare, this cannot be the case. But this indicates that the probability $P(I=1)$ of the illness is kind of important.

Next, let's describe the situation more thoroughly. We have: $$ \begin{align} P(E=1) &= P(E=1|I=0)\;P(I=0) \;+\; P(E=1|I = 1)\;P(I=1)\\ &= P(E=1|I=0)\;(1-P(I=1)) \;+\; P(E=1|I = 1)\;P(I=1)\\ P(E=1) - P(E=1|I = 1)\;P(I=1) &= P(E=1|I=0)\;(1-P(I=1))\\ P(E=1|I=0) &= \frac{P(E=1) - P(E=1|I = 1)\;P(I=1)}{1-P(I=1)}\\ \end{align} $$ Let's presume that $P(E=1)$ and $P(E=1|I=1)$ are roughly known parameters and consider $P(E=1|I=0)$ as a function of $P(I=1)$. To make this more readable, let's use the abbreviations $$ \begin{align} y &:= P(E=1|I=0)\\ x &:= P(I=1)\\ e &:=P(E=1)\\ a &:= P(E=1|I=1). \end{align} $$ Then, we are interested in the function $y(x)$: $$ y(x) = \frac{e - ax}{1-x}. $$ Here is a plot of this function for $a=0.3$ and for three different values for $e$, $e=0.2, 0.3, 0.4$. Note that, from the formula, it is clear that for $x=0$ we have $y=e$ and for $a=e$ we have $y=a=e=const$:

enter image description here

From this plot you can see, for each illness frequency $P(I=1)$ (the $x$-axis) what frequency of epicanthus is to be expected, depending on $e$ (keep in mind, that $y$ is a probability and thus has to stay in the interval [0,1]). E.g., we see that, if the illness is very rare ($x$ near zero), the frequency of epicanthus among the healthy cannot be much less than that among all people ($e=P(E=1)$), which is pretty obvious anyway. However, for a rather common illness, $y$ can become very small for fixed $e$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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