3 ms·

ADHD and autism are comorbid in one direction (people with autism are more likely than the general population to have ADHD, but people with ADHD are not more li

ADHD and autism are comorbid in one direction (people with autism are more likely than the general population to have ADHD, but people with ADHD are not more likely than the general population to have autism).

Obscurity4340 3y ago
Interesting distinction, thank you
josephcsible 3y ago
> people with autism are more likely than the general population to have ADHD, but people with ADHD are not more likely than the general population to have autism I'm pretty sure that's not mathematically possible. If W people have both, X people have only autism, Y people have only ADHD, and Z people have neither, can you provide values for W, X, Y, and Z that would make that claim true?
- d0mine 3y ago
  p(h)p(a|h)=p(h|a)p(a) therefore if p(a)=p(a|h) then p(h)=p(h|a) too.
- Obscurity4340 3y ago
  We don't need mathematical or conjectural proofs here, it makes intuitive sense but if you find a soecific rebuttal to it we'll be glad to reference that as well
- Twisol 3y ago
  Alright, you've nerd-sniped me. Our assumption is that the ratio of people with autism and ADHD to people with just autism is greater than the ratio of people with ADHD to people in general. Using your variables, that is: W / (W + X) > (W + Y) / (W + X + Y + Z). This simplifies out to the constraint that ZW > XY. We want to show that the ratio of people with autism and ADHD to people with just ADHD is no greater than the ratio of people with just autism to people in general. This is: W / (W + Y) <= (W + X) / (W + X + Y + Z). We can recognize that this is almost the same as our assumption, just exchanging the roles of X and Y. (I guess that's obvious from the problem statement.) That means our goal is to show that ZW <= YX, which is immediately just the negation of what we're assuming. That would be a contradiction. So, indeed, this is mathematically impossible! (EDIT: It's surprisingly hard to write these formulae down in English without getting confused.) I think something that would be possible is that having ADHD if you have autism is more likely than having autism if you have ADHD. For instance, "being older than 20 if you're older than 30 is more likely than being older than 30 if you're older than 20" is just obviously true, because one entails the other.
- reanimus 3y ago
  W (9), X (1), Y (9), Z (1) General chance of being in either group: 50% (10 people with autism, 10 with ADHD, 20 total). Chance of having autism if you have ADHD: 50% (9 with autism, 9 without, 10 total). Chance of having ADHD if you have autism: 90% (9 with ADHD, 9 without, 10 total).
  - josephcsible 3y ago
    > General chance of being in either group: 50% (10 people with autism, 10 with ADHD, 20 total). This is wrong. With your numbers ("W (9), X (1), Y (9), Z (1)"), 18 people would have ADHD, not 10.
- BirdieNZ 3y ago
  You are correct, and my memory failed me: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3876754/ https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3876754/ People with autism are highly likely to have comorbid ADHD. People with ADHD are more likely than normal to have autistic traits, although nowhere near as commonly as the first statement.

Obscurity4340 3y ago

Interesting distinction, thank you

josephcsible 3y ago

> people with autism are more likely than the general population to have ADHD, but people with ADHD are not more likely than the general population to have autism I'm pretty sure that's not mathematically possible. If W people have both, X people have only autism, Y people have only ADHD, and Z people have neither, can you provide values for W, X, Y, and Z that would make that claim true?

d0mine 3y ago
p(h)p(a|h)=p(h|a)p(a) therefore if p(a)=p(a|h) then p(h)=p(h|a) too.
Obscurity4340 3y ago
We don't need mathematical or conjectural proofs here, it makes intuitive sense but if you find a soecific rebuttal to it we'll be glad to reference that as well
Twisol 3y ago
Alright, you've nerd-sniped me. Our assumption is that the ratio of people with autism and ADHD to people with just autism is greater than the ratio of people with ADHD to people in general. Using your variables, that is: W / (W + X) > (W + Y) / (W + X + Y + Z). This simplifies out to the constraint that ZW > XY. We want to show that the ratio of people with autism and ADHD to people with just ADHD is no greater than the ratio of people with just autism to people in general. This is: W / (W + Y) <= (W + X) / (W + X + Y + Z). We can recognize that this is almost the same as our assumption, just exchanging the roles of X and Y. (I guess that's obvious from the problem statement.) That means our goal is to show that ZW <= YX, which is immediately just the negation of what we're assuming. That would be a contradiction. So, indeed, this is mathematically impossible! (EDIT: It's surprisingly hard to write these formulae down in English without getting confused.) I think something that would be possible is that having ADHD if you have autism is more likely than having autism if you have ADHD. For instance, "being older than 20 if you're older than 30 is more likely than being older than 30 if you're older than 20" is just obviously true, because one entails the other.
reanimus 3y ago
W (9), X (1), Y (9), Z (1) General chance of being in either group: 50% (10 people with autism, 10 with ADHD, 20 total). Chance of having autism if you have ADHD: 50% (9 with autism, 9 without, 10 total). Chance of having ADHD if you have autism: 90% (9 with ADHD, 9 without, 10 total).
- josephcsible 3y ago
  > General chance of being in either group: 50% (10 people with autism, 10 with ADHD, 20 total). This is wrong. With your numbers ("W (9), X (1), Y (9), Z (1)"), 18 people would have ADHD, not 10.
BirdieNZ 3y ago
You are correct, and my memory failed me: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3876754/ https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3876754/ People with autism are highly likely to have comorbid ADHD. People with ADHD are more likely than normal to have autistic traits, although nowhere near as commonly as the first statement.