Really enjoying the lecture series - thank you so much for taking the time, I'm learning a lot. Just a small point relating to binary classification terminology, which I think is confusing on the best of days. In your notation: P( + |D) = probability of testing positive if disease is present (ie, the test sensitivity). Then P( + |ND) = 1 - specificity, where test specificity = P( - |ND), or the probability of testing negative if you don't have the disease. Accuracy, as I understand, is commonly defined as the total proportion of correct tests divided by all tests (ie true pos + true neg / (all outcomes)), but here you equate it with sensitivity, and then imply that P(+ |ND) = 1 - P(+ | D). (Which is this particular case it may be! - if P(+|D) = P(-|ND)). Thanks again for the great content.
if the evidence (disease present or absent) part is changing then this does not make sense. What makes sense is: 1 - Pr (+|present) = Pr (-|present). Herein the evidence is kept fixed "disease present".
If I get 9% of chance by having the disease given that I tested positive, does it mean that taking other test I have to update P(B) from 0.001 to 0.09? So It would be: P(B|A) = [(0.99)*(0.09)]/[(0.99)*(0.09) + (0.01)*(0.91)] = 90.7%
I think P(B) will remain like that - the disease is very rare. Rather to improve my belief (reduce false positives), I need to add more evidences (more test results): Pr (B | A, C, D) and so on.
That is belief calculus, not probably. You are calculating the strength of a belief given complete (100%) "belief" in the probabilities. What prob and stat fails to do is calculate the strength of the underlying belief, and assumes that strength is 100% certain. The truth is that, if there is cause and effect, the probabilities are meaningless, and the cause exists, or it doesn't.
A real Master class on Bayes' (really Bayes-Laplace) Theorem. The best comprehensive treatment of the concept. Wow!
Really enjoying the lecture series - thank you so much for taking the time, I'm learning a lot.
Just a small point relating to binary classification terminology, which I think is confusing on the best of days. In your notation: P( + |D) = probability of testing positive if disease is present (ie, the test sensitivity). Then P( + |ND) = 1 - specificity, where test specificity = P( - |ND), or the probability of testing negative if you don't have the disease.
Accuracy, as I understand, is commonly defined as the total proportion of correct tests divided by all tests (ie true pos + true neg / (all outcomes)), but here you equate it with sensitivity, and then imply that P(+ |ND) = 1 - P(+ | D). (Which is this particular case it may be! - if P(+|D) = P(-|ND)).
Thanks again for the great content.
thanks , i was confused on how P(+ |ND) = 1 - P(+ | D) was calculated , thanks
outstanding presentation. Too bad this didn't exist at the start of the semester (finals are a week away).
Amazing explanation of this concept.
Thank you for the informed video. I have one point, it may be obvious to you but it’s not obvious to someone who is new to stats.
I have a bit of an issue with your coin example because those are all independent tests... Otherwise great video, thanks!
Thanks a lot professor for this very useful series.
I have question , is correct to have: P(+|disease absent)=1-p(+|disease present) ?
if the evidence (disease present or absent) part is changing then this does not make sense. What makes sense is: 1 - Pr (+|present) = Pr (-|present). Herein the evidence is kept fixed "disease present".
perfect
For those of you taking this course; Is there a textbook that you're supposed to use with it?
There was a book?😮 smh I took it in may
I had a similar course in our uni, followed a book named "probability and stochastic process" by Roy Yates
@@dominicwhitfield943 Was that in a classroom and given credits for, or just online?
@@michealscott6198 Thank you. I will check it out.
Russell and Norvig Artificial Intelligence A Modern Approach
If I get 9% of chance by having the disease given that I tested positive, does it mean that taking other test I have to update P(B) from 0.001 to 0.09?
So It would be:
P(B|A) = [(0.99)*(0.09)]/[(0.99)*(0.09) + (0.01)*(0.91)] = 90.7%
I think P(B) will remain like that - the disease is very rare. Rather to improve my belief (reduce false positives), I need to add more evidences (more test results): Pr (B | A, C, D) and so on.
@@ashekatmvlg makes sense! Thank you
5:15 that distribution must be bimodal
That is belief calculus, not probably. You are calculating the strength of a belief given complete (100%) "belief" in the probabilities.
What prob and stat fails to do is calculate the strength of the underlying belief, and assumes that strength is 100% certain.
The truth is that, if there is cause and effect, the probabilities are meaningless, and the cause exists, or it doesn't.