Q19
I was wondering if the problem assumes that some learning has been done to determine P(Dh=f) for some population or if the person with the heart attack is the only person in D. Obviously, I may not understand Bayesian analysis.

Re: Q20
Thank you very much for the quick reply. This has been an amazing course!!

Re: Q20
I am still a bit confused about the setup of the problem. Is it correct to assume that what we are trying to determine is the underlying probability of somebody picked at random from the population to have a heart attack out of a single sample? If so, shouldn't be relevant? If a single point is all we have, call it the binary variable  equal to 1 if the patient has a heart attach; 0 if he doesn't, that would be the probability of generating a single point with a patient having a heart attack, given the underlying probability that a person has a heart attack, right? In that case, the posterior is going to have two cases and . The question refers only to case right?

Re: Q20
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It should be based on rater than out of. A source of confusion here is that is a probability, but then we have a probability distribution over . Let us just call the fraction of heart attacks in the population. Then the problem is addressing the probability distribution of that fraction  Is the fraction more likely to be 0.1 or 0.5 or 0.9 etc. The prior is that that fraction is equally likely to be anything (uniform probability). The problem then asks how this probability is modified if we get a sample of a single patient and they happen to have a heart attack. If I have not answered your question, please ask again perhaps in those terms. 
Re: Q20
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Re: Q20

Re: Q20
If I understand the problem correctly, P(X=1) is independent on P(h=f). Correct?

Re: Q20
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The reason why you can ignore P(X=1) is because in Bayesian analysis you usually don't care about the absolute probability of the dataset since it is just a constant that all of your hypotheses are divided by, equally, so it doesn't affect which hypothesis is aposteriori most probable. 
Re: Q20
These are some fine points here. We have to use with exact meaning of terms, using them loosely can create misunderstandings.
At a first reading I thought that Haowen's answer was not correct, and also ilya's remark was not correct too. On a second reading, Haowen's answer is correct, but I am not sure that it answers the initial question, since the initial question/remark by ilya was ambiguous. Let me explain. When we talk about independency we can talk about it in probability terms, where we have specific rules on random variables being independent, or we can talk about it in more loose/everyday terms when we want to express that something affects something else. Ilya's question is expressed loosely. It talks about independence of Probabilities not random events. It can be taken with several different meanings. 1) If you are really asking whether X=1 and h=f are independent events, then we can clearly say they are not. The choice of h clearly affects the probability of X=1. More specifically, the choice of h is the probability of X=1. 2) If you are asking whether the distribution of h=f affects P(X=1) for all possible h, (which can be taken as the more literal interpretation of what you are asking) then again: yes there is a connection and Haowen gives you the formula. 3) If you are asking in general "should we care about calculating the value for P(X=1)", then Haowen gives you the answer again. 4) If you are asking whether the event X=1 affects the probability of h=f, it depends whether you are really referring to the apriori or the aposteriori. It does not affect the apriori and it does affect the aposteriori (and Q20 asks how). 
Re: Q20
Need help in verifying if below understanding is correct ?
The Bayesian: P(h=f  D) = P(D  h=f) * P(h=f) / P(D) For this Q, we are given: P(h=f) is uniform in [0,1] D: onepersonwithheartattack Pick f = c (constant) To simplify, I assume that h and f are a discrete randomvariables with 10 possible values from (0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0) and each is equally likely with P=1/10. Essentially simplifying here to make P(h=f) a pmf which is actually a pdf. Now: P (D  h=f) = Pr( onepersonwithheartattack  h=f ) = Probability of onepersonwithheartattack, given (h=f) = c ( because if h=f were given, then the Prob of one picked person getting heartattack is c, as defined by f ) Plug in above to get: P(h=f  D) = c * P(h=f) / P(D) Does above sound correct ? Also P(D) =1 in this case ? Thanks. 
Re: Q20
I find this exercise simple but very useful. If one thinks of the series of following measurements (1s and 0s for heart attack or not) one can clearly form an idea how this transforms stepbystep from a uniform distribution to a Bernoulli one.
Does this mean that this example represents one of those cases where the initial prior is irrelevant and we can safely use it for learning? Also, is this some form of reinforcement learning? thanks, Dorian. 
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Re: Q20
I am very confused by this problem. Perhaps this questions will help:
Is P(Dh=f) a function of D or h or both? It looks to me like it's a function of D, but we need to convert it to a function of h to get the posterior...:clueless: Is this correct? 
Re: Q20
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Since is fixed, this leaves as a function of only. 
Re: Q20
There is an analogy that may be enlightening, which I thought of because of the presentation of the first part of this course.
Suppose you have a large number of urns each containing a large number of black and white balls in varying proportions. You are told how many urns there are with each proportion. Then you go up to one of the urns and take out a ball which you find is black. The question is how likely it is that specific urn has each particular fraction of black balls. 
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