22 votes

A and B are independent. Does P(A ∩ B|C) = P(A|C) · P(B|C) hold?

No this is not in general true, as you can see from a simple counter example: Toss two independent coins. Event $A$ is coin 1 head. $P(A)=0.5$ Event $B$ is coin 2 head. $P(B)=0.5$ Event $C$ is either ...
George Savva's user avatar
  • 2,054
4 votes
Accepted

How to incorporate prior knowledge after ML training?

You can either use an ensemble of multiple classifiers that will improve performance when your ML classifier breaks down[1], or implement boosting for your ML classifier assuming that it's a "...
Leif Peterson's user avatar
4 votes
Accepted

Can a ML classifier's prediction be understood as a probability?

That would be desirable, but it is not guaranteed to make as much sense as we might like. First, you could make an argument that any predicted $p(\mathcal C_k|\mathbf x_i)\in[0,1]$ is a probability in ...
Dave's user avatar
  • 60.9k
2 votes

A seeming paradox regarding estimation of the number of buttons

This, however, seems quite counter-intuitive to me, because (for small 𝑘) our inference only depends on our will regarding how many data points we want to generate! Yes, but only because we're ...
Cliff AB's user avatar
  • 20.7k
2 votes

A and B are independent. Does P(A ∩ B|C) = P(A|C) · P(B|C) hold?

A related question is If $X, Y$ are independent of $Z$, is $P(X|Y, Z) = P(X|Y)$? The example there, $C = XOR(A,B)$, is a simple counter example to this question as well ...
Sextus Empiricus's user avatar
1 vote
Accepted

Exercise involving Bayes' Theorem

The introduction of event $C$, which describes the inspector entering the room and observing rule-breaking, does not fundamentally change the calculation. Since the observation of students asking ...
ADAM's user avatar
  • 721
1 vote

Probability Question on Dice with condition

Assuming there is at most one pearl per box and the values are uniformly distributed between {4,5,6,7,8}. A general R function to calculate the probability: ...
jblood94's user avatar
  • 1,436
1 vote
Accepted

A seeming paradox regarding estimation of the number of buttons

You could sketch this situation more easily when $N \leq 2$ and we can model the two potential means $\mu_1, \mu_2$ with a uniform prior on a square. An equality $\mu_1 = \mu_2$ corresponds to the ...
Sextus Empiricus's user avatar
1 vote
Accepted

Understanding three prisoners (Statistical Inference - Cassella and Berger)

The wikipedia page for the problem helps pinpoint the reason for the 1/6 probability which I quote below. It also articulates the problem setting in a bit of a more verbose way which helps understand ...
giorgio's user avatar
  • 126
1 vote

A seeming paradox regarding estimation of the number of buttons

This one already has a few correct answers, but I think the intuition could be stronger. To start, let's set aside the noise temporarily, and assume that we can observe the $\mu_i$ values directly. ...
Eoin's user avatar
  • 8,867
1 vote

A seeming paradox regarding estimation of the number of buttons

We can predictively manipulate our inference regardless of what data we will get. If I decide to bug the person in the room one more time, I know beforehand that my best estimate would be N=k+1 . So I ...
Spätzle's user avatar
  • 3,559

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