Logic

[tibilicus]

Maths is completely alien to me. I understand foreign languages better than maths.

[The Stranger]

same... thats why im struggling, even tho tellos prolly explained it as clearly as possible im still finding it difficult to understand.

[The Stranger]

Sigh... its only getting more complicated, i dont even know why i should learn this -_-

if anyone knows anything about Bayesian probability please HELP!

prove the following axiomas, H and E are not contradictions (value of 0) or logical constructions such as the bachelor is unmarried (value of 1) (Pr means probablity)

a. Pr(H ∧ E) ≤ Pr(H) , and then ofcourse also Pr(H ∧ E) ≤ Pr(E)

b. If H → ¬E, then Pr(H | E) = 0

c. Pr(H | E) + Pr(¬H | E) = 1

[Kival]

Did you have some degree of set theory? Do you now that Pr(H ∧ E)=Pr(H ∩ E)? a and b are pretty simple then, c might be easy to, I did not check.

If not I need to know your definitions, because I've only made probability theory based on set theory.

[The Stranger]

no i dont have a degree tho i did know Pr(H ∧ E)=Pr(H ∩ E). i do understand the implication of all these things but i dont know how to write down the proof.

i guess the definitions are the same.

[Kival]

Oh, sorry, I did not want to ask if you have a degree, only if you know enough set theory. Okay, ignore my mistakes in english, I hope you can understand the ideas:

Okay let's work with set theory then. a is actually quite trivial but I'll try to give you an idea how to prove it:


Pr(E | H) = Pr(H ∩ E)/Pr(H) (Definition of baysian probability)
=> Pr(H ∩ E) = Pr(E | H)*Pr(H) (just simple transformation)

As Pr(E | H) has a value range in the interval [0;1] (it's a probability function), obviously it follows:

Pr(H ∩ E) <= Pr(E | H)*Pr(H)

Because of Pr(H | E)=Pr(H ∩ E)/Pr(E), the second part of a) follows trivially.

for b) you just need the equations and knew two things:

H → ¬E -> (H ∩ E) = {} and Pr ( {} )=0 ({} being the empty set)

c) Obviously (H ∩ E) and (H^(c) ∩ E) are disjoint sets, so you can use Sigma additivity (H^(c) being the complement of H) . Do you know where to go from there?

[The Stranger]

i get B and C.

I still dont get A however, why is the probability of (H&E) smaller or the same as the probability of (H)?

[Kival]

Pr(E | H) = Pr(H ∩ E)/Pr(H) (Definition of baysian probability)
=> Pr(H ∩ E) = Pr(E | H)*Pr(H) (just simple transformation)

As Pr(E | H) has a value range in the interval [0;1] (it's a probability function), obviously it follows:

Pr(H ∩ E) <= Pr(E | H)*Pr(H)

Because of Pr(H | E)=Pr(H ∩ E)/Pr(E), the second part of a) follows trivially.

From the above you can agree on:

Pr(H ∩ E) = Pr(E | H)*Pr(H) ?

Now just remember, what you know about the Pr-function. It's a function which gives you the probabilty of an event. The event is described as a set or as a logical combination of events. What do you know in general about the probability function? It has only values between 0 and 1. So 0<=Pr(A)<=1 for all possible events A. In mathematical detail, Pr(. | B), the baysian probability function is an own function but it's still a probability function and so it still has only values from 0 to 1. If you multiply Pr(H) with a value between 0 and 1, it can only be Pr(H) or smaller.