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Final

Answers to chapters 4 and 5 for final examination.

11 Pages
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Department
Philosophy
Course Code
PHL246H1
Professor
Colin Howson

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Answers provided by Charles Dalrymple-Fraser.
NOTE: Im going to recap a little bit of chapter 4 here in my own words and understanding.
These two pages are in no way a substitute for the actual chapter. This is me trying to clarify
things and create something I can refer back to as I study. I thought you may find it useful when
you hit a term in a question youre unfamiliar with, or as a footnote to the chapter. This summary
only includes up to page five of chapter four so that you can dig into the questions better
(especially if you missed that lecture like I did). This part may or may not help you it may even
confuse you, though I hope not. The answers begin shortly into this .pdf if you want to skip this.
=====================================================================================================
=====================================================================================================
Induction is the process of evaluating hypotheses on the basis of evidence obtained by
observation and experiment. This is why we use Prob(H) and Prob(E) primarily, rather than
Prob(A) and Prob(B).
Prob is a personal probability. Prob(H) is different than P(H). When writing this upcoming test
double check that you are writing Prob(H) instead of P(H)
We want to evaluate hypotheses on the basis of evidence. In chapter three, we interpreted
Prob(A|B) as a conditional fair betting quotient in a bet on A. We can apply this to hypotheses
and evidence to get Prob(H|E) being our conditional fair betting quotient in a bet on H; that is, we
are betting that the hypothesis H is correct and Prob(H|E) is our personal probability or fair
betting quotient that H is correct, given the evidence E.
So now we introduce Prob(H|E) into Bayes Theorem as you would for Prob(A|B).
If Prob(E)>0 then Prob(H|E) = Prob(E|H)Prob(H)
Prob(E)
Prob(E|H) is called the ‘likelihood
Prob(H) is called the prior probability of H. Basically, it is the probability of H before or prior to
the evidence.
Prob(H|E) is called the posterior probability of H. Simply, its the probability of H after/posterior
to the evidence.
Note that we can keep adding evidence over time. As this happens, we have to keep evaluating
for a current Prob(H|E). This also means that last weeks Prob(H|E) can be today’s Prob(P) as it
is now the probability of H prior to the new evidence.
We always assume that Prob(H) is nonzero because it seems ridiculous that you would be
evaluating a hypothesis that you dont think can ever be true based on evidence.
www.notesolution.com
Answers provided by Charles Dalrymple-Fraser.
If a hypothesis predicts an effect which can be described by E, then Prob(E|H)=1. If we sub that
into the theorem above, we get that Prob(H|E)=Prob(H)/Prob(E). If Prob(E) is less than 1, then
Prob(H) is less than Prob(H|E).
So, if H predicts something which can be described by E, and it occurs, then that successful
prediction increases the probability of H. If H predicts E, then Prob(E|H)=1 and Prob(H|E) is
stringent upon Prob(H) and Prob(E). If Prob(E) is less than one, then Prob(H|E)>Prob(H) and so,
when E is observed, the probability of H increases. Note that the less likely that E will occur [e.g.
Prob(E)], the greater the probability of H increases. If Prob(E)=1, then it was determined to occur
absolutely and the probability of H doesnt change since E was 100% likely to occur anyways.
For example, if H predicts that the watermelon I bought at Sobeys will explode when I finish
typing this sentence, then, if the watermelon does explode, I will say that H is a lot more
probable because it is very unlikely that my watermelon will explode.1 Likewise, if H predicts that
something will get wet if it rains, and it rains and something gets wet, we wouldnt really say that
H is that much more probable because, hey, stuff gets wet when it rains already.
This stuff seems odd at times with all the Probs floating about, but try to think of it in real life
terms and it should help clear things up.
The empirical support of a hypothesis given evidence is represented as S(E,H) and is calculated
via the equation S(E,H)=Prob(H|E)-Prob(H). The support is positive when Prob(H|E)>Prob(H),
negative when Prob(H)>Prob(H|E), and zero when they are equal. In words, the support is a
positive support when, evidence considered, the probability of H increases. The support is
negative when the probability of H decreases given the evidence (for example, if my H predicts
with absolute certainty that my watermelon will explode after that sentence up there, but my
watermelon did not explode, then the probability of my H being true is obviously knocked down a
fair bit: the support is negative to my H). The support for the hypothesis is zero when the
evidence doesnt change our probability of H (for example, if my H predicts that when a coin is
flipped on the moon, there will be half heads and half tails and I flip a coin on the moon once and
get a heads, that one flip doesnt really support my hypothesis at all, though two flips might).
Now onto some questions…
1 For those concerned, I note that my watermelon is fine and dandy having concluded that sentence. Watermelon > H.
www.notesolution.com
Answers provided by Charles Dalrymple-Fraser.
1i. Show that if H => E, Prob(E|H) = 1
We know if H=>E then Prob(E|H)=1from chapter and common sense
Prob(E|H) = Prob(E^H) from rule five
Prob(H)
Now, if H=>E, then HH^Efrom chapter one
Prob(E^H)=Prob(H) from rule one
Prob(E|H) = Prob(H)= 1subbing into rule five
Prob(H)QED
1ii. Show that if H => ~E then Prob(E|H) = 0
If H=>E then Prob(~E|H)=1from 1i
Prob(~E|H) = 1 Prob(E|H) from chapter two (see proof in other emails)
1 = 1 Prob(E|H)substitution
0 = Prob(E|H) rearranging and solving
QED
www.notesolution.com

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Description
Answers provided by Charles Dalrymple-Fraser. NOTE: Im going to recap a little bit of chapter 4 here in my own words and understanding. These two pages are in no way a substitute for the actual chapter. This is me trying to clarify things and create something I can refer back to as I study. I thought you may find it useful when you hit a term in a question youre unfamiliar with, or as a footnote to the chapter. This summary only includes up to page five of chapter four so that you can dig into the questions better (especially if you missed that lecture like I did). This part may or may not help you it may even confuse you, though I hope not. The answers begin shortly into this .pdf if you want to skip this. ===================================================================================================== ===================================================================================================== Induction is the process of evaluating hypotheses on the basis of evidence obtained by observation and experiment. This is why we use Prob(H) and Prob(E) primarily, rather than Prob(A) and Prob(B). Prob is a personal probability. Prob(H) is different than P(H). When writing this upcoming test double check that you are writing Prob(H) instead of P(H) We want to evaluate hypotheses on the basis of evidence. In chapter three, we interpreted Prob(AB) as a conditional fair betting quotient in a bet on A. We can apply this to hypotheses and evidence to get Prob(HE) being our conditional fair betting quotient in a bet on H; that is, we are betting that the hypothesis H is correct and Prob(HE) is our personal probability or fair betting quotient that H is correct, given the evidence E. So now we introduce Prob(HE) into Bayes Theorem as you would for Prob(AB). If Prob(E)>0 then Prob(HE) = Prob(EH)Prob(H) Prob(E) Prob(EH) is called the likelihood Prob(H) is called the prior probability of H. Basically, it is the probability of H before or prior to the evidence. Prob(HE) is called the posterior probability of H. Simply, its the probability of H afterposterior to the evidence. Note that we can keep adding evidence over time. As this happens, we have to keep evaluating for a current Prob(HE). This also means that last weeks Prob(HE) can be todays Prob(P) as it is now the probability of H prior to the new evidence. We always assume that Prob(H) is nonzero because it seems ridiculous that you would be evaluating a hypothesis that you dont think can ever be true based on evidence. www.notesolution.comAnswers provided by Charles Dalrymple-Fraser. If a hypothesis predicts an effect which can be described by E, then Prob(EH)=1. If we sub that into the theorem above, we get that Prob(HE)=Prob(H)Prob(E). If Prob(E) is less than 1, then Prob(H) is less than Prob(HE). So, if H predicts something which can be described by E, and it occurs, then that successful prediction increases the probability of H. If H predicts E, then Prob(EH)=1 and Prob(HE) is stringent upon Prob(H) and Prob(E). If Prob(E) is less than one, then Prob(HE)>Prob(H) and so, when E is observed, the probability of H increases. Note that the less likely that E will occur [e.g. Prob(E)], the greater the probability of H increases. If Prob(E)=1, then it was determined to occur absolutely and the probability of H doesnt change since E was 100% likely to occur anyways. For example, if H predicts that the watermelon I bought at Sobeys will explode when I finish typing this sentence, then, if the watermelon does explode, I will say that H is a lot more probable because it is very unlikely that my watermelon will explode. Likewise, if H predicts that something will get wet if it rains, and it rains and something gets wet, we wouldnt really say that H is that much more probable because, hey, stuff gets wet when it rains already. This stuff seems odd at times with all the Probs floating about, but try to think of it in real life terms and it should help clear things up. The empirical support of a hypothesis given evidence is represented as S(E,H) and is calculated via the equation S(E,H)=Prob(HE)Prob(H). The support is positive when Prob(HE)>Prob(H), negative when Prob(H)>Prob(HE), and zero when they are equal. In words, the support is a positive support when, evidence considered, the probability of H increases. The support is negative when the probability of H decreases given the evidence (for example, if my H predicts with absolute certainty that my watermelon will explode after that sentence up there, but my watermelon did not explode, then the probability of my H being true is obviously knocked down a fair bit: the support is negative to my H). The support for the hypothesis is zero when the evidence doesnt change our probability of H (for example, if my H predicts that when a coin is flipped on the moon, there will be half heads and half tails and I flip a coin on the moon once and get a heads, that one flip doesnt really support my hypothesis at all, though two flips might). Now onto some questions 1For those concerned, I note that my watermelon is fine and dandy having concluded that sentence. Watermelon > H. www.notesolution.com
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