Once again, I find that the only way I can get my graphs to work out to anything sensible is if I "fudge" the value of alpha over and over again in order to make the sum of all the probabilities add up to one. Must say though, I don't like this very much, even if the graph comes out fine, it somehow doesn't feel very "mathematical", it's like I am basically massaging the data to look the way I expect it to look.

I don't think this book is very good as a guide to distance education. It seems to be very thorough, but a lot of information does seem to arrive "out of thin air", with no background to it, which doesn't make me feel very confident in my understanding. I also wish it had a heck of a lot more examples as to how to practically implement all the abstract theory that is thrown around.]]>

Although at this point it doesn't matter anyway, having left the other half of the assignment out etc..]]>

In the second scenario, the ML probably wouldn't break disease B down into 2 seperate hypotheses (keeping it simple), but the Bayesian would probably split the hypotheses into 3 seperate hypotheses.

I'm afraid this is not based on anything more concrete than my very vague understanding of Bayesian statistics, so take anything I say with a pinch of salt!

Anyone else have any better ideas?]]>

eg

S2=(0.25*O2) where O2 is a P(h|d).

T2=...

..

W2

You then sum them all.

eg =SUM(S2:W2)

And this sum is what you graph.

]]>

I'm not sure how correct it is, but somehow, whether it is by hook or by crook, the graphs are coming out beautifully, heck knows how!]]>

I think I have cracked 20.1 (a), my graphs are correctly predicting the hypotheses for which I generated data. But, I am getting stuck on 20.1 (b), can't seem to make head or tail of this one :(

Anyway, it's late and I am sick of this question, suspect I am going to boot it and move onto the question, else I might end up circling around this forever :(]]>

alpha is worked out based on each sweet; for example if the number of sweets are 10, then for this instance, the probs for h1, h2, h3, h4, h5 must add up to one, and so on...

it will be better to use excel...

BTW did you guys get your assignment mark for cos492? I had to submit mine a day later due to the public holiday (only got internet connected a month ago); i haven't received mine yet; the lecture is supposedly not in.]]>

Maybe whats getting me is the alpha value would be different for each of the 10 selections from the bag?

Is that right.

It says on p713

QuoteAIMA

For example, suppose the bag is really an all-lime bag (h5) and the first 10 candies are all

lime; then P(d|h3) is 0.5^10, because half the candies in an h3 bag are lime

I am just guessing deperately aloud so if anyone can correct me, please do! Im struggling here looking for a straw.

If that's right then is:

P(d|h1) is 0.00^10

P(d|h2) is 0.25^10

P(d|h3) is 0.50^10

P(d|h4) is 0.75^10

P(d|h5) is 1.00^10

Thus if

P(hi|d) = ?P(d|hi)P(hi) then

after one lime sweet is picked

P(h1|d) = alpha * 0.00 * 0.1

P(h2|d) = alpha * 0.25 * 0.2

P(h3|d) = alpha * 0.50 * 0.4

P(h4|d) = alpha * 0.75 * 0.2

P(h5|d) = alpha * 1.00 * 0.1

and after 10 then

P(h1|d) = alpha * 0.00^10 * 0.1

P(h2|d) = alpha * 0.25^10 * 0.2

P(h3|d) = alpha * 0.50^10 * 0.4

P(h4|d) = alpha * 0.75^10 * 0.2

P(h5|d) = alpha * 1.00^10 * 0.1

But the graph must be putting in the alpha to get them all to add to 1.0?

Do with one sweet then alpha appears to be 2.0408163265306122448979591836735

h1:- 0.00 --> 0.00 (read off graph)

h2:- 0.04 --> 0.10 "

h3:- 0.20 --> 0.40 "

h4:- 0.15 --> 0.30 "

h5:- 0.10 --> 0.20 "

And for 10: it looks like alpha is 8.9562784214176628754360370763906

h1:- 0.00000000000000000000 --> 0.0? (read off graph)

h2:- 0.00000019073486328125 --> 0.0? "

h3:- 0.00039062500000000000 --> 0.0? "

h4:- 0.01126270294189453125 --> 0.10 "

h5:- 0.10000000000000000000 --> 0.90 "

I would thus assume to answer the question 20.1 we need to calculate what alpha would be for all ten different times we pick the sweet for all other 4 hypotheses?

That would be 10 * 5 * 4 = 20 * 10 times = 200 times (which involves solving for alpha a lot). In order to plot the graph? Well not just alpha. But all five points as well. Thats a lot. I plan to do this tonight.

This seems like a lot of work. So before I begin... am I right, or have I lost the plot.

Any comments, thoughts, suggestions... please don't hesitate to tell me I will appreciate it. Especially before I attack this question tonight.

:)o]]>

your help would be greatly appreciated.]]>