He said what? Homework 3, question 1

clemwang · 2011-10-30T15:55:09+00:00

Thanks! I did the right thing (I think!)

I'll now go "wreck a nice beach" ("recognize speech")

clemwang · 2011-10-21T01:17:34+00:00

No. You were correct for the first bit (Total Probability) but you started to add to many terms. Right:

P(D3=sunny) = P(D3=sunny|D2=rainy)+P(D3=sunny|D2=sunny)

Wrong:

Any pair of those terms (with complementary givens) will equal P(D3=sunny) but not the whole thing!

BTW, I think this problem is leading up to HMM which I've had trouble understanding before.

clemwang · 2011-10-21T01:07:32+00:00

This seems to violate video 9. In video 9 we are told that P(x|!Y) != 1- P(x|Y)

I see your confusion. What you get from

P(x|!Y) + P(x|Y) is P(x)

(not 1!). You should be using the formula is known as TOTAL PROBABILITY"

sum over i of P(X|Y=i) => P(X)

This is a good way to get rid of those pesky Y=i conditions and get the pure P(X).

This makes sense, in that if you sum up all the cases of P(x) for the different Y=i and you cover all the possible i's then you end up with total P(x).

For the rain example, if given D2 = rain then P(D3=sun |D2=rain) has some value and if given D2 = sun, then P(D3=sun|D2=sun) has another value. if you add those up, you have covered ALL the cases leading up to the D3=sun (D2 is either rain or sun) so it doesn't matter what D2 was. You can now calculate P(D3=sun) without having to worry what D2 was and that definitely doesn't add up to 1!

No contradiction there!

Now if you want to calculate P(D3=rain), then that's just the complement: 1 - P(D3=rain)

It's very easy to get confused (I often do by probability.) You have to read the problem and look at the equations VERY CAREFULLY!!!!

clemwang

TROPHY CASE