day1 day2 day3
working .63 ----> .1134 ---> .0102
\
\
\
broken .01 > .0252 ---> .0126
Adit Dalvi wrote:
> And I'm getting Working, Working, Broken for the Viterbi sequence :\
>
> In article <gifcg2$1m8d$1...@agate.berkeley.edu>, cs18...@imail.EECS.Berkeley.EDU
> (Adit Dalvi) writes:
>> I'm getting 0.419 for broken on day 2 (given slow, slow) so I think you're
>> right ... it should be day 3 in my opinion.
>>
>> In article <gif019$1ii8$1...@agate.berkeley.edu>, gar...@imail.EECS.Berkeley.EDU
>> (Gary Mindlin Miguel) writes:
>>> In article <giev6b$1ibe$1...@agate.berkeley.edu>,
> gar...@imail.EECS.Berkeley.EDU
>>> (Gary Mindlin Miguel) writes:
>>>> To whoever is still checking this thing:
>>>> is the solution to written assignment 9 (which incorrectly has "written
>>>> assignment 8" at the top of the page) question 2 (b) correct?
>>>>
>>>> Namely,
>>>> it doesn't seem to incorporate the new evidence (O2 = slow) at all.
> Should
>> we
>>>> be using the forward algorithm to solve this? If so, shouldn't the
>>> calculation
>>>> be:
>>>> for the case where X2 = working:
>>>> P(O2=slow | X2 = working)*(sum over all x1)[P(X2=working | x1) * P(x1,
>>>> O1=slow)]
>>>>
>>>> ?
>>>>
>>>> Many thanks if someone could clarify this.
>>>>
>>>> Gary
>>> Update:
>>> I worked out the arithmetic using the forward algorithm, and I got pretty
>> much
>>> the same answer as in the solution. I guess they're equivalent. I just
> don't
>>> see how immediately.
>>>
>>> I'm pretty sure the answer to (c) is wrong, though. The P(broken |
> slow,slow)
>> <
>>> 0.5, as the solutions show. So it must be day 3 or 4.