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no point product with f, no normalization. This is only the backward message, implements the updates shown on pp. 545 eq (15.7).
You multiply the b with the corresponding f in each time slice to obtained the smoothed vectors sv, as shown on pp. 544 eq (15.6). So as written, there are two multiplications by f taking place! Eq 15.7 shows how the RHS of 15.6 breaks down, NOT the LHS. (BB)
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Thanks for the sanity check Mayerwin.
If someone cares to do the formatting, this is what my implementation yields for this scenario:
i: 1 fv: {s:Rain=0.8181818181818181, s:No Rain=0.18181818181818182}
i: 2 fv: {s:Rain=0.883357041251778, s:No Rain=0.1166429587482219}
i: 3 fv: {s:Rain=0.19066793972352525, s:No Rain=0.8093320602764746}
i: 4 fv: {s:Rain=0.7307940045849821, s:No Rain=0.2692059954150179}
i: 5 fv: {s:Rain=0.8673388895754848, s:No Rain=0.13266111042451526}
i: 5 B: {s:Rain=1.0, s:No Rain=1.0}
i: 4 B: {s:Rain=0.69, s:No Rain=0.41000000000000003}
i: 3 B: {s:Rain=0.4593, s:No Rain=0.2437}
i: 2 B: {s:Rain=0.090639, s:No Rain=0.15025100000000002}
i: 1 B: {s:Rain=0.06611763, s:No Rain=0.04550767}
i: 5 sv: {s:Rain=0.8673388895754848, s:No Rain=0.13266111042451526}
i: 4 sv: {s:Rain=0.8204190536236754, s:No Rain=0.17958094637632463}
i: 3 sv: {s:Rain=0.3074835760066177, s:No Rain=0.6925164239933823}
i: 2 sv: {s:Rain=0.8204190536236753, s:No Rain=0.17958094637632466}
i: 1 sv: {s:Rain=0.8673388895754848, s:No Rain=0.13266111042451528}
The interpretation of these smoothed vectors depends on the case usage, but for this example one conclusion we could draw is the best explanation of {umbrella, umbrella, no umbrella, umbrella, umbrella} seems to be {Rain, Rain, No Rain, Rain, Rain} (the Viterbi alg would explicitly output this). (BB)
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The word 'umbrella' appears suddenly in the middle of the example.
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