<p>I think the HMM-based <a href="http://www.coli.uni-saarland.de/~thorsten/tnt/" rel="nofollow">TnT tagger</a> provides a better approach to handle unknown words (see the approach in <a href="http://www.coli.uni-saarland.de/~thorsten/publications/Brants-ANLP00.pdf" rel="nofollow">TnT tagger's paper</a>). </p>

<p>The accuracy results (for known words and unknown words) of TnT and other two POS and morphological taggers on 13 languages including Bulgarian, Czech, Dutch, English, French, German, Hindi, Italian, Portuguese, Spanish, Swedish, Thai and Vietnamese, can be found in <a href="http://arxiv.org/abs/1412.4021" rel="nofollow">this article</a>.</p>
Part of speech tagging : tagging unknown words
<p>In the part of speech tagger, the best probable tags for the given sentence is determined using HMM by</p>

<pre><code>    P(T*) = argmax P(Word/Tag)*P(Tag/TagPrev)
              T
</code></pre>

<p>But when 'Word' did not appear in the training corpus, P(Word/Tag) produces ZERO for given all possible tags, this leaves no room for choosing the best. </p>

<p>I have tried few ways, </p>

<p>1) Assigning small amount of probability for all unknown words, P(UnknownWord/AnyTag)~Epsilon... means this completely ignores the P(Word/Tag) for unknowns word by assigning the constant probability.. So decision making on unknown word is by prior probabilities.. As expected it is not producing good result. </p>

<p>2) Laplace Smoothing 
I confused with this. I don't know what is difference between (1) and this.   My way of understanding Laplace Smoothing adds the constant probability(lambda) to all unknown &amp; Known words.. So the All Unknown words will get constant probability(fraction of lambda) and Known words probabilities will be the same relatively since all word's prob increased by Lambda.
Is the Laplace Smoothing same as the previous one ?</p>

<p>*)Is there any better way of dealing with unknown words ?</p>