Show that the Laplace smoothing for bigrams is a valid probability distribution

Question

If we consider any smoothing technique like laplace or delta smoothing. Intuitively we can see that the we are stealing from sequences with non zero probablity and re distribute to sequences with zero probablity and there by it should be a probability distribution function. But how do we prove this mathematically say for example laplace smoothing?

Answer 1

First, let me rewrite your formula: $$ \Pr_L[w_j|w_i] = \frac{\#(w_i,w_j)+1}{\#w_i+|\Sigma|}. $$ The quantity $\#w_i$ should be defined so that $\sum_j \#(w_i,w_j) = \#w_i$ (so we should ignore the final character when calculating $\#w_i$). Therefore $$ \sum_{w_i \in \Sigma} \Pr_L[w_j|w_i] = \frac{\sum_{w_i \in \Sigma} \#(w_i,w_j) + \sum_{w_i \in \Sigma} 1}{\#w_i + |\Sigma|} = \frac{\#w_i + |\Sigma|}{\#w_i + |\Sigma|} = 1. $$