sequences of different length I have two sequences of coin tosses, which have different lengths. I want to compare their likelihoods under a model. But since the likelihood is a product of probabilities, the longer sequence ends up having a smaller likelihood. Should we normalise for the length of the sequence somehow?
Example:
Sequence 1: $1,0$
Sequence 2: $1,0,0$
$p_1(1)p_0(0) > p_1(1)p_0(0)p_0(0)$ since $p_0(0)<1$
ADDED: Purpose of the comparison: Actually I only have one sequence, but it consists of a concatenation of two sequences generated from two different models. The point of concatenation is not known. The model that generated the first sequence is known, the second model is not. I was trying to find the concatenation point by comparing the likelihood of the sub-sequence from start to index i with the likelihood of the sub-sequence from start to index i+1.
EDIT: The appended sequence is too short to estimate the second model from it simultaneously with finding the switching point.
 A: Assuming you have estimated the parameters of the second model somehow, you can then compute the likelihood of the entire sequence, start to finish, for a given concatenation point. Repeat this computation for every concatenation point and choose the one that maximizes the overall sequence likelihood.
This is actually very similar to a classic problem for a hidden markov model. In this context (a sequence of coin flip results, each coming from one of two possible coins with different biases), an HMM can be used to estimate the bias of each coin, as well as the probability of switching from each coin to the other. Then you would use something called the viterbi algorithm to find the most likely sequence of coins that produced the result sequence.
In this case, you would use a few extra constraints. You know the bias of the first coin, so during model training you would only reestimate the bias of the second coin and the transition probability. You also know that the sequence has to start with the first coin, and that the probability of transitioning from the second coin back to the first is zero. The viterbi algorithm would then output the most likely point where the switch occurred.
