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I am looking for a comparison of different regression tree node splitting approaches within the random forest framework. I am looking at the trade-off between ensemble accuracy/reliability (holding forest size constant) and computational complexity of the split since I deal with large datasets.

The standard approach is to minimise $\sum_{i \in Region_1} (y_i - mean(y_i)_{Region1})^2$ +$\sum_{i \in Region_2} (y_i - mean(y_i)_{Region2})^2$. But another approach I have seen in a PhD thesis which drastically reduces the computational complexity is to look at the average of $(\sum_{i \in Region_1} y_i)^2/(Cardinality of Region_1)$ versus the same for $Region_2$; this allows me to use a running sum of $y_i$.

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up vote 4 down vote accepted

You can compute variance with the same cost as the averages. Simply put you have to use an on-line version of algorithm for computing variance. It is crystal clear explained on an article of John D. Cook. I used myself this online computing for a regression tree. I used to compute 2 running statistics, one starting from left and another one starting from right. It is possible to do that in a single step, I considered however that multiplication by a constant, gives also linear time and that does not bother me.

For more information you can take a look on Wikipedia also.

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Very nice. What do you mean by 'constant time'. Wouldn't it be linear time? –  Jase Feb 5 at 7:34
    
Sorry, I was meaning multiplication by a constant time, since I compute 2 running statistics. I will correct that in the answer. –  rapaio Feb 5 at 7:39
    
How do you handle the large amounts of noise in the variance estimate in the first few iterations (whether starting from front or back of the vector)? –  Jase Feb 6 at 15:12
    
@Jase I do not have an answer. What I usually do in practice is to adjust the parameter which decides the minimum number of instances in a node. If there is plenty of data, then is no problem to have a large value for this minimum. If it is not, then I try to take a look at my data and understand from the underlying distribution which would be a proper value for this parameter. I imagine that one could automate that by trying to estimate a confidence interval for the variance estimation, but I never tried to work on that. –  rapaio Feb 6 at 16:07

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