I don't believe such a cutoff exists, although the variable importance plots can be informative. Carry out two experiments. Rerun the random forest and see how the list changes. Delete an observation and also observe. In my experience, the answer relates to what is the goal of feature selection. For example, why not use every variable to make predictions. The random forest can easily do that. We usually use feature selection for a reason, for example, seeking a rule using just a small number of features that can easily be measured in the future. In my case, the number was set by the technology I plan to use the diagnostic on. More importantly, if you are using feature selection, this has to be repeated at each iteration of cross-validation. Searching this site for "cross-validation", "feature selection", and "stepwise regression" will give you a start.
You simply did not used the target class variable at all. Gini impurity as all other impurity functions, measures impurity of the outputs after a split. What you have done is to measure something using only sample size.
I try to derive formula for your case.
Suppose for simplicity you have have a binary classifier.
Denote with $A$ the test attribute, with $C$ the class attribute which have $c_+, c_-$ values.
The initial gini index before split is given by
$$I(A) = 1 - P(A_+)^2 - P(A_-)^2$$
where $P(A_+)$ is the proportion of data points which have $c_+$ value for class variable.
Now, impurity for left node would be
$$I(Al) = 1 - P(Al_+)^2-P(Al_-)^2$$
$$I(Ar) = 1 - P(Ar_+)^2-P(Ar_-)^2$$
where $P(Al_+)$ is proportion of data points from left subset of $A$ which have value $c_+$ in the class variable, etc.
Now the final formula for GiniGain would be
$$GiniGain(A) = I(A) - p_{left}I(Al) - p_{right}I(Ar)$$
where $p_{left}$ is the proportion of instances for the left subset, or $\frac{\#|Al|}{\#|Al|+\#|Ar|}$ (how many instances are in left subset divided by the total number of instances from $A$.
I feel my notation could be improved, I will watch later when I will have more time.
Conclusion
Using only number of data points is not enough, impurity mean how well one feature (test feature) is able to reproduce the distribution of another feature (class feature). Test feature distribution produces the number you used (how to left, how to right), but distribution of the class feature is not used in your formulas.
Later edit - proove why it decrease
Now I noticed that I missed the part which proves why it always the gini index on child node is less than on parent node. I do not have a complete proove or a verified one, but I am thinking is a valid proof. For other interenting thing related with the topic you might check Technical Note: Some Properties of Splitting Criteria - Leo Breiman. Now it will follow my proof.
Suppose that we are in the binary case, and all the values in a node could be completely described by a pair $(a,b)$ with the meaning of $a$ instances of the first class, and $b$ instances of the second class. We can state than that in the parent node we have $(a,b)$ instances.
In order to find the best split we sort the instances according with a test feature and we try all the binary possible splits. Sorted by a given feature is actually a permutation of instances, in which classes starts with an instance of the first class or of the second class. Without loosing the generality, we will suppose that it starts with an instance of the first class (if this is not the case we have a mirror proof with the same calculation).
The first split to try is in the left $(1,0)$ and in the right $(a-1,b)$ instances. How the gini index for those possible candidates for left and right child nodes are compared with the parent node? Obviously in the left we have $h(left) = 1 - (1/1)^2 - (0/1)^2 = 0$. So on the left side we have a smaller gini index value. How about the right node?
$$h(parent) = 1 - (\frac{a}{a+b})^2 - (\frac{b}{a+b})^2$$
$$h(right) = 1 - (\frac{a-1}{(a-1)+b})^2 - (\frac{b}{(a-1)+b})^2$$
Considering that $a$ is greater or equal than $0$ (since otherwise how could we separate an instance of the first class in the left node?) and after simplification it's simple to see that the gini index for the right node has a smaller value than for the parent node.
Now the final stage of the proof is to node that while considering all the possible split points dictated by the data we have, we keep the one which has the smallest aggregated gini index, which means that the optimum we choose is less or equal than the trivial one which I prooved that is smaller. Which concludes that in the end the gini index will decrease.
As a final conclusion we have to note even if various splits can give values bigger that parent node, the one that we choose will be the smallest among them and also smaller that the parent gini index value.
Hope it helps.
Best Answer
My understanding is that the sum is over all nodes where the predictor variable is used. In fact, a predictor variable can be used more than once in a given tree to split a node or not at all.
My understanding is that at each node, a random subset of predictors is selected. From these predictors, the one that most reduces the impurity is then selected. ``Impurity'' can be quantified by different measures. Two that I have come across are the Gini impurity and the entropy measure.
I believe that there is a chance that a predictor is never used (although very small chance if you build a lot of trees). It that case, the importance of type 2 is zero.