Pessimistic Error Pruning example of C4.5

This example is from 《An Empirical Comparison of Pruning Methods
for Decision Tree Induction》
Pessimistic Error Pruning example of C4.5
How to read these node and leaves?
For example:
node 30:
15 are classified as “class1”
2 are mis-classified as “class1”
you can reduce the rest nodes or leaves from above

criterion :
$n'(T_t)+SE(n'(Tt))<n'(t)①$
where
$SE(n'(Tt))=\sqrt{\frac{n'(T_t)·（N(t)-n'(T_t)）}{N(t)}}$
Be short :
Errors when unpruned<Errors after pruned

when ① is satisfied ,the current tree remains,
otherwise, it will be pruned.

The principle why above Algorithm always take effect
B(n,p)->N( np,np(1-p) )
Pessimistic Error Pruning example of C4.5
Picture Reference
:https://stats.stackexchange.com/questions/213966/why-does-the-continuity-correction-say-the-normal-approximation-to-the-binomia/213995

when in reverse,we set a continuity corretion for binomial distribution:
we use “x+0.5” to make these two curse closer(of course this is not accurate enough)，then you can use theory of Normal distribution with x+0.5
of course 0.5 is not rigorous,here is just approximation

Why the standard error occur in the criterion?
$n'(T_t)+SE(n'(Tt))<n'(t)$
$<=>n'(T_t)+\sqrt{\frac{n'(T_t)·（N(t)-n'(T_t)）}{N(t)}}<n'(t)$
Let’s see an example:
$Y=X_1+X_2+X_3+X_4$
$X_i$ will fluctuate and Y will fluctuate(I mean they are all variables,Not Constant).
then ,when does Y reach maximum?
Now if we have 4 values Y ever have produced.
1,2,1,1 ②
then average Y̅= $\frac{1}{4}(1+2+1+1)=1.25$
Standard Deviation= $\sqrt{\frac{1}{4}\{(1-1.25)^2+(2-1.25)^2+(1-1.25)^2+(1-1.25)^2\}}$ =0.43
so when
Y̅+Standard Deviation=1.25+0.43=1.68≈2.0

Conclusion 1：
All above means that when Y̅+Standard Deviation,we’ll get a value nearest to the maximum in②
------------------------------------------
Let’s come back to Errors we focus just now:
regard Y as the total number of Errors of un-pruned Tree:
Assume（Such Assumption is of course Not rigorous～！）:
Y̅= $n'(T_t)$
$X_i$ :Error number of the $i_{th}$ leaf
Standard Deviation: $SE(n'(Tt))$

just like the conclusion 1:
$n'(T_t)+SE(n'(Tt))$ means that:
we’ll get a value nearest to the maximum number among possible values of “errors of un-pruned tree”.
Attention please that we assume “errors of un-pruned tree” as a variable,Not constant,
which is used to get the " maximum possible error numbers".
The reason why we call it"pessimistic" is just from $SE(n'(Tt))$
this item means:“pessimistic Error counts”

Note:
There’s a complaint from part2.2.5 of《An Empirical Comparison of Pruning Methods for Decision Tree Induction》for PEP that:
"The statistical justification of this method is somewhat dubious"☺
So the principle of PEP is Not rigorous.

After Principle ，Computation comes:
For pruned-tree,Error counts: $n'(t)=15+0.5$
For un-pruned-tree,Error counts:
$n'(T_t)+SE(n'(Tt))$
$n'(T_t)=2(node 30)+0(node 31)+6(node 28)+2(node 29)+continuation\ Errors=10+4（node 30,node31,node28,node29）·0.5=12$
$pessmistic\ error \ counts=SE(n'(Tt))=\sqrt{\frac{12·（35-12）}{35}}=2.8$
then
$n'(T_t)+SE(n'(Tt))=12+2.8=14.8<15.5=n'(t)$
So,this tree should be kept and Not pruned

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