when calculating
$U_{CF}$(e,N)
CF: Confidence Level(here is 25%)
e:misclassifying counts of current subtree we focus on
N:counts of sub-datasets relevant to current subtree who is under judgment whether to be pruned or not.

“Pr” is short for “Pessimistic error rate”.

$U_{CF}$(e,N)=Pr= $\frac{e}{}$

The method to get $\ Coeff$ is as follows:

$\frac{Coeff-Deviation[i-1]}{Deviation[i]-Deviation[i-1]}=\frac{Confidence \ Level-val[i-1]}{val[i]-val[i-1]}②$

Now Let’s analyse the meaning of ② and how to get “i” in formula ②.

What is Deviation and Val?
Val[] = { 0, 0.001, 0.005, 0.01, 0.05, 0.10, 0.20, 0.40, 1.00},
Deviation[] = {4.0, 3.09, 2.58, 2.33, 1.65, 1.28, 0.84, 0.25, 0.00};

for Normal Distribution：
P{X>Deviation[i]}=val[i],
for example:
P{X>2.58}=0.005
--------------------

Now we analysis the whole meaning of formula ②

the red line is $P(X>x)=\int_x^{+∞}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$

the blue line is $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$
and the two points are:
(deviation[i],val[i])=(0.25,0.4)
(deviation[i-1,val[i-1])=(0.84,0.2)

Let’s amplify the above figure to the following picture:

so tan C= $\frac{Val[i]-Val[i-1]}{Deviation[i-1]-Deviation[i]}=\frac{Confidence\ Level-Val[i-1]}{Deviation[i-1]-Coeff}$

then we can get ②and compute $Coeff$ and $Coeff^2$

but how to get “i” in above formula?
when $Val[i-1]＜Confidence \ level≤Val[i]$
is satisfied,
we get “i”.
when Confidence Level=0.25
i=7
val[i-1]=0.2
val[i]=0.4
deviation[i-1]=0.84
deviation[i]=0.25
substitute the above values in ②
we get Coeff=0.6925
then,
$Coeff^2=0.479556$
--------------------

**Now Let’s calculate **
$U_{0.25}(1,16)$
CF=0.25
e=1
N=16
$Coeff^2=0.479556$
We’ll get :
Pr=

$\frac{1+0.5+ \frac{0.479556}{2}+ \sqrt{ 0.479556·\{ (1+0.5）[1- \frac{1+0.5}{16} ]+ \frac{0.479556}{4} \} } } {16+0.479556}$