**Proof**

The total profit of the credit process will be determined by the formula (4):

$$
\hat{H} = \mathbf{M} \cdot (\mathbf{1} - \hat{\mathbf{x}} - \boldsymbol{\beta} \cdot (\mathbf{1} - \text{CAP}(\hat{\mathbf{x}}, \text{AR}, \text{ p})),
\tag{10}
$$

where *x*^ – optimal cut-off point. Obviously, the point *x*^ will be calculated from the condition as follows:

$$\frac{\partial \Pi}{\partial \mathbf{x}} = M \cdot (\boldsymbol{\beta} \cdot \mathbf{C} A P'\_{\mathbf{x}}(\hat{\mathbf{x}}, \ A R, \ p) - \mathbf{1}) = \mathbf{0},\tag{11}$$

Which is equivalent to the identity *EL*ð Þ¼ *x*^ *M* and corresponds **Figure 3**.

If the discrimination power AR is increased by a small amount Δ*AR* , then the formula for marginal income should be obtained from the expression:

<sup>Δ</sup>*<sup>Π</sup>* <sup>¼</sup> *<sup>d</sup>Π*^ *dAR* � Δ*AR* þ *o*ð Þ Δ*AR* . From (10) follows that

$$\pi = \frac{d\hat{\Pi}}{dAR} = M \cdot \left( -\frac{d\hat{\varkappa}}{dAR} + \beta \cdot \text{CAP}\_{\text{x}}' \cdot \frac{d\hat{\varkappa}}{dAR} \right) + \text{EL} \cdot \text{CAP}\_{AR}'(\hat{\varkappa}, AR, p).$$

The first part of this expression is equal to zero due to the condition (11), means remains as:

$$
\pi = \mathbf{E} \mathbf{L} \cdot \mathbf{C} \mathbf{A} P'\_{\mathcal{AR}}(\hat{\mathbf{x}}, \mathbf{A} \mathbf{R}, p) .
$$

Taking into account the volumes of placed funds *E*^, which are less than potential ones by 1 � *x*^, the resulting formula will be rewritten in the form as shown below:

$$\pi = \frac{\hat{E}}{1 - \hat{\mathfrak{x}}} \cdot \text{EL} \cdot \text{CAP}'\_{AR}(\hat{\mathfrak{x}}, AR, p). \tag{12}$$

To estimate the guaranteed value of the desired marginal profit π, it is necessary to carry out more transformations (12). Namely, if formula (7) is rewritten taking into account all arguments, *AR* <sup>¼</sup> <sup>2</sup>� Ð 1 <sup>0</sup> *CAP x*ð Þ , *AR*, *<sup>p</sup> dx*�<sup>1</sup> <sup>1</sup>�*<sup>p</sup>* ,

Then, after differentiation with respect to AR, we get the identity as:

2 <sup>1</sup>�*<sup>p</sup>* � Ð 1 0*CAP*<sup>0</sup> *AR*ð Þ *x*, *AR*, *p dx* ¼ 1 . Which can be used in formula (12) and it will be rewritten in the following form:

$$\pi = \hat{\mathbf{E}} \cdot \frac{\mathbf{E} \mathbf{L}}{2} \cdot \frac{\mathbf{1} - p}{\mathbf{1} - \hat{\mathbf{x}}} \cdot \frac{\mathbf{C} \mathbf{A} \mathbf{P}'\_{AR}(\hat{\mathbf{x}}, AR, p)}{\int\_0^1 \mathbf{C} \mathbf{A} \mathbf{P}'\_{AR}(\mathbf{x}, AR, p) d\mathbf{x}} \,. \tag{13}$$

Taking into account the mean value theorem (for example, [18], Theorem 5.19.), we can state that there are closed subsets *x*^ of the interval [0,1] on which the function of the right fractional part of the expression: (13)

*Risk Management Tools to Improve the Efficiency of Lending to Retail Segments DOI: http://dx.doi.org/10.5772/intechopen.108527*

$$\frac{\text{CAP}\_{AR}'(\hat{\mathbf{x}}, AR, p)}{\int\_0^1 \text{CAP}\_{AR}'(\mathbf{x}, AR, p)d\mathbf{x}} \ge 1\tag{14}$$

If we require a fairly simple property of the CAP-curve, namely that there be a unique root *x AR* <sup>~</sup>ð Þ∈ð Þ 0, 1 of equations *<sup>∂</sup>*<sup>2</sup> *<sup>∂</sup>x∂ARCAP x*ð Þ¼ , *AR* 0, then these subsets are the only segment<sup>6</sup> ½ � *<sup>x</sup>*1, *<sup>x</sup>*<sup>2</sup> <sup>∈</sup> ½ � 0, 1 *:*

Then, if *x*^ ∈ ½ � *x*1, *x*<sup>2</sup> , then we get a guaranteed estimate as:

$$
\pi \ge \hat{E} \cdot \frac{EL}{2} \cdot \frac{1-p}{1-\hat{x}} \cdot \hat{\pi}
$$

The cut-off level x exceeds the default probability level p. Indeed, the optimal cut- ^ off for an ideal model (AR ¼ 1) should be at a minimum level of x^ ¼ p, in this case, ð Þ 1 � *p =*ð Þ¼ 1 � *x*^ 1. For a real model, if it has 0 < AR<1, the cut-off level must be greater than for the ideal one equal to p, which means 1ð Þ � *p =*ð Þ 1 � *x*^ >1*:* Therefore, it can be argued that the segment ½ � *<sup>x</sup>*1, *<sup>x</sup>*<sup>2</sup> guaranteed level *<sup>π</sup>* <sup>&</sup>gt;*π*<sup>~</sup> occurs when *<sup>π</sup>*<sup>~</sup> <sup>¼</sup> *<sup>E</sup>*^ � EL <sup>2</sup> . ∎
