On the model updating operators in univariate estimation of distribution algorithms

Abstract

The role of the selection operation—that stochastically discriminate between individuals based on their merit—on the updating of the probability model in univariate estimation of distribution algorithms is investigated. Necessary conditions for an operator to model selection in such a way that it can be used directly for updating the probability model are postulated. A family of such operators that generalize current model updating mechanisms is proposed. A thorough theoretical analysis of these operators is presented, including a study on operator equivalence. A comprehensive set of examples is provided aiming at illustrating key concepts, main results, and their relevance.

Appendix

Proof of Theorem 1

Under conditions (a)–(e) function $\gamma _1$ is differentiable in $( - \infty ,0]$ and its convexity implies that

$$\begin{aligned} \frac{d}{{dy}}\gamma _1 (y) = \frac{{\varphi '\left( {e^y } \right) e^y }}{{\varphi \left( {e^y } \right) }} \end{aligned}$$

is increasing in $( - \infty ,0]$. This is obviously equivalent to have $F(x) = \frac{{\varphi '(x)x}}{{\varphi (x)}}$ as an increasing function in $(0,1]$. Solving this differential equation w.r.t. $\varphi$ we see that the function $\varphi$ can be expressed as $\varphi (x) = Cx^a e^{\int \limits _0^x {\frac{{F_0 (y)}}{y}} dy}$, where an arbitrary constant $C$ should be chosen such that $\varphi (1) = 1$, i.e., $C = e^{ - \int \limits _0^1 {\frac{{F_0 (y)}}{y}} dy}$.

Let us show that the value $a \geqslant 1$(clearly $a > 0$ in order to have $\varphi (0) = 0)$. Denote $\varphi _0 (x) = Ce^{\int \limits _0^x {\frac{{F_0 (y)}}{y}} dy}$. Then

$$\begin{aligned} \varphi (x)= & {} x^a \varphi _0 (x);\\ \varphi '(x)= & {} x^a \varphi '_0 (x) + ax^{a - 1} \varphi _0 (x) = x^{a - 1} \varphi _0 (x)F_0 (x) + ax^{a - 1} \varphi _0 (x) \\= & {} x^{a - 1} \varphi _0 (x)\left( {F_0 (x) + a} \right) ;\\ \varphi ''(x)= & {} \varphi '(x)\left( {\frac{{a - 1}}{x} + \frac{{\varphi '_0 (x)}}{{\varphi _0 (x)}} + \frac{{F'_0 (x)}}{{F_0 (x) + a}}} \right) \\= & {} \varphi '(x)\left( {\frac{{a - 1}}{x} + \frac{{F_0 (x)}}{x} + \frac{{F'_0 (x)}}{{F_0 (x) + a}}} \right) . \end{aligned}$$

Therefore, if $a - 1 < 0$ then $\varphi ''(x) < 0$ for some values that are close to 0, since $\mathop {\lim }\limits _{x \rightarrow + 0} \left( {{{\frac{{F_0 (x)}}{x}} / {\frac{{a - 1}}{x}}}} \right) = 0$. This means that $a \geqslant 1$.

For the second part of the theorem, if we choose the function $F_0$ as stated by the theorem, then conditions (a)–(c), and (e) are obviously satisfied. It remains to show that (d) is also satisfied. Function $\gamma _2 (y) = \ln \left( {1 - \varphi \left( {1 - e^y } \right) } \right)$ is differentiable and the function

$$\begin{aligned} \frac{d}{{dy}}\gamma _2 (y) = - \frac{{\varphi '\left( {1 - e^y } \right) e^y }}{{1 - \varphi \left( {1 - e^y } \right) }} \end{aligned}$$

should be decreasing. This is equivalent to

$$\begin{aligned} g(x) = \frac{{\varphi '(x)(1 - x)}}{{1 - \varphi (x)}} \end{aligned}$$

be an increasing function. Substituting $\varphi (x) = x^a \varphi _0 (x)$, $\varphi '(x) = x^{a - 1} \varphi _0 (x)\left( {F_0 (x) + a} \right)$, we get

$$\begin{aligned} g(x) = \frac{{x^{a - 1} \varphi _0 (x)\left( {F_0 (x) + a} \right) (1 - x)}}{{1 - x^a \varphi _0 (x)}}. \end{aligned}$$

Then

$$\begin{aligned} g'(x) = g(x)\left( {\frac{{a - 1}}{x} + \frac{{\varphi _0 (x)}}{{\varphi '_0 (x)}} + \frac{{F'_0 (x)}}{{F_0 (x) + a}} + \frac{{x^{a - 1} \varphi _0 (x)\left( {F_0 (x) + a} \right) }}{{1 - x^a \varphi _0 (x)}} - \frac{1}{{1 - x}}} \right) . \end{aligned}$$

Notice that

$$\begin{aligned} \frac{{\varphi _0 (x)}}{{\varphi '_0 (x)}}= & {} \frac{{F_0 (x)}}{x};\\ \frac{{x^{a - 1} \varphi _0 (x)\left( {F_0 (x) + a} \right) }}{{1 - x^a \varphi _0 (x)}}= & {} \frac{{F_0 (x) + a}}{{x\left( {1 - x^a \varphi _0 (x)} \right) }} - \frac{{F_0 (x) + a}}{x}. \end{aligned}$$

Therefore,

$$\begin{aligned} g'(x) = g(x)\left( {\frac{{F'_0 (x)}}{{F_0 (x) + a}} + \frac{{F_0 (x) + a}}{{x\left( {1 - x^a \varphi _0 (x)} \right) }} - \frac{1}{{x(1 - x)}}} \right) . \end{aligned}$$

Because $\varphi _0 (x) \leqslant 1$ and $F_0 (x) \geqslant 0$, ${{F'_0 (x)} / {\left( {F_0 (x) + a} \right) }} \geqslant 0$, we get

$$\begin{aligned} g'(x) \geqslant g(x)\left( {\frac{a}{{x\left( {1 - x^a } \right) }} - \frac{1}{{x(1 - x)}}} \right) . \end{aligned}$$

We see that

$$\begin{aligned} \frac{a}{{x\left( {1 - x^a } \right) }} - \frac{1}{{x(1 - x)}} = \frac{{x^a + a(1 - x) - 1}}{{x\left( {1 - x^a } \right) (1 - x)}} \geqslant 0, \end{aligned}$$

where the inequality $x^a + a(1 - x) - 1 \geqslant 0$ for $a \geqslant 1$ and $x \in [0,1]$ is proved as in Example 3, i.e., $g'(x) \geqslant 0$ and the condition d) is also fulfilled. $\square$

Proof of Proposition 3

We will use the notation from Theorem 1. Computing the function $g$ for this case, we get

$$\begin{aligned} g(x) = \frac{{\beta \varPhi '\left( {\alpha + \beta \varPhi ^{ - 1} (x)} \right) x}}{{\varPhi '\left( {\varPhi ^{ - 1} (x)} \right) \varPhi \left( {\alpha + \beta \varPhi ^{ - 1} (x)} \right) }} \end{aligned}$$

Let us compute the limit $b = \mathop {\lim }\limits _{x \rightarrow + 0} g(x)$, involving a new variable $y = \varPhi ^{ - 1} (x)$:

$$\begin{aligned} b = \mathop {\lim }\limits _{y \rightarrow - \infty } \frac{{\beta \varPhi '\left( {\alpha + \beta y} \right) \varPhi (y)}}{{\varPhi '\left( {y} \right) \varPhi \left( {\alpha + \beta y} \right) }} = \beta \mathop {\lim }\limits _{y \rightarrow - \infty } \frac{{e^{ - \frac{{\left( {\alpha + \beta y} \right) ^2 - y^2 }}{2}} \varPhi (y)}}{{\varPhi \left( {\alpha + \beta y} \right) }} \end{aligned}$$

Obviously, $b = 1$ for $\beta = 1$. Applying the L’Hospital rule for the general case, one can obtain that $b = \beta ^2$, i.e., we conclude that $\beta \geqslant 1$.

Let us check now when $g$ is an increasing function. This condition is equivalent to the non-negativity of the derivative

$$\begin{aligned} f(y) = \frac{d}{{dy}}\ln \left( {\frac{{\beta \varPhi '\left( {\alpha + \beta y} \right) \varPhi (y)}}{{\varPhi '\left( {y} \right) \varPhi \left( {\alpha + \beta y} \right) }}} \right) \end{aligned}$$

for any $y \in ( - \infty , + \infty )$. Making simple calculations, we get

$$\begin{aligned} f(y)= & {} \beta \frac{{\varPhi ''\left( {\alpha + \beta y} \right) }}{{\varPhi '\left( {\alpha + \beta y} \right) }} + \frac{{\varPhi '(y)}}{{\varPhi \left( {y} \right) }} - \frac{{\varPhi '(y)}}{{\varPhi ''\left( {y} \right) }} - \beta \frac{{\varPhi '\left( {\alpha + \beta y} \right) }}{{\varPhi \left( {\alpha + \beta y} \right) }} \\= & {} - \alpha \beta - \left( {\beta ^2 - 1} \right) y + \frac{{\varPhi '(y)}}{{\varPhi \left( {y} \right) }} - \beta \frac{{\varPhi '\left( {\alpha + \beta y} \right) }}{{\varPhi \left( {\alpha + \beta y} \right) }}. \end{aligned}$$

Notice that $\mathop {\lim }\limits _{y \rightarrow + \infty } f(y) = - \infty$ for $\beta > 1$ and $\mathop {\lim }\limits _{y \rightarrow + \infty } f(y) = - \alpha$ for $\beta = 1$, therefore, the function $\varphi$ can satisfy the conditions of Theorem 1 if $\alpha < 0$ and $\beta = 1$. For this case we can represent the function $f$ in the form $f(y) = w(y) - w(y + \alpha )$, where $w(y) = y + \frac{\displaystyle {\varPhi '(y)}}{\displaystyle {\varPhi \left( {y} \right) }}$ and $f(y) \geqslant 0$ for all $y \in {\mathbb {R}}$ and $\alpha < 0$, if $\frac{\displaystyle {dw}}{\displaystyle {dy}} \geqslant 0$. Let us denote $u(y) = \frac{\displaystyle {\varPhi '(y)}}{\displaystyle {\varPhi \left( {y} \right) }}$. The function $u$ can be conceived as a partial solution of the Bernoulli differential equation $\frac{\displaystyle {du}}{\displaystyle {dy}} = - uy - u^2$. This equation and the condition $\frac{\displaystyle {dw}}{\displaystyle {dy}} = \frac{\displaystyle {du}}{\displaystyle {dy}} + 1 \geqslant 0$ imply that the function $f$ is increasing iff $\frac{\displaystyle {du}}{\displaystyle {dy}} = - uy - u^2 \geqslant - 1$ or, taking in account that the function $u$ is non-negative, that

$$\begin{aligned} u \leqslant \frac{{ - y + \sqrt{y^2 + 4} }}{2} \end{aligned}$$

By expressing the last inequality in terms of cdf for the standard normal distribution one gets:

$$\begin{aligned} \varPhi (y) \geqslant \frac{2}{{\left( { - y + \sqrt{y^2 + 4} } \right) \sqrt{2\pi } }}e^{ - \frac{{y^2 }}{2}} = \frac{{\left( {y + \sqrt{y^2 + 4} } \right) }}{{2\sqrt{2\pi } }}e^{ - \frac{{y^2 }}{2}} \end{aligned}$$

(18)

which implies

$$\begin{aligned} \frac{d}{{dy}}\varPhi (y) \geqslant \frac{d}{{dy}}\frac{{\left( {y + \sqrt{y^2 + 4} } \right) }}{{2\sqrt{2\pi } }}e^{ - \frac{{y^2 }}{2}} \text{ for } \text{ all } y \in {\mathbb {R}}, \end{aligned}$$

or

$$\begin{aligned} \frac{1}{{\sqrt{2\pi } }}e^{ - \frac{{y^2 }}{2}} \geqslant \frac{1}{{2\sqrt{2\pi } }}e^{ - \frac{{y^2 }}{2}} \left( {1 + \frac{y}{{\sqrt{y^2 + 4} }} - y^2 - y\sqrt{y^2 + 4} } \right) \end{aligned}$$

or

$$\begin{aligned} 1 \geqslant \frac{1}{2}\left( {1 + \frac{y}{{\sqrt{y^2 + 4} }} - y^2 - \frac{{y\left( {y^2 + 4} \right) }}{{\sqrt{y^2 + 4} }}} \right) \end{aligned}$$

Now denote $v = {y / {\sqrt{y^2 + 4} }}$. Then $y^2 = {{4v^2 } / {\left( {1 - v^2 } \right) }}$ and we get:

$$\begin{aligned} 1 \geqslant \frac{1}{2}\left( {1 + v - \frac{{4v^2 }}{{1 - v^2 }} - v\left( {4 + \frac{{4v^2 }}{{1 - v^2 }}} \right) } \right) = \frac{1}{2}\left( {1 - 3v - \frac{{4v^2 }}{{1 - v^2 }}(1 + v)} \right) \end{aligned}$$

We proceed taking in to account that $\left| v \right| < 1$:

$$\begin{aligned} 1 \geqslant - 3v - \frac{{4v^2 }}{{1 - v}} \end{aligned}$$

or

$$\begin{aligned} v^2 + 2v + 1 \geqslant 0 \end{aligned}$$

As this inequality is satisfied for all real $v$ we conclude that $g$ is an increasing function and that the function $\varphi$ satisfies the conditions of Theorem 1 for $\alpha < 0$ and $\beta = 1$. $\square$