Granular approach to data processing under probabilistic uncertainty

Abstract

The existing algorithms for data processing under probabilistic uncertainty often require too much computation time. Sometimes, we can speed up the corresponding computations if we take into account the fact that in many real-life situations, uncertainty can be naturally described as a combination of several components, components which are described by different granules. In such situations, to process this uncertainty, it is often beneficial to take this granularity into account by processing these granules separately and then combining the results. In this paper, we show that granular computing can help even in situations when there is no such natural decomposition into granules, namely we can often speed up processing of uncertainty if we first (artificially) decompose the original uncertainty into appropriate granules.

Appendix: Non-uniform distribution of \(\alpha _j\) is better

1.1 Idea

If we select two values $\alpha _j$ too close to each other, there will be too much correlation between them, so adding the function corresponding to the second value does not add much information to what we know from a function corresponding to the first value.

We are approximating a general function (logarithm of a characteristic function) as a linear combination of functions $|t|^{\alpha _j}$. If two values $\alpha _j$ and $\alpha _{j+1}$ are close, then the function $|t|^{\alpha _{j+1}}$ can be well approximated by a term linear in $|t|^{\alpha _j}$; thus, the term proportional to $|t|^{\alpha _{j+1}}$ is not needed.

It, therefore, makes sense to select the values $\alpha _j$ in such as way that for each j, the part of $|t|^{\alpha _{j+1}}$ that cannot be approximated by terms proportional to $|t|^{\alpha _j}$ should be the largest possible.

1.2 Let us reformulate this idea in precise terms

For every two functions f(t) and g(t), the part of g(t) which cannot be represented by terms $a\cdot f(t)$ (proportional to f(t)) can be described as follows. It is reasonable to describe the difference between the two functions f(t) and g(t) by the least squares ($L^2$) metric $\int (f(t)-g(t))^2\,{\text {d}}t$. In these terms, the value of a function itself itself can be described as its distance from 0, i.e., as $\int (f(t))^2\,{\text {d}}t$.

When we approximate a function g(t) by a term $a\cdot f(t)$, then the remainder $g(t)-a\cdot f(t)$ has the value

$$\begin{aligned} \int (g(t)-a\cdot f(t))^2\,{\text {d}}t. \end{aligned}$$

The best approximation occurs when this value is the smallest, i.e., when it is equal to $\min \nolimits _a \int (g(t)-a\cdot f(t))^2\,{\text {d}}t$. Out of the original value $\int (g(t))^2\,{\text {d}}t$, we have unrepresented the part equal to $\min \nolimits _a \int (g(t)-a\cdot f(t))^2\,{\text {d}}t$. Thus, the relative size of what cannot be represented by terms $a\cdot f(t)$ can be defined as a ratio

$$\begin{aligned} R(f(t),g(t))=\frac{\min \limits _a \int (g(t)-a\cdot f(t))^2\,{\text {d}}t}{\int (g(t))^2\,{\text {d}}t}. \end{aligned}$$

1.3 Let us simplify the resulting expression

This expression can be simplified if we find the explicit expression for a for which the value $\int (g(t)-a\cdot f(t))^2\,{\text {d}}t$ is the smallest possible. Differentiating the minimized expression with respect to a and equating the derivative to 0, we conclude that

$$\begin{aligned} -\int (g(t)-a\cdot f(t))\cdot f(t)\,{\text {d}}t=0, \end{aligned}$$

i.e., that

$$\begin{aligned} a\cdot \int (f(t))^2\,dt=\int f(t)\cdot g(t)\,{\text {d}}t, \end{aligned}$$

and

$$\begin{aligned} a=\frac{\int f(t)\cdot g(t)\,{\text {d}}t}{\int (f(t))^2\,{\text {d}}t}. \end{aligned}$$

For this a, the value $\int (g(t)-a\cdot f(t))^2\,{\text {d}}t$ takes the form

$$\begin{aligned}&\int (g(t)-a\cdot f(t))^2\,{\text {d}}t\\&\quad =\int (g(t))^2\,{\text {d}}t-2a\cdot \int f(t)\cdot g(t)\,{\text {d}}t+a^2\cdot \int (f(t))\,{\text {d}}t. \end{aligned}$$

Substituting the above expression for a into this formula, we conclude that

$$\begin{aligned}&\int (g(t)-a\cdot f(t))^2\,{\text {d}}t\\&\quad =\int (g(t))^2\,{\text {d}}t-\frac{2(\int f(t)\cdot g(t)\,{\text {d}}t)^2}{\int (f(t))^2\,{\text {d}}t}+\frac{(\int f(t)\cdot g(t)\,{\text {d}}t)^2}{\int (f(t))^2\,{\text {d}}t}, \end{aligned}$$

i.e., that

$$\begin{aligned} \int (g(t)-a\cdot f(t))^2\,{\text {d}}t=\int (g(t))^2\,{\text {d}}t-\frac{(\int f(t)\cdot g(t)\,{\text {d}}t)^2}{\int (f(t))^2\,{\text {d}}t}. \end{aligned}$$

Thus, the desired ratio takes the form

$$\begin{aligned}&R(f(t),g(t)){\mathop {=}\limits ^{\mathrm{def}}}\frac{\min \limits _a \int (g(t)-a\cdot f(t))^2\,{\text {d}}t}{\int (g(t))^2\,{\text {d}}t}\\&\quad =1-\frac{(\int f(t)\cdot g(t)\,{\text {d}}t)^2}{(\int (f(t))^2\,{\text {d}}t)\cdot (\int (g(t))^2\,{\text {d}}t)}. \end{aligned}$$

Thus, we arrive at the following optimization problem.

1.4 Resulting optimization problem

To make sure that the above remainders are as large as possible, it makes sense to find the values $\alpha ^{\mathrm{opt}}_1<\cdots <\alpha ^{\mathrm{opt}}_k$ that maximize the smallest of the remainders between the functions $f(t)=|t|^{\alpha _j}$ and $g(t)=|t|^{\alpha _{j+1}}$:

$$\begin{aligned} \min \limits _j R\left( |t|^{\alpha ^{\mathrm{opt}}_j},|t|^{\alpha ^{\mathrm{opt}}_{j+1}}\right) = \max _{\alpha _1<\ldots <\alpha _k} \min \limits _j R(|t|^{\alpha _j},|t|^{\alpha _{j+1}}). \end{aligned}$$

1.5 Solving the optimization problem

Let us consider an interval $[-T,T]$ for some T. Since the function is symmetric, it is sufficient to consider the values from [0, T].

For $f(t)=t^\alpha$ and $g(t)=t^\beta$, the integral in the numerator of the ratio is equal to

$$\begin{aligned} \int _0^T f(t)\cdot g(t)\,{\text {d}}t=\int _0^T t^\alpha \cdot t^\beta \,{\text {d}}t=\int _0^T t^{\alpha +\beta }\,{\text {d}}t=\frac{T^{\alpha +\beta +1}}{\alpha +\beta +1}. \end{aligned}$$

Similarly, the integrals in the denominator take the form

$$\begin{aligned} \int _0^T f^2(t)\,{\text {d}}t=\int _0^T t^{2\alpha }\,{\text {d}}t=\frac{T^{2\alpha +1}}{2\alpha +1} \end{aligned}$$

and

$$\begin{aligned} \int _0^T g^2(t)\,{\text {d}}t=\int _0^T t^{2\beta }\,{\text {d}}t=\frac{T^{2\beta +1}}{2\beta +1}, \end{aligned}$$

so

$$\begin{aligned} R=1-\frac{\displaystyle \frac{T^{2(\alpha +\beta +1)}}{(\alpha +\beta +1)^2}}{\displaystyle \frac{T^{2\alpha +1}}{2\alpha +1}\cdot \displaystyle \frac{T^{2\beta +1}}{2\beta +1}}. \end{aligned}$$

One can see that the powers of T cancel each other, and we get

$$\begin{aligned} R=1-\frac{(2\alpha +1)\cdot (2\beta +1)}{(\alpha +\beta +1)^2}, \end{aligned}$$

or, equivalently, if we denote $r{\mathop {=}\limits ^\mathrm{def}}\displaystyle \frac{\beta +0.5}{\alpha +0.5}$, we get

$$\begin{aligned} R=R(r){\mathop {=}\limits ^\mathrm{def}}1-4\cdot \frac{r}{(1+r)^2}. \end{aligned}$$

The derivative of the function R(r) is equal to

$$\begin{aligned} \frac{{\text {d}}R}{{\text {d}}r}&=-4\cdot \frac{(1+r)^2-2\cdot (1+r)}{(1+r)^4}=-4\cdot \frac{(1+r)\cdot (1+r-2)}{(1+r)^4}\\&=4\cdot \frac{(1+r)\cdot (r-1)}{(1+r)^4}=4\cdot \frac{r-1}{(1+r)^3}. \end{aligned}$$

So this derivative is positive for all $r>1$. Thus, the function R(r) is monotonically increasing, and looking for the values $\alpha ^\mathrm{opt}_j$ for which $\min \limits _j R(|t|^{\alpha _j},|t|^{\alpha _{j+1}})$ is the largest if equivalent to looking for the values $\alpha ^{\mathrm{opt}}_j$ for which the smallest $\min \limits _j \displaystyle \frac{\alpha _{j+1}+0.5}{\alpha _j+0.5}$ of the ratios $r=\displaystyle \frac{\alpha _{j+1}+0.5}{\alpha _j+0.5}$ attains the largest possible value:

$$\begin{aligned} \min \limits _j \displaystyle \frac{\alpha ^{\mathrm{opt}}_{j+1}+0.5}{\alpha ^\mathrm{opt}_j+0.5}=\max _{\alpha _1<\ldots <\alpha _k} \min \limits _j \displaystyle \frac{\alpha _{j+1}+0.5}{\alpha _j+0.5}. \end{aligned}$$

One can check that this happens when

$$\begin{aligned} \alpha _j+0.5=1.5\cdot \left( \displaystyle \frac{5}{3}\right) ^{(j-1)/(k-1)}. \end{aligned}$$

Indeed, in this case, $\min \nolimits _j \frac{\alpha _{j+1}+0.5}{\alpha _j+0.5}=\left( \frac{5}{3}\right) ^{1/(k-1)}.$ We cannot have it larger: if we had $\min \nolimits _j \frac{\alpha _{j+1}+0.5}{\alpha _j+0.5}>\left( \frac{5}{3}\right) ^{k-1}$, then we would have $\displaystyle \frac{\alpha _{j+1}+0.5}{\alpha _j+0.5}>\left( \displaystyle \frac{5}{3}\right) ^{k-1}$ for all j. Here

$$\begin{aligned} \alpha _k+0.5=(\alpha _1+0.5)\cdot \frac{\alpha _2+0.5}{\alpha _1+0.5}\cdot \frac{\alpha _3+0.5}{\alpha _2+0.5}\cdot \ldots \cdot \frac{\alpha _k+0.5}{\alpha _{k-1}+0.5}. \end{aligned}$$

The first factor $\alpha _1+0.5$ is $\ge 1.5$, each of the other $k-1$ terms is greater than $(\frac{5}{3})^{1/(k-1)}$, so for their product, we get

$$\begin{aligned} \alpha _k+0.5>1.5\cdot \left( \left( \displaystyle \frac{5}{3}\right) ^{1/(k-1)}\right) ^{k-1}=1.5\cdot \displaystyle \frac{5}{3}=2.5, \end{aligned}$$

while we assumed that all the values $\alpha _j$ are from the interval [1, 2], and so, we should have $\alpha _k+0.5\le 2.5$.

1.6 Resulting optimal values of $\alpha _j$

Thus, the optimal way is not to take the values uniformly distributed in the interval [1, 2], but rather take the values

$$\begin{aligned} \alpha ^{\mathrm{opt}}_j=1.5\cdot \left( \displaystyle \frac{5}{3}\right) ^{(j-1)/(k-1)}-0.5 \end{aligned}$$

for which the logarithms $\ln (\alpha ^\mathrm{opt}_j+0.5)= \frac{j-1}{k-1}\cdot \ln (\frac{5}{3})$ are uniformly distributed.

Comment It is worth mentioning that there is intriguing connection between these values $\alpha _j$ and music: for example, the twelve notes on a usual Western octave correspond to the following frequencies:

• the first note corresponds to the frequency $f_1$,

• the second note corresponds to the frequency

  $$\begin{aligned} f_2=f_1\cdot 2^{1/12}, \end{aligned}$$

• the third note correspond to the frequency

  $$\begin{aligned} f_3=f_1\cdot 2^{2/12}, \end{aligned}$$

• ...,

• the last note corresponds to the frequency

  $$\begin{aligned} f_{12}=f_1\cdot 2^{11/12}, \end{aligned}$$

  and

• the first note of the next octave corresponds to the frequency $f_{13}=f_1\cdot 2$.

For these frequencies, the logarithms $\ln (f_j)$ are uniformly distributed.

Similar formulas exist for five-note and other octaves typical for some Oriental musical traditions.