Graph of uniform distribution of a continuous random variable. Uniform and exponential laws of distribution of a continuous random variable

This issue has long been studied in detail, and the most widely used method is the polar coordinate method, proposed by George Box, Mervyn Muller and George Marsaglia in 1958. This method allows you to obtain a pair of independent normally distributed random variables with mathematical expectation 0 and variance 1 as follows:

Where Z 0 and Z 1 are the desired values, s = u 2 + v 2, and u and v are random variables uniformly distributed on the interval (-1, 1), selected in such a way that condition 0 is satisfied< s < 1.
Many people use these formulas without even thinking, and many do not even suspect their existence, since they use ready-made implementations. But there are people who have questions: “Where did this formula come from? And why do you get a couple of quantities at once?” Next, I will try to give a clear answer to these questions.

To begin with, let me remind you what probability density, distribution function of a random variable and inverse function are. Suppose there is a certain random variable, the distribution of which is specified by the density function f(x), which has the following form:

This means that the probability that the value of a given random variable will be in the interval (A, B) is equal to the area of ​​the shaded area. And as a consequence, the area of ​​the entire shaded area must be equal to one, since in any case the value of the random variable will fall into the domain of definition of the function f.
The distribution function of a random variable is the integral of the density function. And in this case, its approximate appearance will be like this:

The meaning here is that the value of the random variable will be less than A with probability B. And as a consequence, the function never decreases, and its values ​​lie in the interval.

An inverse function is a function that returns an argument to the original function if the value of the original function is passed into it. For example, for the function x 2 the inverse is the function of extracting the root, for sin(x) it is arcsin(x), etc.

Since most pseudorandom number generators produce only a uniform distribution as output, there is often a need to convert it to some other one. In this case, to normal Gaussian:

The basis of all methods for transforming a uniform distribution into any other is the inverse transformation method. It works as follows. A function is found that is inverse to the function of the required distribution, and a random variable uniformly distributed on the interval (0, 1) is passed into it as an argument. At the output we obtain a value with the required distribution. For clarity, I provide the following picture.

Thus, a uniform segment is, as it were, smeared in accordance with the new distribution, projected onto another axis through an inverse function. But the problem is that the integral of the density of a Gaussian distribution is not easy to calculate, so the above scientists had to cheat.

There is a chi-square distribution (Pearson distribution), which is the distribution of the sum of squares of k independent normal random variables. And in the case where k = 2, this distribution is exponential.

This means that if a point in a rectangular coordinate system has random X and Y coordinates distributed normally, then after converting these coordinates to the polar system (r, θ), the square of the radius (the distance from the origin to the point) will be distributed according to the exponential law, since the square of the radius is the sum of the squares of the coordinates (according to Pythagorean law). The distribution density of such points on the plane will look like this:

Since it is equal in all directions, the angle θ will have a uniform distribution in the range from 0 to 2π. The converse is also true: if you define a point in the polar coordinate system using two independent random variables (an angle distributed uniformly and a radius distributed exponentially), then the rectangular coordinates of this point will be independent normal random variables. And it is much easier to obtain an exponential distribution from a uniform one using the same inverse transformation method. This is the essence of the polar Box-Muller method.
Now let's derive the formulas.

(1)

To obtain r and θ, we need to generate two random variables uniformly distributed on the interval (0, 1) (let’s call them u and v), the distribution of one of which (let’s say v) needs to be converted to exponential to obtain the radius. The exponential distribution function looks like this:

Its inverse function is:

Since the uniform distribution is symmetrical, the transformation will work similarly with the function

From the chi-square distribution formula it follows that λ = 0.5. Substitute λ, v into this function and get the square of the radius, and then the radius itself:

We obtain the angle by stretching the unit segment to 2π:

Now we substitute r and θ into formulas (1) and get:

(2)

These formulas are already ready to use. X and Y will be independent and normally distributed with a variance of 1 and a mathematical expectation of 0. To obtain a distribution with other characteristics, it is enough to multiply the result of the function by the standard deviation and add the mathematical expectation.
But it is possible to get rid of trigonometric functions by specifying the angle not directly, but indirectly through the rectangular coordinates of a random point in the circle. Then, through these coordinates, it will be possible to calculate the length of the radius vector, and then find the cosine and sine by dividing x and y by it, respectively. How and why does it work?
Let us choose a random point from those uniformly distributed in a circle of unit radius and denote the square of the length of the radius vector of this point by the letter s:

The selection is made by specifying random rectangular coordinates x and y, uniformly distributed in the interval (-1, 1), and discarding points that do not belong to the circle, as well as the central point at which the angle of the radius vector is not defined. That is, condition 0 must be met< s < 1. Тогда, как и в случае с Гауссовским распределением на плоскости, угол θ будет распределен равномерно. Это очевидно - количество точек в каждом направлении одинаково, значит каждый угол равновероятен. Но есть и менее очевидный факт - s тоже будет иметь равномерное распределение. Полученные s и θ будут независимы друг от друга. Поэтому мы можем воспользоваться значением s для получения экспоненциального распределения, не генерируя третью случайную величину. Подставим теперь s в формулы (2) вместо v, а вместо тригонометрических функций - их расчет делением координаты на длину радиус-вектора, которая в данном случае является корнем из s:

We get the formulas as at the beginning of the article. The disadvantage of this method is that it discards points that are not included in the circle. That is, using only 78.5% of the generated random variables. On older computers, the lack of trigonometry functions was still a big advantage. Now, when one processor command calculates both sine and cosine in an instant, I think these methods can still compete.

Personally, I still have two questions:

• Why is the value of s distributed evenly?
• Why is the sum of the squares of two normal random variables distributed exponentially?

If a similar transformation is made for a normal random variable, then the density function of its square will turn out to be similar to a hyperbola. And the addition of two squares of normal random variables is a much more complex process associated with double integration. And the fact that the result will be an exponential distribution, I personally only have to check using a practical method or accept as an axiom. And for those who are interested, I suggest that you take a closer look at the topic, gaining knowledge from these books:

• Ventzel E.S. Probability theory
• Knut D.E. The Art of Programming, Volume 2

In conclusion, here is an example of implementing a normally distributed random number generator in JavaScript:

Function Gauss() ( var ready = false; var second = 0.0; this.next = function(mean, dev) ( mean = mean == undefined ? 0.0: mean; dev = dev == undefined ? 1.0: dev; if ( this.ready) ( this.ready = false; return this.second * dev + mean; ) else ( var u, v, s; do ( u = 2.0 * Math.random() - 1.0; v = 2.0 * Math. random() - 1.0; s = u * u + v * v; while (s > 1.0 || s == 0.0); this.second = r * u; this.ready = true; return r * v * dev + mean ) ) g = new Gauss(); // create an object a = g.next(); // generate a pair of values ​​and get the first one b = g.next(); // get the second c = g.next(); // generate a pair of values ​​again and get the first one
The parameters mean (mathematical expectation) and dev (standard deviation) are optional. I draw your attention to the fact that the logarithm is natural.

As an example of a continuous random variable, consider a random variable X uniformly distributed over the interval (a; b). The random variable X is said to be evenly distributed on the interval (a; b), if its distribution density is not constant on this interval:

From the normalization condition we determine the value of the constant c. The area under the distribution density curve should be equal to unity, but in our case it is the area of ​​a rectangle with base (b - α) and height c (Fig. 1).

Rice. 1 Uniform distribution density
From here we find the value of the constant c:

So, the density of a uniformly distributed random variable is equal to

Let us now find the distribution function using the formula:
1) for
2) for
3) for 0+1+0=1.
Thus,

The distribution function is continuous and does not decrease (Fig. 2).

Rice. 2 Distribution function of a uniformly distributed random variable

We'll find mathematical expectation of a uniformly distributed random variable according to the formula:

Dispersion of uniform distribution is calculated by the formula and is equal to

Example No. 1. The scale division value of the measuring device is 0.2. Instrument readings are rounded to the nearest whole division. Find the probability that an error will be made during the counting: a) less than 0.04; b) large 0.02
Solution. The rounding error is a random variable uniformly distributed over the interval between adjacent integer divisions. Let us consider the interval (0; 0.2) as such a division (Fig. a). Rounding can be carried out both towards the left border - 0, and towards the right - 0.2, which means that an error less than or equal to 0.04 can be made twice, which must be taken into account when calculating the probability:



P = 0.2 + 0.2 = 0.4

For the second case, the error value can also exceed 0.02 on both division boundaries, that is, it can be either more than 0.02 or less than 0.18.


Then the probability of an error like this:

Example No. 2. It was assumed that the stability of the economic situation in the country (absence of wars, natural disasters, etc.) over the past 50 years can be judged by the nature of the population distribution by age: in a calm situation it should be uniform. As a result of the study, the following data were obtained for one of the countries.

We carry out the solution using a calculator. Testing hypotheses. Table for calculating indicators.

Solution:
1. Find estimates of the parameters a * and b * of the uniform distribution using the formulas:


2. Find the density of the assumed uniform distribution:
f(x) = 1/(b * - a *) = 1/(84.42 - 1.58) = 0.0121
3. Let's find the theoretical frequencies:
n 1 = n*f(x)(x 1 - a *) = 1 * 0.0121(10-1.58) = 0.1
n 8 = n*f(x)(b * - x 7) = 1 * 0.0121(84.42-70) = 0.17
The remaining n s will be equal to:
n s = n*f(x)(x i - x i-1)