Distribution function of a continuous value. random variable

Let a continuous random variable X be given by the distribution function f(x). Let us assume that all possible values of the random variable belong to the interval [ a,b].

Definition. mathematical expectation continuous random variable X, the possible values of which belong to the segment , is called a definite integral

If the possible values of a random variable are considered on the entire number axis, then the mathematical expectation is found by the formula:

In this case, of course, it is assumed that the improper integral converges.

Definition. dispersion continuous random variable is called the mathematical expectation of the square of its deviation.

By analogy with the variance of a discrete random variable, the following formula is used for the practical calculation of the variance:

Definition. Standard deviation is called the square root of the variance.

Definition. Fashion M 0 of a discrete random variable is called its most probable value. For a continuous random variable, the mode is the value of the random variable at which the distribution density has a maximum.

If the distribution polygon for a discrete random variable or the distribution curve for a continuous random variable has two or more maxima, then such a distribution is called bimodal or multimodal. If a distribution has a minimum but no maximum, then it is called antimodal.

Definition. median M D of a random variable X is its value, relative to which it is equally likely to obtain a larger or smaller value of the random variable.

Geometrically, the median is the abscissa of the point at which the area bounded by the distribution curve is divided in half. Note that if the distribution is unimodal, then the mode and median coincide with the mathematical expectation.

Definition. Starting moment order k random variable X is called the mathematical expectation of X k.

The initial moment of the first order is equal to the mathematical expectation.

Definition. Central point order k random variable X is called the mathematical expectation of the value

For a discrete random variable: .

For a continuous random variable: .

The first order central moment is always zero, and the second order central moment is equal to the dispersion. The central moment of the third order characterizes the asymmetry of the distribution.

Definition. The ratio of the central moment of the third order to the standard deviation in the third degree is called asymmetry coefficient.

Definition. To characterize the sharpness and flatness of the distribution, a quantity called kurtosis.

In addition to the quantities considered, the so-called absolute moments are also used:

Absolute starting moment: . Absolute central moment: . The absolute central moment of the first order is called arithmetic mean deviation.

Example. For the example considered above, determine the mathematical expectation and variance of the random variable X.

Example. An urn contains 6 white and 4 black balls. A ball is removed from it five times in a row, and each time the ball taken out is returned back and the balls are mixed. Taking the number of extracted white balls as a random variable X, draw up the law of distribution of this quantity, determine its mathematical expectation and variance.

Because balls in each experiment are returned back and mixed, then the trials can be considered independent (the result of the previous experiment does not affect the probability of occurrence or non-occurrence of an event in another experiment).

Thus, the probability of a white ball appearing in each experiment is constant and equal to

Thus, as a result of five consecutive trials, the white ball may not appear at all, appear once, twice, three, four or five times. To draw up a distribution law, you need to find the probabilities of each of these events.

1) The white ball did not appear at all:

2) The white ball appeared once:

3) The white ball will appear twice: .

The distribution function of a random variable X is the function F(x), expressing for each x the probability that the random variable X takes the value, smaller x

Example 2.5. Given a series of distribution of a random variable

Find and graphically depict its distribution function. Decision. According to the definition

F(jc) = 0 for X X

F(x) = 0.4 + 0.1 = 0.5 at 4 F(x) = 0.5 + 0.5 = 1 at X > 5.

So (see Fig. 2.1):

Distribution function properties:

1. The distribution function of a random variable is a non-negative function enclosed between zero and one:

2. The distribution function of a random variable is a non-decreasing function on the entire number axis, i.e. at X 2 >x

3. At minus infinity, the distribution function is equal to zero, at plus infinity, it is equal to one, i.e.

4. Probability of hitting a random variable X in the interval is equal to the definite integral of its probability density ranging from a before b(see Fig. 2.2), i.e.

Rice. 2.2

3. The distribution function of a continuous random variable (see Fig. 2.3) can be expressed in terms of the probability density using the formula:

F(x)= Jp(*)*. (2.10)

4. Improper integral in infinite limits of the probability density of a continuous random variable is equal to one:

Geometric properties / and 4 probability densities mean that its plot is distribution curve - lies not below the x-axis, and the total area of the figure, limited distribution curve and x-axis, is equal to one.

For a continuous random variable X expected value M(X) and variance D(X) are determined by the formulas:

(if the integral converges absolutely); or

(if the reduced integrals converge).

Along with the numerical characteristics noted above, the concept of quantiles and percentage points is used to describe a random variable.

q level quantile(or q-quantile) is such a valuex qrandom variable, at which its distribution function takes the value, equal to q, i.e.

100The q%-ou point is the quantile X~ q .
? Example 2.8.

According to example 2.6 find the quantile xqj and 30% random variable point x.

Decision. By definition (2.16) F(xo t3)= 0.3, i.e.

~Y~ = 0.3, whence the quantile x 0 3 = 0.6. 30% random variable point X, or quantile Х)_о,з = xoj» is found similarly from the equation ^ = 0.7. whence *,= 1.4. ?

Among the numerical characteristics of a random variable, there are initial v* and central R* k-th order moments, determined for discrete and continuous random variables by the formulas:

4. Density of the probability distribution of a continuous random variable

A continuous random variable can be specified using the distribution function F(x) . This way of setting is not the only one. A continuous random variable can also be specified using another function called the distribution density or probability density (sometimes called the differential function).

Definition 4.1: Distribution density of a continuous random variable X call the function f (x) - the first derivative of the distribution function F(x) :

f ( x ) = F "( x ) .

It follows from this definition that the distribution function is the antiderivative of the distribution density. Note that to describe the probability distribution of a discrete random variable, the distribution density is not applicable.

Probability of hitting a continuous random variable in a given interval

Knowing the distribution density, we can calculate the probability that a continuous random variable will take a value that belongs to a given interval.

Theorem: The probability that a continuous random variable X will take values belonging to the interval (a, b), is equal to a certain integral of the distribution density, taken in the range fromabeforeb :

Proof: We use the ratio

P(a ≤ Xb) = F(b) – F(a).

According to the Newton-Leibniz formula,

Thus,

As P(a ≤ X b)= P(a X b) , then we finally get

Geometrically, the result can be interpreted as follows: the probability that a continuous random variable takes a value belonging to the interval (a, b), is equal to the area of the curvilinear trapezoid bounded by the axisOx, distribution curvef(x) and directx = aandx = b.

Comment: In particular, if f(x) is an even function and the ends of the interval are symmetrical with respect to the origin, then

Example. Given the probability density of a random variable X

Find the probability that as a result of the test X will take values belonging to the interval (0.5; 1).

Decision: Desired probability

Finding the distribution function from a known distribution density

Knowing the distribution density f(x) , we can find the distribution function F(x) according to the formula

Really, F(x) = P(X x) = P(-∞ X x) .

Hence,

Thus, knowing the distribution density, you can find the distribution function. Of course, from the known distribution function, one can find the distribution density, namely:

f(x) = F"(x).

Example. Find the distribution function for a given distribution density:

Decision: Let's use the formula

If a x ≤ a, then f(x) = 0 , hence, F(x) = 0 . If a a , then f(x) = 1/(b-a),

hence,

If a x > b, then

So, the desired distribution function

Comment: We have obtained the distribution function of a uniformly distributed random variable (see uniform distribution).

Distribution Density Properties

Property 1: The distribution density is a non-negative function:

f ( x ) ≥ 0 .

Property 2: The improper integral of the distribution density in the range from -∞ to ∞ is equal to one:

Comment: The plot of the distribution density is called distribution curve.

Comment: The distribution density of a continuous random variable is also called the distribution law.

Example. The distribution density of a random variable has the following form:

Find constant parameter a.

Decision: The distribution density must satisfy the condition , so we require that the equality

From here
. Let's find the indefinite integral:

We calculate the improper integral:

Thus, the required parameter

Probable meaning of distribution density

Let be F(x) is the distribution function of a continuous random variable X. By definition of the distribution density, f(x) = F"(x) , or

Difference F(x+∆х) -F(x) determines the probability that X will take the value belonging to the interval (x, x+∆х). Thus, the limit of the ratio of the probability that a continuous random variable takes a value belonging to the interval (x, x+∆х), to the length of this interval (at ∆х→0) is equal to the value of the distribution density at the point X.

So the function f(x) determines the probability distribution density for each point X. It is known from differential calculus that the increment of a function is approximately equal to the differential of the function, i.e.

As F"(x) = f(x) and dx = ∆ x, then F(x+∆ x) - F(x) ≈ f(x)∆ x.

The probabilistic meaning of this equality is as follows: the probability that a random variable takes a value belonging to the interval (x, x+∆ x) , is approximately equal to the product of the probability density at the point x and the length of the interval ∆х.

Geometrically, this result can be interpreted as: the probability that a random variable takes a value belonging to the interval (x, x+∆ x), approximately equal to the area of a rectangle with base ∆х and heightf(x).

5. Typical distributions of discrete random variables

5.1. Bernoulli distribution

Definition 5.1: Random value X, which takes two values 1 and 0 with probabilities (“success”) p and (“failure”) q, is called Bernoulli:

, where k=0,1.

5.2. Binomial distribution

Let it be produced n independent trials, in each of which an event A may or may not appear. The probability of an event occurring in all trials is constant and equal to p(hence the probability of non-appearance q = 1 - p).

Consider a random variable X– number of occurrences of the event A in these tests. Random value X takes values 0,1,2,… n with probabilities calculated by the Bernoulli formula: , where k = 0,1,2,… n.

Definition 5.2: Binomial is called the probability distribution determined by the Bernoulli formula.

Example. Three shots are fired at the target, and the probability of hitting each shot is 0.8. We consider a random variable X- the number of hits on the target. Find its distribution series.

Decision: Random value X takes values 0,1,2,3 with probabilities calculated by the Bernoulli formula, where n = 3, p = 0,8 (probability of hit), q = 1 - 0,8 = = 0,2 (probability of missing).

Thus, the distribution series has the following form:

Use the Bernoulli formula for large values n rather difficult, therefore, to calculate the corresponding probabilities, the local Laplace theorem is used, which allows one to approximately find the probability of an event occurring exactly k once a n trials if the number of trials is large enough.

Local Laplace theorem: If probability p occurrence of an event A
that the event A will appear in n tests exactly k times, approximately equal (the more accurate, the more n) function value
, where
,
.

Note1: Tables containing function values
, are given in Appendix 1, and
. Function is the density of the standard normal distribution (see normal distribution).

Example: Find the probability that the event A comes exactly 80 once a 400 trials if the probability of occurrence of this event in each trial is equal to 0,2.

Decision: By condition n = 400, k = 80, p = 0,2 , q = 0,8 . Let us calculate the value determined by the problem data x:
. According to the table in Appendix 1, we find
. Then the desired probability will be:

If you want to calculate the probability that an event A will appear in n tests at least k 1 once and no more k 2 times, then you need to use the Laplace integral theorem:

Laplace integral theorem: If probability p occurrence of an event A in each test is constant and different from zero and one, then the probability
that the event A will appear in n tests from k 1 before k 2 times, approximately equal to the definite integral

, where
and
.

In other words, the probability that an event A will appear in n tests from k 1 before k 2 times, approximately equal to

where
, and .

Remark2: Function
called the Laplace function (see normal distribution). Tables containing function values , are given in Appendix 2, and
.

Example: Find the probability that among 400 randomly selected parts will be unchecked from 70 to 100 parts, if the probability that the part did not pass the quality control check is equal to 0,2.

Decision: By condition n = 400, p = 0,2 , q = 0,8, k 1 = 70, k 2 = 100 . Let us calculate the lower and upper limits of integration:

;
.

Thus, we have:

According to the table in Appendix 2, we find that
and . Then the required probability is:

Remark3: In a series of independent trials (when n is large, p is small), the Poisson formula is used exactly k times to calculate the probability of an event occurring (see Poisson distribution).

5.3. Poisson distribution

Definition 5.3: A discrete random variable is called poisson, if its distribution law has the following form:

, where
and
(constant value).

Examples of Poisson random variables:

Number of calls to an automatic station in a time interval T.

The number of decay particles of some radioactive substance over a period of time T.

The number of TVs that enter the workshop in a period of time T in the big city .

The number of cars that will arrive at the stop line of an intersection in a large city .

Note1: Special tables for calculating these probabilities are given in Appendix 3.

Remark2: In a series of independent trials (when n great, p small) to calculate the probability of an event occurring exactly k once the Poisson formula is used:
, where
, that is, the average number of occurrences of events remains constant.

Remark3: If there is a random variable that is distributed according to the Poisson law, then there is necessarily a random variable that is distributed according to the exponential law and vice versa (see the exponential distribution).

Example. The factory sent to the base 5000 good quality products. The probability that the product will be damaged in transit is equal to 0,0002 . Find the probability that exactly three unusable items will arrive at the base.

Decision: By condition n = 5000, p = 0,0002, k = 3. Let's find λ: λ = np= 5000 0.0002 = 1.

According to the Poisson formula, the desired probability is equal to:

, where random variable X- the number of defective products.

5.4. Geometric distribution

Let independent trials be made, in each of which the probability of occurrence of an event BUT is equal to p(0p

q = 1 - p. Trials end as soon as the event appears BUT. Thus, if an event BUT appeared in k-th test, then in the previous k – 1 It didn't show up in the tests.

Denote by X discrete random variable - the number of trials to be carried out before the first occurrence of the event BUT. Obviously, the possible values X are natural numbers x 1 \u003d 1, x 2 \u003d 2, ...

Let the first k-1 test event BUT did not come, but k th test appeared. The probability of this “complex event”, according to the theorem of multiplication of probabilities of independent events, P (X = k) = q k -1 p.

Definition 5.4: A discrete random variable has geometric distribution if its distribution law has the following form:

P ( X = k ) = q k -1 p , where
.

Note1: Assuming k = 1,2,… , we get a geometric progression with the first term p and denominator q (0q. For this reason, the distribution is called geometric.

Remark2: Row
converges and its sum is equal to one. Indeed, the sum of the series is
.

Example. The gun fires at the target until the first hit. Probability of hitting the target p = 0,6 . Find the probability that the hit will occur on the third shot.

Decision: By condition p = 0,6, q = 1 – 0,6 = 0,4, k = 3. The desired probability is equal to:

P (X = 3) = 0,4 2 0.6 = 0.096.

5.5. Hypergeometric distribution

Consider the following problem. Let the party out N products available M standard (MN). randomly selected from the party n products (each product can be removed with the same probability), and the selected product is not returned to the batch before the selection of the next one (therefore, the Bernoulli formula is not applicable here).

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Continuous random variables have an infinite number of possible values. Therefore, it is impossible to introduce a distribution series for them.

Instead of the probability that the random variable X will take on a value equal to x, i.e. p(X = x), consider the probability that X will take on a value less than x, i.e. P(X< х).

We introduce a new characteristic of random variables - the distribution function and consider its properties.

The distribution function is the most universal characteristic of a random variable. It can be defined for both discrete and continuous random variables:

F(x) = p(X< x).

Distribution function properties.

The distribution function is a non-decreasing function of its argument, i.e. if:

At minus infinity, the distribution function is zero:

At plus infinity, the distribution function is equal to one:

The probability of a random variable falling into a given interval is determined by the formula:

The function f(x), which is equal to the derivative of the distribution function, is called the probability density of a random variable X or the distribution density:

Let's express the probability of hitting the section b to c in terms of f(x). It is equal to the sum of the probability elements in this section, i.e. integral:

From here, we can express the distribution function in terms of the probability density:

Probability density properties.

The probability density is a non-negative function (since the distribution function is a non-decreasing function):

Density probably

sti is a continuous function.

The integral in infinite limits of the probability density is equal to 1:

The probability density has the dimension of a random variable.

Mathematical expectation and dispersion of a continuous random variable

The meaning of the mathematical expectation and variance remains the same as in the case of discrete random variables. The form of formulas for finding them changes by replacing:

Then we obtain formulas for calculating the mathematical expectation and dispersion of a continuous random variable:

Example. The distribution function of a continuous random variable is given by:

Find the value of a, the probability density, the probability of hitting the site (0.25-0.5), the mathematical expectation and the variance.

Since the distribution function F(x) is continuous, then for x = 1 ax2 = 1, hence a = 1.

The probability density is found as a derivative of the distribution function:

The calculation of the probability of hitting a given area can be done in two ways: using the distribution function and using the probability density.

1st way. We use the formula for finding the probability through the distribution function:
2nd way. We use the formula for finding the probability through the probability density:

Finding the mathematical expectation:

Finding the variance:

Uniform distribution

Consider a continuous random variable X, the possible values of which lie in a certain interval and are equally probable.

The probability density of such a random variable will be:

where c is some constant.

The probability density graph will be displayed as follows:

We express the parameter c in terms of b and c. To do this, we use the fact that the integral of the probability density over the entire region must be equal to 1:

Distribution density of a uniformly distributed random variable

Find the distribution function:

Distribution function of a uniformly distributed random variable

Let's plot the distribution function:

Let us calculate the mathematical expectation and variance of a random variable obeying a uniform distribution.

Then the standard deviation will look like:

Normal (Gaussian) distribution

A continuous random variable X is called normally distributed with parameters a, y > 0 if it has a probability density:

The distribution curve of a random variable has the form:

Test 2

Task 1. Compose the law of distribution of a discrete random variable X, calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 1

QCD checks products for standardization. The probability that the item is standard is 0.7. 20 items tested. Find the law of distribution of the random variable X - the number of standard products among the tested ones. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 2

There are 4 balls in the urn, on which points 2 are indicated; 4; 5; 5. A ball is drawn at random. Find the law of distribution of a random variable X - the number of points on it. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 3

The hunter shoots the game until it hits, but can fire no more than three shots. The probability of hitting each shot is 0.6. Compose the law of distribution of the random variable X - the number of shots fired by the shooter. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 4

The probability of exceeding the specified accuracy in the measurement is 0.4. Compose the law of distribution of a random variable X - the number of errors in 10 measurements. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 5

The probability of hitting the target with one shot is 0.45. 20 shots fired. Compose the law of distribution of a random variable X - the number of hits. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 6

Products of a certain factory contain 5% of the marriage. Make a distribution law for a random variable X - the number of defective products among five taken for good luck. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 7

The parts needed by the assembler are in three of the five boxes. The assembler opens the boxes until he finds the right parts. Compose the law of distribution of a random variable X - the number of opened boxes. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 8

An urn contains 3 black and 2 white balls. Sequential extraction of balls without return is carried out until black appears. Compose the law of distribution of a random variable X - the number of extracted balls. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 9

The student knows 15 questions out of 20. There are 3 questions in the ticket. Compose the law of distribution of a random variable X - the number of questions known to the student in the ticket. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Option 10

There are 3 light bulbs, each of which has a defect with a probability of 0.4. When turned on, the defective light bulb burns out and is replaced by another. Make a distribution law for a random variable X - the number of lamps tested. Calculate the mathematical expectation, variance and standard deviation of a random variable.

Task 2. The random variable X is given by the distribution function F(X). Find the distribution density, mathematical expectation, variance, and also the probability of a random variable falling into the interval (b, c). Construct graphs of the functions F(X) and f(X).

Option 1

Option 2

Option 3

Option 4

Option 5

Option 6

Option 7

Option 8

Option 9

Option 10

Questions for the exam

The classic definition of probability.

Elements of combinatorics. Accommodation. Examples.

Elements of combinatorics. Permutation. Examples.

Elements of combinatorics. Combinations. Examples.

Theorem on the sum of probabilities.

Probability multiplication theorem.

Operations on events.

Total Probability Formula.

Bayes formula.

Repetition of tests. Bernoulli formula.

Discrete random variables. Distribution range. Example.

Mathematical expectation of a discrete random variable.

Dispersion of a discrete random variable.

Binomial distribution of a random variable.

Poisson distribution.

Distribution according to the law of geometric progression.

Continuous random variables. Distribution function and its properties.

Probability density and its properties.

Mathematical expectation of a continuous random variable.

Dispersion of a continuous random variable.

Uniform distribution of a continuous random variable.

Normal distribution law.

Concepts of mathematical expectation M(X) and dispersion D(X) introduced earlier for a discrete random variable can be extended to continuous random variables.

· Mathematical expectation M(X) continuous random variable X is defined by the equality:

provided that this integral converges.

· Dispersion D(X) continuous random variable X is defined by the equality:

· Standard deviationσ( X) continuous random variable is defined by the equality:

All the properties of mathematical expectation and dispersion considered earlier for discrete random variables are also valid for continuous ones.

Problem 5.3. Random value X given by the differential function f(x):

To find M(X), D(X), σ( X), as well as P(1 < X< 5).

Decision:

M(X)= =

+ = 8/9 0+9/6 4/6=31/18,

D(X)=

= = /

P 1 =

Tasks

5.1. X

f(x), as well as

R(‒1/2 < X< 1/2).

5.2. Continuous random variable X given by the distribution function:

Find the differential distribution function f(x), as well as

R(2π /9< X< π /2).

5.3. Continuous random variable X

Find: a) number with; b) M(X), D(X).

5.4. Continuous random variable X given by the distribution density:

Find: a) number with; b) M(X), D(X).

5.5. X:

Find: a) F(X) and plot its graph; b) M(X), D(X), σ( X); c) the probability that in four independent trials the value X takes exactly 2 times the value belonging to the interval (1;4).

5.6. Given the probability distribution density of a continuous random variable X:

Find: a) F(X) and plot its graph; b) M(X), D(X), σ( X); c) the probability that in three independent trials the value X will take exactly 2 times the value belonging to the interval .

5.7. Function f(X) is given as:

with X; b) distribution function F(x).

5.8. Function f(x) is given as:

Find: a) the value of the constant with, at which the function will be the probability density of some random variable X; b) distribution function F(x).

5.9. Random value X, concentrated on the interval (3;7), is given by the distribution function F(X)= X takes the value: a) less than 5, b) not less than 7.

5.10. Random value X, concentrated on the interval (-1; 4), is given by the distribution function F(X)= . Find the probability that the random variable X takes the value: a) less than 2, b) less than 4.

5.11.

Find: a) number with; b) M(X); c) probability R(X > M(X)).

5.12. The random variable is given by the differential distribution function:

Find: a) M(X); b) probability R(X ≤ M(X)).

5.13. The Time distribution is given by the probability density:

Prove that f(x) is indeed a probability density distribution.

5.14. Given the probability distribution density of a continuous random variable X:

Find a number with.

5.15. Random value X distributed according to Simpson's law (isosceles triangle) on the segment [-2; 2] (Fig. 5.4). Find an analytical expression for the probability density f(x) on the whole number line.

Rice. 5.4 Fig. 5.5

5.16. Random value X distributed according to the "right triangle" law in the interval (0; 4) (Fig. 5.5). Find an analytical expression for the probability density f(x) on the whole number line.

Answers

P (-1/2<X<1/2)=2/3.

P(2π /9<X< π /2)=1/2.

5.3. a) with=1/6, b) M(X)=3 , c) D(X)=26/81.

5.4. a) with=3/2, b) M(X)=3/5, c) D(X)=12/175.

b) M(X)= 3 , D(X)= 2/9, σ( X)= /3.

b) M(X)=2 , D(X)= 3 , σ( X)= 1,893.

5.7. a) c = ; b)

5.8. a) with=1/2; b)

5.9. a) 1/4; b) 0.

5.10. a) 3/5; b) 1.

5.11. a) with= 2; b) M(X)= 2; in 1- ln 2 2 ≈ 0,5185.