The law of distribution of the minimum (maximum) of two random variables. Law of distribution of order statistics
In this section we will consider first of all such a functional transformation c. c., which consists in choosing the maximum (minimum) of two values.
Problem 1. The law of distribution of the minimum of two random variables. A continuous system is given. V. (X and X 2) with p.r./(*!, x 2). Find the distribution function of r.v. Y:
Solution. Let us first find P ( Y> y) = P (Xi > y; X 2 > y). Region D(y), where X> y and X 2 > y shown in Fig. 9.6.1. Probability of hitting a point (X[, X 2) to the region D(y) is equal
Where F (x b x 2) - system distribution function c. V. (Хь Х 2), F x(jq), F 2 (x 2) - distribution functions c. V. X And X 2 respectively. Hence,
To determine p.r. g (y) you need to find the derivative of the right side (9.6.1):
If with. V. X x, X 2 independent and distributed identically with p.r. Fi(X) =/ 2 (x) =f(x), That
Example 1. We consider the operation of a device consisting of two blocks Bi and B 2, the joint operation of which is absolutely necessary for the operation of the device. Block B operating times! and B 2 represent independent s. V. X And X 2, distributed according to exponential laws with parameters X And X 2. It is required to find the distribution law c. V. U- operating time of the technical unit.
Solution. It's obvious that
Using formulas (9.6.4) we find:
i.e. at least two independent random variables, distributed according to exponential laws with parameters X x and X 2, also distributed according to exponential laws with parameter X x + X 2. ?
Problem 2. Law of distribution of the minimum of P independent random variables. Given the system P independent villages V. (X x, X 2, ..., X p) with p.r .f (x x),f 2 (x 2), ...,f n (x n). Find f. R. and density c. V. Y= min (X X,.... X p).
Solution. A-priory
Example 2. We consider the operation of an automated system (AS), consisting of P subsystems For the speakers to work, everyone needs to work P subsystems; uptime of the /th subsystem 7} distributed according to the exponential law with the parameter (/ = 1, 2, P) and does not depend on the operating time of other subsystems. Determine the law of time distribution D i) of failure-free operation of the AS.
Solution. It's obvious that
Using formula (9.6.6) we find the r.v. distribution function. D l)
Thus, the distribution law c. V. - the minimum of P independent villages c., distributed according to exponential laws, is also exponential; while its parameter i)S n)) is equal to the sum of the parameters of these exponential distributions. It follows that
It can be shown that the distribution law c. V. D i) when large enough P will converge to the exponential law, even if s. V. 7) (/= 1, 2, ..., P) are not distributed according to exponential laws. Let us demonstrate this using the example of equally uniformly distributed s. V.:
In this case
and this is f. R. demonstrative law.
Thus, we can draw a conclusion that is widely used in engineering applications: if any device consists of a sufficiently large number of elements n, the operation of which is absolutely necessary for the operation of the device, then the law of time distribution F p) of failure-free operation of the device is close to exponential with the parameter, determined by the formula
where M[ Tj- average failure-free operation time of the i-th element.
The failure flow of such a device will be close to Poisson with the parameter )S n ?
Problem 3. The law of distribution of the maximum of two random variables. A continuous system is given. V. (Хь X 2) with density/(lbs x 2). It is required to find the r.v. distribution law.
Solution. A-priory,
Where F(x x, x 2) - system distribution function (X and X 2).
Differentiating this expression as we did before, we get:
If random variables X and X2 are equally distributed, then
If random variables X x 2 are independent, then
If random variables X x 2 independent and equally distributed, then
Example 3. The operation of a technical device cannot begin before the assembly of its two blocks Bi and B2 is completed. The assembly time of blocks Bi and B 2 is a system of independent s. V. X x And X 2, distributed according to exponential laws with parameters X x And X 2. Y- time of completion of assembly of both technical specification blocks.
Solution. It's obvious that Y= max (X ъ X 2). Distribution density c. V. ^is determined by formula (9.6.12)
This law is not indicative. ?
Problem 4. The law of distribution of the maximum of P independent random variables. A continuous system is given. V. (X x, X 2 , ..., X p) with density f(x x, x 2,
Find the distribution law of a random variable
Solution. A-priory
Where F(x 1, X 2 ,..., x p) - system distribution function (X x, X 2, ..., X p). By differentiating, we find the distribution density:
Where Fj (Xj) - f. R. With. V. Xjfj(xj) - its density.
If with. V. x b ..., X p independent and equally distributed (Fi(y) = F(y);f (y) =f(y) (/"= 1,P)), That
If random variables X and ..., X p are independent, then
Example 4. The work of technical equipment cannot begin before the assembly of all P its blocks: B b Bg, ..., B„. The assembly times of blocks B b..., B l represent a system P independent villages V. (Huh..., X p), distributed according to exponential laws with parameters A.1,..., A, p.
We need to find the density c. V. U- completion time for all assembly P TU blocks.
Solution. Obviously y = max (X,..., X p). According to formula (9.6.16) we have 
Problem 5. Law of distribution of order statistics. Let us consider a continuous system of identically distributed, independent s. V. (X v X 2, ..., X p) with f. R. F(x) and p.r./(x). Let us arrange the values assumed by the random variables X v X 2, ..., X p, in ascending order and denote:
X (1) - random variable that takes the smallest of values: (X (1) = min (X v X 2, ..., X p));
X(2) - second largest accepted value of the random variables X v X 2, ..., X p;
X(T) - y-i by the magnitude of the accepted value from random variables X x, X 2, ..., X p;
X(P) - the largest random variable according to the accepted value X, X 2, x„ (X (n) = Shah (X and X 2, ..., X p)).
Obviously,
Random variables X(i), X@),..., X(") are called ordinal statistics.
Formulas (9.6.8) and (9.6.17) give the laws of distribution of extreme terms X(i), And X(") systems (*).
Let's find the distribution function F^m)(x)s. V. X^t y Event (X^x) is that T With. V. from the system P With. V. (X ( , X 2 ,..., X n) will be less than x and (p - t) With. V. will be greater than x. Since s. V. Xt (/" = 1, 2,..., P) are independent and identically distributed, then P (X t x) = F(x) R (Xj > x) = 1 - F(x). We need to find the probability that in P independent experiments event (Xj x) will appear exactly T once. Applying the binomial distribution, we get
Draw up a distribution law for the number of defective parts produced during a shift on both machines, and calculate the mathematical expectation and standard deviation of this random variable.
192. The probability that the watch needs additional adjustment is 0.2. Draw up a law for the distribution of the number of watches that need additional adjustment among three randomly selected watches. Using the resulting distribution law, find the mathematical expectation and variance of this random variable. Check the result using the appropriate formulas for the mathematical expectation and dispersion of a random variable distributed according to the binomial law.
193. From the available six lottery tickets, of which four are non-winning, one ticket is drawn at random until a winning ticket is encountered. Draw up a distribution law for the random variable X - the number of tickets taken out, if each ticket taken out is not returned back. Find the mathematical expectation and standard deviation of this random variable.
194. A student can take the exam no more than four times. Draw up a distribution law for the random variable X - the number of attempts to pass the exam, if the probability of passing it is 0.75 and subsequently increases by 0.1 with each subsequent attempt. Find the variance of this random variable.
195. The laws of distribution of two independent random variables X and Y are given:
| X | – 6 | Y | – 3 | – 1 | ||||
| P | 0,3 | 0,45 | 0,25 | 0,75 | 0,25 |
Draw up a distribution law for the random variable X–Y and check the dispersion property D(X–Y) = D(X) + D(Y).
196. Among the five clocks of the same type available in the workshop, only one has a misaligned pendulum. The master checks a randomly chosen watch. The review ends as soon as a clock with a displaced pendulum is detected (the checked clocks are not viewed again). Draw up a distribution law for the number of hours watched by the master and calculate the mathematical expectation and dispersion of this random variable.
197. Independent random variables X and Y are specified by distribution laws:
| X | Y | – 2 | ||||||
| P | 0,1 | 0,3 | ? | 0,4 | 0,6 |
Draw up the law of distribution of the random variable X 2 + 2Y and check the property of the mathematical expectation: M(X 2 + 2Y) = M(X 2) + 2M(Y).
198. It is known that a random variable X, taking two values x 1 = 1 and x 2 = 2, has a mathematical expectation equal to 7/6. Find the probabilities with which the random variable X takes its values. Draw up a distribution law for a random variable 2 X 2 and find its variance.
199. Two independent random variables X and Y are specified by the distribution laws:
Find P(X= 3) and P(Y= 4). Draw up the law of distribution of the random variable X – 2Y and check the properties of the mathematical expectation and dispersion: M(X – 2Y) = M(X) – 2M(Y); D(X – 2Y) = D(X) + 4D(Y).
In problems 201–210, random variables are given that are distributed according to the normal law
201. The random variable ξ is normally distributed. Find P(0< ξ<10), если Мξ= 10 и Р(10< ξ<20)= 0,3.
202. The random variable ξ is normally distributed. Find P(35< ξ<40), если Мξ= 25 и Р(10< ξ<15)= 0,2.
203. The random variable ξ is normally distributed. Find P(1< ξ<3), если Мξ= 3 и Р(3< ξ<5)= 0,1915.
204. <σ).
205. For a random variable ξ distributed according to the normal law, find Р(|ξ–а|<2σ).
206. For a random variable ξ distributed according to the normal law, find Р(|ξ–а|<4σ).
207. Independent random variables ξ and η are normally distributed,
Мξ= –1; Dξ= 2; Мη= 5; Dη= 7. Write down the probability density and the distribution function of their sum. Find Р(ξ+η<5) и Р(–1< ξ+η<3).
208. Independent random variables ξ, η, ζ are distributed according to the normal law and Мξ= 3; Dξ= 4; Мη= –2; Dη= 0.04; Мζ= 1; Dζ= 0.09. Write down the probability density and distribution function for their sum. Find Р(ξ+η+ζ<5) и Р(–1< ξ+η+ζ<3).
209. Independent random variables ξ, η, ζ are normally distributed and Мξ= –1; Dξ= 9; Мη= 2; Dη= 4; Мζ= –3; Dζ= 0.64. Write down the probability density and distribution function for their sum. Find Р(ξ+η+ζ<0) и
Р(–3< ξ+η+ζ<0).
210. The automatic machine produces rollers, controlling their diameters ξ. Assuming that ξ is normally distributed and a = 10 mm, σ = 0.1 mm, find the interval in which the diameters of the manufactured rollers will be contained with a probability of 0.9973.
In problems 211–220, a sample X of volume n = 100 is given by the table:
| x i | x 1 | x 2 | x 3 | x 4 | x 5 | x 6 | x 7 |
| n i | 20+(a+b) | 30–(a+b) |
where the measurement results x i = 0.2·a +(i –1)·0.3·b; n i – frequencies with which values x i occur.
1) construct a polygon of relative frequencies w i =n i /n;
2) calculate the sample mean, sample variance D B and standard deviation σ B;
3) calculate theoretical frequencies. Construct a graph on the same drawing as the polygon;
4) using the χ 2 criterion, test the hypothesis about the normal distribution of the population at a significance level of α = 0.05.
211. a = 4; b = 3; 212 . a = 3; b = 2; 213. a = 5; b = 1; 214. a = 1; b = 4;
215. a = 3; b = 5; 216. a=2; b = 3; 217. a = 4; b = 1; 218. a = 2; b = 5; 219. a = 1; b = 2; 220. a = 5; b = 4.
In problems 221–230, a two-dimensional sample of results of joint measurements of characteristics X and Y with a volume of n = 100 is specified by a correlation table:
| X Y | y 1 | y 2 | y 3 | y 4 | y 5 | n xi |
| x 1 | – | – | – | |||
| x 2 | – | – | ||||
| x 3 | – | 8+a | 12+b | – | – | 20+(a+b) |
| x 4 | – | – | 16–a | 14–b | – | 30–(a+b) |
| x 5 | – | – | – | |||
| x 6 | – | – | ||||
| x 7 | – | – | – | |||
| n yi | 19+a | 42+b–a | 31–b | n = 100 |
where x i = 0.2·a +(i –1)·0.3·b; y i = 0.5·a +(j – 1)·0.2·b.
1) Find and σ y. Take the values of and σ x from the previous problem.
2) Calculate the correlation coefficient r B . Draw a conclusion about the nature of the relationship between characteristics X and Y.
3) Construct the equation of a straight line of regression of Y on X in the form.
4) Draw the correlation field on the graph, i.e. plot the points (xi, yi) and construct a straight line.
221. a = 4; b = 3; 222. a = 3; b = 2; 223. a = 5; b = 1;
224. a = 1; b = 4; 225. a = 3; b = 5; 226. a = 2; b = 3;
227. a = 4; b = 1; 228. a = 2; b = 5; 229. a = 1; b = 2
230. a = 5; b = 4
In problems 231–240, find the maximum value of the function 
under conditions 
.
Take the values from the table
| Options | Options | |||||||||
| A 1 | ||||||||||
| A 2 | ||||||||||
| A 3 | ||||||||||
| B 1 | ||||||||||
| B 2 | ||||||||||
| B 3 | ||||||||||
| T 1 | ||||||||||
| T 2 | ||||||||||
| T 3 | ||||||||||
| C 1 | ||||||||||
| C 2 |
required:
1) solve a linear programming problem using a graphical method;
2) solve the problem using the tabular simplex method;
3) show the correspondence between the support solutions and the vertices of the region of feasible solutions;
In problems 241–250, some homogeneous cargo concentrated among three suppliers A i () must be delivered to five consumers B j (). The cargo inventories of suppliers a i and the needs of consumers b j , as well as the cost of transporting a unit of cargo from the i-th supplier to the j-th consumer C ij are given in the table.
| Suppliers | Consumers | Reserves | ||||
| B 1 | B 2 | B 3 | B 4 | B 5 | ||
| A 1 | From 11 | From 12 | From 13 | From 14 | From 15 | a 1 |
| A 2 | From 21 | From 22 | From 23 | From 24 | From 25 | a 2 |
| A 3 | C 31 | C 32 | C 33 | C 34 | From 35 | a 3 |
| Needs | b 1 | b 2 | b 3 | b 4 | b 5 |
Need to determine an optimal transportation plan that allows all cargo to be removed from suppliers and satisfies the needs of all consumers in such a way that this plan has a minimum cost. Find the first support plan using the “northwest” angle method. Find the optimal plan using the potential method. Calculate the shipping costs for each plan.
| Options | Options | |||||||||
| a 1 | ||||||||||
| a 2 | ||||||||||
| a 3 | ||||||||||
| b 1 | ||||||||||
| b 2 | ||||||||||
| b 3 | ||||||||||
| b 4 | ||||||||||
| b 5 | ||||||||||
| From 11 | ||||||||||
| From 12 | ||||||||||
| From 13 | ||||||||||
| From 14 | ||||||||||
| From 15 | ||||||||||
| From 21 | ||||||||||
| From 22 | ||||||||||
| From 23 | ||||||||||
| From 24 | ||||||||||
| From 25 | ||||||||||
| C 31 | ||||||||||
| C 32 | ||||||||||
| C 33 | ||||||||||
| C 34 | ||||||||||
| From 35 |
In tasks 251-260, the industry makes capital investments in four objects. Taking into account the characteristics of the contribution and local conditions, the profit of the industry, depending on the amount of financing, is expressed by the elements of the payment matrix. To simplify the problem, assume that the industry loss is equal to the industry profit. Find optimal industry strategies. Required:
1) summarize the initial data in a table and find a solution to the matrix game in pure strategies, if it exists (otherwise, see the next step 2);
2) simplify the payment matrix;
3) create a pair of mutually dual problems equivalent to the given matrix game;
4) find the optimal solution to the direct problem (for industry B) using the simplex method;
5) using the correspondence of variables, write out the optimal solution to the dual problem (for industry A);
6) give a geometric interpretation of this solution (for industry A);
7) using the relationship between optimal solutions to a pair of dual problems, optimal strategies and the cost of the game, find a solution to the game in mixed strategies;
option 1 option 2 option 3
1. Analytical geometry and vector algebra……………….. 4
2. Systems of linear equations and complex numbers………….. 5
3. Plotting function graphs, calculating limits
and identifying breakpoints of functions.…………….……………. 6
4. Derivatives of functions, greatest and least values
on the segment..…………………………………………………….… 9
5. Research of functions and construction of graphs,
functions of several variables, least squares method..… 11
6. Indefinite, definite and improper integral….. 12
7. Solving differential equations and systems
differential equations…………….……….…….….…… 14
8. Multiple and curvilinear integrals …………………………… 15
9. Study of numerical and power series, approximate
solutions to differential equations………………...……… 17
10. Probability theory……………….……………………...……… 18
Petr Alekseevich Burov
Anatoly Nikolaevich Muravyov
Collection of tasks
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Two random variables $X$ and $Y$ are called independent if the distribution law of one random variable does not change depending on what possible values the other random variable takes. That is, for any $x$ and $y$ the events $X=x$ and $Y=y$ are independent. Since the events $X=x$ and $Y=y$ are independent, then by the theorem of the product of probabilities of independent events $P\left(\left(X=x\right)\left(Y=y\right)\right)=P \left(X=x\right)P\left(Y=y\right)$.
Example 1 . Let the random variable $X$ express the cash winnings from tickets of one lottery “Russian Lotto”, and the random variable $Y$ express the cash winnings from tickets of another lottery “Golden Key”. It is obvious that the random variables $X,\Y$ will be independent, since the winnings from tickets of one lottery do not depend on the law of distribution of winnings from tickets of another lottery. In the case where the random variables $X,\Y$ would express the winnings of the same lottery, then, obviously, these random variables would be dependent.
Example 2 . Two workers work in different workshops and produce various products that are unrelated to each other by manufacturing technologies and the raw materials used. The distribution law for the number of defective products manufactured by the first worker per shift has the following form:
$\begin(array)(|c|c|)
\hline
Number of \ defective \ products \ x & 0 & 1 \\
\hline
Probability & 0.8 & 0.2 \\
\hline
\end(array)$
The number of defective products produced by the second worker per shift obeys the following distribution law.
$\begin(array)(|c|c|)
\hline
Number of \ defective \ products \ y & 0 & 1 \\
\hline
Probability & 0.7 & 0.3 \\
\hline
\end(array)$
Let us find the distribution law for the number of defective products produced by two workers per shift.
Let the random variable $X$ be the number of defective products produced by the first worker per shift, and $Y$ the number of defective products produced by the second worker per shift. By condition, the random variables $X,\Y$ are independent.
The number of defective products produced by two workers per shift is a random variable $X+Y$. Its possible values are $0,\ 1$ and $2$. Let us find the probabilities with which the random variable $X+Y$ takes its values.
$P\left(X+Y=0\right)=P\left(X=0,\ Y=0\right)=P\left(X=0\right)P\left(Y=0\right) =0.8\cdot 0.7=0.56.$
$P\left(X+Y=1\right)=P\left(X=0,\ Y=1\ or\ X=1,\ Y=0\right)=P\left(X=0\right )P\left(Y=1\right)+P\left(X=1\right)P\left(Y=0\right)=0.8\cdot 0.3+0.2\cdot 0.7 =0.38.$
$P\left(X+Y=2\right)=P\left(X=1,\ Y=1\right)=P\left(X=1\right)P\left(Y=1\right) =0.2\cdot 0.3=0.06.$
Then the law of distribution of the number of defective products manufactured by two workers per shift:
$\begin(array)(|c|c|)
\hline
Number of \ defective \ products & 0 & 1 & 2 \\
\hline
Probability & 0.56 & 0.38 & 0.06\\
\hline
\end(array)$
In the previous example, we performed an operation on random variables $X,\Y$, namely, we found their sum $X+Y$. Let us now give a more rigorous definition of operations (addition, difference, multiplication) over random variables and give examples of solutions.
Definition 1. The product $kX$ of a random variable $X$ by a constant variable $k$ is a random variable that takes values $kx_i$ with the same probabilities $p_i$ $\left(i=1,\ 2,\ \dots ,\ n\ right)$.
Definition 2. The sum (difference or product) of random variables $X$ and $Y$ is a random variable that takes all possible values of the form $x_i+y_j$ ($x_i-y_i$ or $x_i\cdot y_i$), where $i=1 ,\ 2,\dots ,\ n$, with probabilities $p_(ij)$ that the random variable $X$ will take the value $x_i$, and $Y$ the value $y_j$:
$$p_(ij)=P\left[\left(X=x_i\right)\left(Y=y_j\right)\right].$$
Since the random variables $X,\Y$ are independent, then according to the probability multiplication theorem for independent events: $p_(ij)=P\left(X=x_i\right)\cdot P\left(Y=y_j\right)= p_i\cdot p_j$.
Example 3 . Independent random variables $X,\ Y$ are specified by their probability distribution laws.
$\begin(array)(|c|c|)
\hline
x_i & -8 & 2 & 3 \\
\hline
p_i & 0.4 & 0.1 & 0.5 \\
\hline
\end(array)$
$\begin(array)(|c|c|)
\hline
y_i & 2 & 8 \\
\hline
p_i & 0.3 & 0.7 \\
\hline
\end(array)$
Let's formulate the law of distribution of the random variable $Z=2X+Y$. The sum of random variables $X$ and $Y$, that is, $X+Y$, is a random variable that takes all possible values of the form $x_i+y_j$, where $i=1,\ 2,\dots ,\ n$ , with probabilities $p_(ij)$ that the random variable $X$ will take the value $x_i$, and $Y$ the value $y_j$: $p_(ij)=P\left[\left(X=x_i\right )\left(Y=y_j\right)\right]$. Since the random variables $X,\Y$ are independent, then according to the probability multiplication theorem for independent events: $p_(ij)=P\left(X=x_i\right)\cdot P\left(Y=y_j\right)= p_i\cdot p_j$.
So, it has distribution laws for the random variables $2X$ and $Y$, respectively.
$\begin(array)(|c|c|)
\hline
x_i & -16 & 4 & 6 \\
\hline
p_i & 0.4 & 0.1 & 0.5 \\
\hline
\end(array)$
$\begin(array)(|c|c|)
\hline
y_i & 2 & 8 \\
\hline
p_i & 0.3 & 0.7 \\
\hline
\end(array)$
For the convenience of finding all values of the sum $Z=2X+Y$ and their probabilities, we will compose an auxiliary table, in each cell of which we will place in the left corner the values of the sum $Z=2X+Y$, and in the right corner - the probabilities of these values obtained as a result multiplying the probabilities of the corresponding values of random variables $2X$ and $Y$.
As a result, we obtain the distribution $Z=2X+Y$:
$\begin(array)(|c|c|)
\hline
z_i & -14 & -8 & 6 & 12 & 10 & 16 \\
\hline
p_i & 0.12 & 0.28 & 0.03 & 0.07 & 0.15 & 0.35 \\
\hline
\end(array)$