A.V.: Yulia, you did your diploma at Sergei Tchoban's studio "Movement Coordination", where your design object was the D-1 block in Skolkovo. As far as I can tell, your work was probably the most specific: you were designing for a place whose context had not yet been formed. What is it like?
Yu.A.: Working without an existing context was indeed a bit strange. In that area of Skolkovo, the master plan for which was developed by Sergey Tchoban's Speech bureau together with David Chipperfield's company, we were given a site, and we had to figure out what to do with it. In the first semester, we were divided into 3 groups of 4 people each and a competition was announced between us for a planning solution for one quarter. We had to place on the piece of land that we got, twelve, on average - five-story, houses, according to the number of students in the group. It so happened that our team won the competition: Anya Shevchenko, Dima Stolbovoy, Artem Slizunov and me. We ended up with a fairly rigid plan, which was limited not only by some cadastral parameters, but also terms of reference, and design code.
What is your master plan?
We changed the structure that was in the original version of the master plan: in order to reduce the scale of the environment, we divided our quarter into 4 sub-quarters with public space inside each. In addition, each sub-block had its own function: housing, start-ups, a sub-block with a sports function and a major building, and a site with a hostel, a hotel, a museum, and the main square is also located right there.
What restrictions did you write in the design code?
The quarter is very small, and the intentions of each of the participants could greatly influence the others. Therefore, we did not prescribe specific materials, but regulated possible shape changes by setting the "footprint" and FAR. For example, if you make a “gnaw”, your number of storeys grows, which in turn is also limited to a certain level.
What was the next step?
Further, each of us had to develop one of the buildings on the site, but which one, with what function - was determined by lot, we pulled pieces of paper with “lots”. That was Sergei Tchoban's plan. And this situation is fundamentally different from the one when you yourself choose the topic of the diploma and design a building with a specific function, which, perhaps, you dreamed of designing for all six years of study. Here we had to come to terms with the fact that we got by lot, and, on the one hand, it was quite painful, but on the other, this is a situation close to life.
What did you get?
I'm lucky, in my opinion. I designed the startup building. With certain dimensions, which could not be changed. The most important principle from which I proceeded was both ideological and functional: today it is a start-up, and tomorrow, it is quite likely that it will no longer exist.
After all, what is Skolkovo in general? No one can reasonably answer this question. After studying the materials, I came to the conclusion that own strategy development of Skolkovo is quite flexible. For me, this was the main condition that my project had to meet. Therefore, with a building width of 12 meters, it was important for me that there were no extra walls in my building. I left nothing but the stiffening cores, which are mandatory from the point of view of the design. Inside is an open, open floor plan. As for the external appearance, I tried to design my building in such a way that it was quite modest, but at the same time expressive.
The main facade turned out to be a 12-meter butt overlooking the boulevard. So I decided to sharpen its shape. The gable roof, which has become the visual accent of the entire building, plays an important role. It is an intermediate link between the two "neighbors" of my object, different in height and expressiveness.
Have you formed your own attitude towards the very idea of the Skolkovo Innovation Center in the process of work?
It changed during the course of the work. At first, the ideological context dominated a little. And then we began to perceive Skolkovo no longer as a phenomenon on the scale of Russia, but to carefully consider the problems of the place itself. 'Cause today it could be Innovation Center and tomorrow is something else. And what, your building therefore should be demolished? Good architecture can live longer than its original context. She creates a new one.
Was it difficult to work in a group? How was the relationship within the studio built when each of you started your own project?
Yes, of course it's hard. After all, it worked out in such a way that the situation as a whole could radically change from the wishes of each person. The plot is quite small, and someone's idea to make, say, a console or something else, could affect, for example, insolation standards. And then we all sat down and started discussing whether it was right or not.
The final version pleasantly surprised me. At first, it seemed to me that in everyone the desire to make a wow-thesis project, rather than a harmonious group work, would outweigh. But the master plan in the end turned out to be quite balanced. It seems to me that we have managed to find a "golden mean" between personal ambitions and the need to follow certain rules of the game.
What were the peculiarities of studying with Sergey Tchoban?
It was a pleasure to work with all the heads of our studio. In addition to Sergei, these are Alexei Ilyin and Igor Chlenov from the Speech bureau, and allied specialists also came to help deal with certain nodes. The educational process was built in a delightfully precise manner, literally to the minute. Although Sergei to some extent, probably, it was difficult with us. It seems to me that he was counting on the fact that we were already almost professionals. And we, I can’t say that we are still children, but the difference between an employee of the bureau and a student is still incredibly great. He shared his knowledge with us not as a teacher, but as a practicing architect and managed to make us work more independently and with each other than with teachers. It really was "coordination of movements".
What did two years of study at MARCH give you in general?
I cannot say that the third eye has opened. But some doubts were resolved, some positions were strengthened. Now I am more responsible for what I do and what I say. Perhaps a big thank you for this MARCH, perhaps a big thank you for this time. I can say that the most valuable thing in MARCH is main resource schools are people and some special atmosphere. Basically, I went there for the people. I went to Sergei Sitar, to Kiril Ass, to Evgeny Viktorovich, to Narine Tyutcheva. In addition, I had you, comrades, who inspired and supported me. I hope we will continue to communicate, I hope we will do something together.
Where did you study before?
I defended my bachelor's degree at the Moscow Institute of Architecture from the most excellent teacher Irina Mikhailovna Yastrebova. And I can add that I treat the Moscow Architectural Institute very well and do not think that this is some kind of Soviet relic. He gives the academic basics, and later everyone decides for himself what he wants to do.
What do you want to do now?
For all the years of my existence in architecture, I wrote about it, read about it, talked about it, but I never created it in the full sense of the word. I was doing, in fact, paper architecture, with such, you know, a claim to conceptual art. And if before I was completely sure that theory determines practice, now I cannot believe in it until I check it. Therefore, I now need to visit a construction site, I need to understand what it is - when you did something on paper, then you fought for it, argued, coordinated, and in the end you stand, look and understand: that's it, it happened! This is my fixed idea. Therefore, for the next two years, I plan to practice and try to make my way to the construction site, to implementation, as short as possible.
Problems in probability theory with solutions
1. Combinatorics
Task 1 . There are 30 students in a group. It is necessary to choose the headman, the deputy headman and the union leader. How many ways are there to do this?
Solution. Any of the 30 students can be chosen as a headman, any of the remaining 29 students as a deputy, and any of the remaining 28 students as a trade union organizer, i.e. n1=30, n2=29, n3=28. According to the multiplication rule, the total number N of ways to choose the headman, his deputy and trade union leader is N=n1´n2´n3=30´29´28=24360.
Task 2 . Two postmen have to deliver 10 letters to 10 addresses. In how many ways can they distribute work?
Solution. The first letter has n1=2 alternatives - either it refers to addressee the first postman, or the second. There are also n2=2 alternatives for the second letter, and so on, i.e. n1=n2=…=n10=2. Therefore, by virtue of the multiplication rule, the total number of ways to distribute letters between two postmen is
Task 3. There are 100 parts in a box, of which 30 are parts of the 1st grade, 50 are of the 2nd grade, and the rest are of the 3rd grade. How many ways are there to extract one part of the 1st or 2nd grade from the box?
Solution. A detail of the 1st grade can be extracted in n1=30 ways, of the 2nd grade – in n2=50 ways. According to the sum rule, there are N=n1+n2=30+50=80 ways to extract one part of the 1st or 2nd grade.
Task 5 . The order of performance of 7 participants of the competition is determined by lot. How many different variants of the draw are possible?
Solution. Each version of the draw differs only in the order of the participants in the competition, i.e. it is a permutation of 7 elements. Their number is
Task 6 . 10 films participate in the competition in 5 nominations. How many options for the distribution of prizes are there, if for all nominations various prizes?
Solution. Each of the prize distribution options is a combination of 5 films out of 10, which differs from other combinations both in composition and in their order. Since each film can receive prizes in one or several nominations, the same films can be repeated. Therefore, the number of such combinations is equal to the number of placements with repetitions of 10 elements by 5:
Task 7 . 16 people participate in a chess tournament. How many games must be played in a tournament if one game is to be played between any two participants?
Solution. Each game is played by two participants out of 16 and differs from the others only in the composition of pairs of participants, i.e., it is a combination of 16 elements by 2. Their number is 
Task 8 . In the conditions of task 6, determine how many options for the distribution of prizes exist, if for all nominations the same prizes?
Solution. If the same prizes are set for each nomination, then the order of films in the combination of 5 prizes does not matter, and the number of options is the number of combinations with repetitions of 10 elements of 5, determined by the formula
Task 9. The gardener must plant 6 trees within three days. In how many ways can he distribute the work among the days if he plants at least one tree a day?
Solution. Suppose a gardener is planting trees in a row and can make different decisions about which tree to stop on the first day and which one to stop on the second. Thus, one can imagine that the trees are separated by two partitions, each of which can stand in one of the 5 places (between the trees). Partitions must stand there one at a time, because otherwise not a single tree will be planted on some day. Thus, it is necessary to choose 2 elements from 5 (without repetitions). Therefore, the number of ways .
Task 10. How many four-digit numbers (possibly starting at zero) are there whose digits sum to 5?
Solution. Let's represent the number 5 as a sum of consecutive ones, divided into groups by partitions (each group in the sum forms the next digit of the number). It is clear that 3 such partitions will be needed. There are 6 places for partitions (before all units, between them and after). Each seat can be occupied by one or more partitions (in the latter case, there are no ones between them, and the corresponding sum is zero). Consider these places as elements of a set. Thus, it is necessary to choose 3 elements out of 6 (with repetitions). Therefore, the desired number of numbers 
Task 11 . In how many ways can a group of 25 students be divided into three subgroups A, B and C of 6, 9 and 10 people respectively?
Solution. Here n=25, k=3, n1=6, n2=9, n3=10..gif" width="160" height="41">
Task 1 . There are 5 oranges and 4 apples in a box. 3 fruits are chosen at random. What is the probability that all three fruits are oranges?
Solution. The elementary outcomes here are sets that include 3 fruits. Since the order of the fruits is indifferent, we will assume that their choice is unordered (and non-repetitive). gif" width="161 height=83" height="83">.
Task 2 . The teacher offers each of the three students to think of any number from 1 to 10. Assuming that the choice of any number from the given ones by each of the students is equally possible, find the probability that one of them will have the same conceived numbers.
Solution. First, let's calculate the total number of outcomes. The first student chooses one of 10 numbers and has n1=10 possibilities, the second one also has n2=10 possibilities, and finally the third one also has n3=10 possibilities. By virtue of the multiplication rule, the total number of ways is: n= n1´n2´n3=103 = 1000, i.e. the whole space contains 1000 elementary outcomes. To calculate the probability of the event A, it is convenient to pass to the opposite event, i.e., count the number of those cases when all three students think of different numbers. The first one still has m1=10 ways to choose a number. The second student now has only m2=9 possibilities, since he has to take care that his number does not coincide with the intended number of the first student. The third student is even more limited in his choice - he has only m3=8 possibilities. Therefore, the total number of combinations of conceived numbers in which there are no matches is equal to m=10×9×8=720. There are 280 cases in which there are matches. Therefore, the desired probability is P=280/1000=0.28.
Task 3 . Find the probability that in an 8-digit number exactly 4 digits are the same and the rest are different.
Solution. Event A=(an eight-digit number contains 4 identical digits). From the condition of the problem it follows that in the number of five different digits, one of them is repeated. The number of ways to choose it is equal to the number of ways to choose one digit from 10 digits..gif" width="21" height="25 src="> . Then the number of favorable outcomes. The total number of ways to compose 8-digit numbers is |W|=108 The desired probability is equal to
Task 4 . Six clients randomly apply to 5 firms. Find the probability that no one applies to at least one firm.
Solution. Consider the opposite event https://pandia.ru/text/78/307/images/image020_10.gif" width="195" height="41">. The total number of ways to distribute 6 clients among 5 firms. Hence 
. Consequently, .
Task 5 . Let an urn contain N balls, of which M are white and N–M are black. n balls are drawn from the urn. Find the probability that there will be exactly m white balls among them.
Solution. Since the order of the elements is not significant here, the number of all possible sets of size n of N elements is equal to the number of combinations of m white balls, n–m black balls, and, therefore, the desired probability is P(A)=https://pandia. ru/text/78/307/images/image031_2.gif" width="167" height="44">.
Task 7 (meeting task) . Two persons A and B agreed to meet at a certain place between 12 and 13 o'clock. The first person to arrive waits for the other for 20 minutes, after which he leaves. What is the probability of meeting persons A and B if the arrival of each of them can happen at random during the specified hour and the moments of arrival are independent?
Solution. Let's denote the arrival time of person A as x and person B as y. In order for the meeting to take place, it is necessary and sufficient that ôx-yô£20. Let's represent x and y as coordinates on the plane, as a unit of scale we will choose a minute. All possible outcomes are represented by the points of a square with a side of 60, and those favorable to the meeting are located in the shaded area. The desired probability is equal to the ratio of the area of the shaded figure (Fig. 2.1) to the area of the entire square: P(A) = (602–402)/602 = 5/9.
3. Basic formulas of probability theory
Task 1 . There are 10 red and 5 blue buttons in a box. Two buttons are taken out at random. What is the probability that the buttons are the same color? ?
Solution. The event A=(buttons of the same color are removed) can be represented as a sum , where the events and mean the choice of red and blue buttons, respectively. The probability of drawing two red buttons is equal, and the probability of drawing two blue buttons https://pandia.ru/text/78/307/images/image034_2.gif" width="19 height=23" height="23">.gif" width="249" height="83">
Task 2 . Among the firm's employees, 28% know English language, 30% - German, 42% - French; English and German - 8%, English and French - 10%, German and French - 5%, all three languages - 3%. Find the probability that a randomly selected employee of the company: a) knows English or German; b) knows English, German or French; c) does not know any of the listed languages.
Solution. Let A, B, and C denote the events that a randomly selected employee of the firm speaks English, German, or French, respectively. Obviously, the shares of the firm's employees who speak certain languages determine the probabilities of these events. We get:
a) P(AÈB)=P(A)+P(B) -P(AB)=0.28+0.3-0.08=0.5;
b) P(AÈBÈC)=P(A)+P(B)+P(C)-(P(AB)+P(AC)+P(BC))+P(ABC)=0.28+0, 3+0.42-
-(0,08+0,1+0,05)+0,03=0,8;
c) 1-P(AÈBÈC)=0.2.
Task 3 . The family has two children. What is the probability that the eldest child is a boy if it is known that there are children of both sexes in the family?
Solution. Let A = (the eldest child is a boy), B = (there are children of both sexes in the family). Let us assume that the birth of a boy and the birth of a girl are equiprobable events. If the birth of a boy is denoted by the letter M, and the birth of a girl is denoted by D, then the space of all elementary outcomes consists of four pairs: . In this space, only two outcomes (MD and MM) correspond to event B. Event AB means that there are children of both sexes in the family. The eldest child is a boy, therefore, the second (youngest) child is a girl. This event AB corresponds to one outcome - MD. Thus |AB|=1, |B|=2 and
Task 4 . The master, having 10 parts, of which 3 are non-standard, checks the parts one by one until he comes across a standard one. What is the probability that he checks exactly two details?
Solution. The event A=(the master checked exactly two parts) means that during such a check, the first part turned out to be non-standard, and the second - standard. Hence, , where =( the first part turned out to be non-standard) and =(the second part is standard). It is obvious that the probability of the event A1 is also equal to 
, since before taking the second part, the master had 9 parts left, of which only 2 are non-standard and 7 are standard. By the multiplication theorem
Task 5 . One box contains 3 white and 5 black balls, and the other box contains 6 white and 4 black balls. Find the probability that a white ball will be drawn from at least one box if one ball is drawn from each box.
Solution. The event A=(a white ball is taken out of at least one box) can be represented as a sum , where the events and mean the appearance of a white ball from the first and second boxes, respectively..gif" width="91" height="23">..gif " width="20" height="23 src=">.gif" width="480" height="23">.
Task 6 . Three examiners take an exam in a certain subject from a group of 30 people, with the first questioning 6 students, the second - 3 students, and the third - 21 students (students are randomly selected from the list). The ratio of the three examiners to the poorly prepared is different: the chances of such students to pass the exam are 40% for the first teacher, only 10% for the second, and 70% for the third. Find the probability that a poorly prepared student will pass the exam .
Solution. Denote by the hypotheses that the poorly prepared student answered the first, second, and third examiners, respectively. According to the task
, 
, 
.
Let the event A=(poorly prepared student passed the exam). Then again, by virtue of the condition of the problem
, 
, 
.
According to the total probability formula, we get:
Task 7 . The company has three sources of supply of components - companies A, B, C. Company A accounts for 50% of the total supply, B - 30% and C - 20%. It is known from practice that among the parts supplied by company A, 10% are defective, by company B - 5% and by company C - 6%. What is the probability that a part chosen at random will be good?
Solution. Let the event G be the appearance of a good part. The probabilities of the hypotheses that the part was supplied by firms A, B, C are respectively P(A)=0.5, P(B)=0.3, P(C)=0.2. The conditional probabilities of the appearance of a good part in this case are P(G|A)=0.9, P(G|B)=0.95, P(G|C)=0.94 (as the probabilities of opposite events to the appearance of a defective part). According to the total probability formula, we get:
P(G)=0.5×0.9+0.3×0.95+0.2×0.94=0.923.
Task 8 (see problem 6). Let it be known that the student did not pass the exam, i.e., received an “unsatisfactory” grade. Which of the three teachers most likely answered ?
Solution. The probability of getting "failed" is . It is required to calculate conditional probabilities. According to Bayes' formulas, we get:
https://pandia.ru/text/78/307/images/image059_0.gif" width="183" height="44 src=">, 
.
It follows that, most likely, the poorly prepared student took the exam to the third examiner.
4. Repeated independent tests. Bernoulli's theorem
Task 1 . A die is thrown 6 times. Find the probability that a six will come up exactly 3 times.
Solution. Rolling a die six times can be viewed as a sequence of independent trials with a probability of success ("six") equal to 1/6 and a probability of failure - 5/6. The desired probability is calculated by the formula 
.
Task 2 . The coin is tossed 6 times. Find the probability that the coat of arms appears at most 2 times.
Solution. The desired probability is equal to the sum of the probabilities of three events, consisting in the fact that the coat of arms does not fall out even once, either once or twice:
P(A) = P6(0) + P6(1) + P6(2) = https://pandia.ru/text/78/307/images/image063.gif" width="445 height=24" height= "24">.
Task 4 . The coin is tossed 3 times. Find the most probable number of successes (coat of arms).
Solution. The possible values for the number of successes in the three trials under consideration are m = 0, 1, 2, or 3. Let Am be the event that, on three tosses of a coin, the coat of arms appears m times. Using the Bernoulli formula, it is easy to find the probabilities of events Am (see table):
This table shows that the most likely values are the numbers 1 and 2 (their probabilities are 3/8). The same result can also be obtained from Theorem 2. Indeed, n=3, p=1/2, q=1/2. Then
, i.e. .
Task 5. As a result of each visit insurance agent the contract is concluded with a probability of 0.1. Find the most likely number of contracts signed after 25 visits.
Solution. We have n=10, p=0.1, q=0.9. The inequality for the most probable number of successes takes the form: 25×0.1–0.9 £m*£25×0.1+0.1 or 1.6 £m*£2.6. This inequality has only one integer solution, namely, m*=2.
Task 6 . It is known that the reject rate for some part is 0.5%. The inspector checks 1000 parts. What is the probability of finding exactly three defective parts? What is the probability of finding at least three defective parts?
Solution. We have 1000 Bernoulli trials with a probability of "success" p=0.005. Applying the Poisson approximation with λ=np=5, we get
2) P1000(m³3)=1-P1000(m<3)=1-»1-,
and P1000(3)»0.14; P1000 (m³3) "0.875.
Task 7 . The probability of a purchase when a customer visits a store is p=0.75. Find the probability that in 100 visits a customer will make a purchase exactly 80 times.
Solution. In this case, n=100, m=80, p=0.75, q=0.25. We find 
, and determine j(x)=0.2036, then the desired probability is P100(80)= 
.
Task 8. The insurance company concluded 40,000 contracts. The probability of an insured event for each of them during the year is 2%. Find the probability that there will be no more than 870 such cases.
Solution. By the condition of the problem n=40000, p=0.02. We find np=800,. To calculate P(m £ 870), we use the integral theorem of Moivre-Laplace:
P(0
We find according to the table of values of the Laplace function:
P(0 Task 9
.
The probability of an event occurring in each of 400 independent trials is 0.8. Find a positive number e such that with a probability of 0.99 the absolute value of the deviation of the relative frequency of occurrence of an event from its probability does not exceed e. Solution. By the condition of the problem p=0.8, n=400. We use the corollary from the Moivre-Laplace integral theorem: 5.
Discrete random variables Task 1
.
In a bunch of 3 keys, only one key fits the door. Keys are sorted until a suitable key is found. Build a distribution law for a random variable x - the number of tested keys .
Solution. The number of tried keys can be 1, 2 or 3. If only one key is tested, this means that this first key came to the door immediately, and the probability of such an event is 1/3. So, Further, if there were 2 tested keys, i.e. x=2, this means that the first key did not fit, and the second did. The probability of this event is 2/3×1/2=1/3..gif" width="100" height="21"> The result is the following distribution series: Task 2
.
Construct the distribution function Fx(x) for the random variable x from Problem 1. Solution. The random variable x has three values 1, 2, 3, which divide the entire numerical axis into four intervals: . If x<1, то неравенство x£x невозможно (левее x нет значений случайной величины x) и значит, для такого x функция Fx(x)=0. If 1£x<2, то неравенство x£x возможно только если x=1, а вероятность такого события равна 1/3, поэтому для таких x функция распределения Fx(x)=1/3. If 2£x<3, неравенство x£x означает, что или x=1, или x=2, поэтому в этом случае вероятность P(x And, finally, in the case of x³3, the inequality x£x holds for all values of the random variable x, so P(x So we got the following function: Task 3.
The joint law of distribution of random variables x and h is given using the table Calculate particular laws of distribution of components x and h. Determine if they are dependent..gif" width="423" height="23 src=">; https://pandia.ru/text/78/307/images/image086.gif" width="376" height="23 src=">. The partial distribution for h is similarly obtained: https://pandia.ru/text/78/307/images/image088.gif" width="229" height="23 src=">. The resulting probabilities can be written in the same table opposite the corresponding values of random variables: Now let's answer the question about the independence of the random variables x and h..gif" width="108" height="25 src="> in this cell. For example, in the cell for the values x=-1 and h=1 there is a probability 1/ 16, and the product of the corresponding partial probabilities 1/4×1/4 is equal to 1/16, i.e. coincides with the joint probability.This condition is also checked in the remaining five cells, and it turns out to be true in all.Therefore, the random variables x and h are independent. Note that if our condition were violated in at least one cell, then the quantities should be recognized as dependent. To calculate the probability mark the cells for which the condition is fulfilled https://pandia.ru/text/78/307/images/image092.gif" width="574" height="23 src="> Task 4
.
Let the random variable ξ have the following distribution law: Calculate the mathematical expectation Mx, dispersion Dx and standard deviation s. Solution. By definition, the expectation of x is Standard deviation https://pandia.ru/text/78/307/images/image097.gif" width="51" height="21">. Solution. Let's use the formula Task 6
.
For a pair of random variables from Problem 3, calculate the covariance cov(x, h). Solution. In the previous problem, the mathematical expectation has already been calculated .
It remains to calculate and .
Using the partial distribution laws obtained in solving Problem 3, we obtain ; ;
and that means which was to be expected due to the independence of the random variables. Task 7.
The random vector (x, h) takes the values (0.0), (1.0), (–1.0), (0.1), and (0,–1) with equal probability. Calculate the covariance of random variables x and h. Show that they are dependent. Solution. Since Р(x=0)=3/5, P(x=1)=1/5, P(x=–1)=1/5; Р(h=0)=3/5, P(h=1)=1/5, P(h=–1)=1/5, then Мx=3/5´0+1/5´1+1 /5´(–1)=0 and Мh=0; М(xh)=0´0´1/5+1´0´1/5–1´0´1/5+0´1´1/5–0´1´1/5=0. We get cov(x, h)=M(xh)–MxMh=0, and the random variables are uncorrelated. However, they are dependent. Let x=1, then the conditional probability of the event (h=0) is equal to Р(h=0|x=1)=1 and is not equal to the unconditional Р(h=0)=3/5, or the probability (ξ=0,η =0) is not equal to the product of probabilities: Р(x=0,h=0)=1/5¹Р(x=0)Р(h=0)=9/25. Hence x and h are dependent. Task 8
. Random increments in the stock prices of two companies on days x and h have a joint distribution given by the table: Find the correlation coefficient. Solution. First of all, we calculate Mxh=0.3-0.2-0.1+0.4=0.4. Next, we find particular distribution laws for x and h: We define Mx=0.5-0.5=0; Mh=0.6-0.4=0.2; Dx=1; Dh=1–0.22=0.96; cov(x, h)=0.4. We get Task 9.
Random increments in the prices of shares of two companies per day have dispersions Dx=1 and Dh=2, and their correlation coefficient is r=0.7. Find the variance of the increment in the price of a portfolio of 5 shares of the first company and 3 shares of the second company. Solution. Using the properties of variance, covariance and the definition of the correlation coefficient, we get: Task 10
.
The distribution of a two-dimensional random variable is given by the table: Find the conditional distribution and the conditional expectation h for x=1. Solution. The conditional expectation is From the condition of the problem, we find the distribution of the components h and x (the last column and the last row of the table).
. Consequently, 
..gif" width="587" height="41">
. Namely, in each cell of the table, we multiply the corresponding values and , multiply the result by the probability pij, and summarize all this over all the cells of the table. As a result, we get:
.
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