Dispersion of a discrete random variable
Although M(X) yields the weighted average of the possible values of X, it does not tell us anything about the variation, or spread, of these values. For instance, although random variables W, Y, and Z, having probability mass functions determined by
W = 0 with probability 1
All have the same expectation – namely, 0 – there is much greater spread in the possible value of Y than in those of W (which is a constant) and in the possible values of Z than in those of Y.
As we expect X to take on values around its mean M(X), it would appear that a reasonable way of measuring the possible variation of X would be to look at how far apart X would be from its mean on the average. In practice it is often required to estimate the dispersion (variation) of possible values of a random variable around of its average value. For example, in artillery it is important to know as far as shells will concentrically lie near to the target which should be struck.
One possible way to measure this would be to consider the quantity M(|X – a|), where a = M(X). However, it turns out to be mathematically inconvenient to deal with this quantity, and so a more tractable quantity is usually considered – namely, the expectation of the square of the difference between X and its mean. We thus have the following definition:
If X is a random variable with expectation M(X), then the dispersion (variance) of X, denoted by D(X), is defined by
D(X) = M[X – M(X)]2
An alternative formula for D(X) is derived as follows:
That is,
In words, the dispersion of X is equal to the expected value of X 2 minus the square of its expected value. This is, in practice, often the easiest way to compute D(X).
Thus, for our example we have D(W) = 0, D(Y) = 1 and D(Z) = 10000.
Example. Calculate D(X) if X represents the outcome when a die is rolled.
Solution: We have the following law of distribution:
| X | ||||||
| p | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
Consequently,
Also,
Hence,
Remark. Analogous to the mean being the center of gravity of a distribution of mass, the dispersion (variance) represents, in the terminology of mechanics, the moment of inertia.
The mean square deviation (the standard deviation) s(X) of a random variable X is the arithmetic value of the square root of its dispersion:
Property 1. The dispersion of a constant is equal to zero: D(C) = 0.
Property 2. A constant multiplier can be taken out from the argument of the dispersion involving it in square:
D(kX) = k 2 D(X)
Property 3. The dispersion of a random variable is equal to the difference between the mathematical expectation of the square of the random variable and the square of its mathematical expectation:
D(X) = M(X 2) – [M(X)]2
Property 4. The dispersion of the algebraic sum of finitely many mutually independent random variables is equal to the sum of their dispersions:
Observe that the dispersion of both the sum and the difference of independent random variables X and Y is equal to the sum of their dispersions, i.e.
D(X + Y) = D(X – Y) = D(X) + D(Y).
The mathematical expectation, the dispersion and the mean square deviation are numerical characteristics of a random variable.
Glossary
shell – артиллерийский снаряд; aim, sight – прицел
exhaustive – исчерпывающий; assignment – задание
ascending order – возрастающий порядок
mathematical expectation, mean value – математическое ожидание
rod –стержень; on the average – в среднем; deviation – отклонение
concentration– кучность; to strike – поражать
dispersion, variance – дисперсия, рассеяние
mean square deviation – среднее квадратическое отклонение
numerical characteristic – числовая характеристика
Exercises for Seminar 7
7.1. Two balls are chosen randomly from an urn containing 8 white, 4 black and 2 orange balls. Suppose that we win $2 for each black ball selected and we lose $1 for each white ball selected. Let X denote our winnings. What are the possible values of X, and what are the probabilities associated with each value?
7.2. The probability of working each of four combines without breakages during a certain time is equal to 0,9. Compose the law of distribution of a random variable X – the number of combines working trouble-free. Find the mathematical expectation, the dispersion and the mean square deviation of the random variable X.
The answer: M(X) = 3,6; D(X) = 0,36; s(X) = 0,6.
7.3. The probability of birth of a boy in a family is equal to 0,515. Compose the law of distribution of a random variable X – the number of boys in families having four children. Find the mathematical expectation, the dispersion and the mean square deviation.
The answer: M(X) = 2,06; D(X) = 0,999; s(X) = 1,0.
7.4. There are 6 masters of sports in a group of 10 sportsmen. One selects (under the circuit without replacement) 3 sportsmen. Compose the law of distribution of a random variable X – the number of masters of sports of the selected sportsmen. Find the mathematical expectation of the random variable X.
The answer: M(X) = 1,8.
7.5. A shooter makes shots in a target before the first hit. The probability of hit in the target at each shot is equal to 0,7. Compose the law of distribution of a random variable X – the number of shots made by the shooter. Find the most probable number of cartridges (patrons) given to the shooter.
The answer: k0 = 1.
7.6. The mathematical expectation of a random variable X is equal to 8. Find the mathematical expectation of the following random variables: a) X – 4; b) 3X + 4.
7.7. The dispersion of a random variable X is equal to 8. Find the dispersion of the following random variables: a) X – 2; b) 3X + 2.
7.8. Independent random variables X and Y have the following distributions:
| X | Y | ||||||
| p | 0,3 | 0,5 | 0,2 | p | 0,4 | 0,6 |
Compose the law of distribution of the random variable Z = X + Y. Find the mathematical expectation, the dispersion and the mean square deviation of the random variable Z.
The answer: M(Z) = 7,4; D(Z) = 2,2.
7.9. Find the mathematical expectation and the dispersion of random variable Z = 4X – 2Y if M(X) = 5, M(Y) = 3, D(X) = 4, D(Y) = 6. The random variables X and Y are independent.
The answer: M(Z) = 14; D(Z) = 88.
7.10. A total of 4 buses carrying 148 students from the same school arrives at a football stadium. The buses carry, respectively, 40, 33, 25, and 50 students. One of the students is randomly selected. Let X denote the number of students that were on the bus carrying this randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on his bus. Which of M(X) or M(Y) do you think is larger? Why? Compute M(X) and M(Y).
Exercises for Homework 7
7.11. Two dice are rolled. Let X equal the sum of the 2 dice. What are the possible values of X, and what are the probabilities associated with each value?
7.12. The probability that a buyer will make a purchase in a shop is equal to 0,4. Compose the law of distribution of a random variable X – the number of buyers who have made a purchase if the shop was visited by 3 buyers. Find the mathematical expectation, the dispersion and the mean square deviation of the random variable X.
The answer: M(X) = 1,2; D(X) = 0,72; s(X) = 0,85.
7.13. A buyer attends shops for purchasing the necessary goods. The probability that the goods are in a certain shop is equal to 0,4. Compose the law of distribution of a random variable X – the number of shops which will be attended by the buyer from four possible. Find the most probable number of shops which will be visited by the buyer.
The answer: 1 £ k0 £ 2.
7.14. A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number (mathematical expectation) of defective items in the sample.
The answer: 0,6.
7.15. A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if they are different colors, then you win – $1.00 (that is, you lose $1.00). Calculate the mathematical expectation and the dispersion of the amount you win (marble – мрамор; to withdraw – извлекать).
The answer: M(X) = – 1/15; D(X) = 49/45.
7.16. The mathematical expectation of a random variable X is equal to 7. Find the mathematical expectation of the following random variables:
a) X + 6; b) 4X – 3.
7.17. The dispersion of a random variable X is equal to 9. Find the dispersion of the following random variables: a) X + 6; b) 2X – 7.
7.18. Independent random variables X and Y have the following distributions:
| X | Y | ||||||
| p | 0,3 | 0,5 | 0,2 | p | 0,4 | 0,6 |
Compose the law of distribution of the random variable V = XY. Find the mathematical expectation, the dispersion and the mean square deviation of the random variable V.
The answer: M(V) = 13,68; D(V) = 29,3376.
7.19. Find the mathematical expectation and the dispersion of random variables:
a) Z = 2X – 4Y;
b) Z = 3X + 5Y
if M(X) = 5, M(Y) = 3, D(X) = 4, D(Y) = 6. The random variables X and Y are independent.
The answer: a) M(Z) = – 2; D(Z) = 112; b) M(Z) = 30; D(Z) = 186.
L E C T U R E 8