Mathematical expectation of the distribution table. Properties of mathematical expectation. Distribution function of a discrete random variable
As is already known, the distribution law completely characterizes a random variable. However, often the distribution law is unknown and one has to limit oneself to less information. Sometimes it is even more profitable to use numbers that describe the random variable in total; such numbers are called numerical characteristics random variable .
One of the important numerical characteristics is the mathematical expectation.
Expected value approximately equal to the average value of the random variable.
Mathematical expectation of a discrete random variable is the sum of the products of all its possible values and their probabilities.
If a random variable is characterized by a finite distribution series:
| X | x 1 | x 2 | x 3 | … | x n |
| R | p 1 | p 2 | p 3 | … | r p |
then the mathematical expectation M(X) determined by the formula:
The mathematical expectation of a continuous random variable is determined by the equality:
where is the probability density of the random variable X.
Example 4.7. Find the mathematical expectation of the number of points that appear when throwing a dice.
Solution:
Random value X takes the values 1, 2, 3, 4, 5, 6. Let’s create the law of its distribution:
| X | ||||||
| R |
Then the mathematical expectation is:
Properties of mathematical expectation:
1. The mathematical expectation of a constant value is equal to the constant itself:
M (S) = S.
2. The constant factor can be taken out of the mathematical expectation sign:
M (CX) = CM (X).
3. The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations:
M(XY) = M(X)M(Y).
Example 4.8. Independent random variables X And Y are given by the following distribution laws:
| X | Y | ||||||
| R | 0,6 | 0,1 | 0,3 | R | 0,8 | 0,2 |
Find the mathematical expectation of the random variable XY.
Solution.
Let's find the mathematical expectations of each of these quantities:
Random variables X And Y independent, therefore the required mathematical expectation is:
M(XY) = M(X)M(Y)=
Consequence. The mathematical expectation of the product of several mutually independent random variables is equal to the product of their mathematical expectations.
4. The mathematical expectation of the sum of two random variables is equal to the sum of the mathematical expectations of the terms:
M (X + Y) = M (X) + M (Y).
Consequence. The mathematical expectation of the sum of several random variables is equal to the sum of the mathematical expectations of the terms.
Example 4.9. 3 shots are fired with probabilities of hitting the target equal to p 1 = 0,4; p2= 0.3 and p 3= 0.6. Find the mathematical expectation of the total number of hits.
Solution.
The number of hits on the first shot is a random variable X 1, which can only take two values: 1 (hit) with probability p 1= 0.4 and 0 (miss) with probability q 1 = 1 – 0,4 = 0,6.
The mathematical expectation of the number of hits on the first shot is equal to the probability of a hit:
Similarly, we find the mathematical expectations of the number of hits for the second and third shots:
M(X 2)= 0.3 and M(X 3)= 0,6.
The total number of hits is also a random variable consisting of the sum of hits in each of the three shots:
X = X 1 + X 2 + X 3.
The required mathematical expectation X We find it using the theorem on the mathematical expectation of the sum.
Magnitude
Basic numerical characteristics of random
The density distribution law characterizes a random variable. But often it is unknown, and one has to limit oneself to less information. Sometimes it is even more profitable to use numbers that describe a random variable in total. Such numbers are called numerical characteristics random variable. Let's look at the main ones.
Definition:The mathematical expectation M(X) of a discrete random variable is the sum of the products of all possible values of this quantity and their probabilities:
If a discrete random variable X takes a countably many possible values, then
Moreover, the mathematical expectation exists if this series is absolutely convergent.
From the definition it follows that M(X) a discrete random variable is a non-random (constant) variable.
Example: Let X– number of occurrences of the event A in one test, P(A) = p. We need to find the mathematical expectation X.
Solution: Let's create a tabular distribution law X:
| X | 0 | 1 |
| P | 1 - p | p |
Let's find the mathematical expectation:
Thus, the mathematical expectation of the number of occurrences of an event in one trial is equal to the probability of this event.
Origin of the term expected value associated with the initial period of the emergence of probability theory (XVI-XVII centuries), when the scope of its application was limited to gambling. The player was interested in the average value of the expected win, i.e. mathematical expectation of winning.
Let's consider probabilistic meaning of mathematical expectation.
Let it be produced n tests in which the random variable X accepted m 1 times value x 1, m 2 times value x 2, and so on, and finally she accepted m k times value x k, and m 1 + m 2 +…+ + m k = n.
Then the sum of all values taken by the random variable X, is equal x 1 m 1 +x 2 m 2 +…+x k m k.
Arithmetic mean of all values taken by a random variable X,equals:
since is the relative frequency of a value for any value i = 1, …, k.
As is known, if the number of tests n is sufficiently large, then the relative frequency is approximately equal to the probability of the event occurring, therefore,
Thus, .
Conclusion:The mathematical expectation of a discrete random variable is approximately equal (the more accurately, the greater the number of tests) to the arithmetic mean of the observed values of the random variable.
Let's consider the basic properties of mathematical expectation.
Property 1:The mathematical expectation of a constant value is equal to the constant value itself:
M(C) = C.
Proof: Constant WITH can be considered , which has one possible meaning WITH and accepts it with probability p = 1. Hence, M(C) =C 1= S.
Let's define product of a constant variable C and a discrete random variable X as a discrete random variable CX, the possible values of which are equal to the products of the constant WITH to possible values X CX equal to the probabilities of the corresponding possible values X:
| CX | C | C | … | C |
| X | … | |||
| R | … |
Property 2:The constant factor can be taken out of the mathematical expectation sign:
M(CX) = CM(X).
Proof: Let the random variable X is given by the law of probability distribution:
| X | … | |||
| P | … |
Let's write the law of probability distribution of a random variable CX:
| CX | C | C | … | C |
| P | … |
M(CX) = C +C =C + ) = C M(X).
Definition:Two random variables are called independent if the distribution law of one of them does not depend on what possible values the other variable took. Otherwise, the random variables are dependent.
Definition:Several random variables are said to be mutually independent if the distribution laws of any number of them do not depend on what possible values the remaining variables took.
Let's define product of independent discrete random variables X and Y as a discrete random variable XY, the possible values of which are equal to the products of each possible value X for every possible value Y. Probabilities of possible values XY are equal to the products of the probabilities of possible values of the factors.
Let the distributions of random variables be given X And Y:
| X | … | |||
| P | … |
| Y | … | |||
| G | … |
Then the distribution of the random variable XY has the form:
| XY | … | |||
| P | … |
Some works may be equal. In this case, the probability of a possible value of the product is equal to the sum of the corresponding probabilities. For example, if = , then the probability of the value is
Property 3:The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations:
M(XY) = M(X) M(Y).
Proof: Let independent random variables X And Y are specified by their own probability distribution laws:
| X | ||
| P |
| Y | ||
| G |
To simplify the calculations, we will limit ourselves to a small number of possible values. In the general case the proof is similar.
Let's create a law of distribution of a random variable XY:
| XY | ||||
| P |
M(XY) =
M(X) M(Y).
Consequence:The mathematical expectation of the product of several mutually independent random variables is equal to the product of their mathematical expectations.
Proof: Let us prove for three mutually independent random variables X,Y,Z. Random variables XY And Z independent, then we get:
M(XYZ) = M(XY Z) = M(XY) M(Z) = M(X) M(Y) M(Z).
For an arbitrary number of mutually independent random variables, the proof is carried out by the method of mathematical induction.
Example: Independent random variables X And Y
| X | 5 | 2 | |
| P | 0,6 | 0,1 | 0,3 |
| Y | 7 | 9 |
| G | 0,8 | 0,2 |
Need to find M(XY).
Solution: Since random variables X And Y are independent, then M(XY)=M(X) M(Y)=(5 0,6+2 0,1+4 0,3) (7 0,8+9 0,2)= 4,4 7,4 = =32,56.
Let's define sum of discrete random variables X and Y as a discrete random variable X+Y, the possible values of which are equal to the sums of each possible value X with every possible value Y. Probabilities of possible values X+Y for independent random variables X And Y are equal to the products of the probabilities of the terms, and for dependent random variables - to the products of the probability of one term by the conditional probability of the second.
If = and the probabilities of these values are respectively equal, then the probability (the same as ) is equal to .
Property 4:The mathematical expectation of the sum of two random variables (dependent or independent) is equal to the sum of the mathematical expectations of the terms:
M(X+Y) = M(X) + M(Y).
Proof: Let two random variables X And Y are given by the following distribution laws:
| X | ||
| P |
| Y | ||
| G |
To simplify the conclusion, we will limit ourselves to two possible values of each quantity. In the general case the proof is similar.
Let's compose all possible values of a random variable X+Y(assume, for simplicity, that these values are different; if not, then the proof is similar):
| X+Y | ||||
| P |
Let's find the mathematical expectation of this value.
M(X+Y) = + + + +
Let's prove that + = .
Event X = ( its probability P(X = ) entails the event that the random variable X+Y will take the value or (the probability of this event, according to the addition theorem, is equal to ) and vice versa. Then = .
The equalities = = = are proved in a similar way
Substituting the right-hand sides of these equalities into the resulting formula for the mathematical expectation, we obtain:
M(X + Y) = + ) = M(X) + M(Y).
Consequence:The mathematical expectation of the sum of several random variables is equal to the sum of the mathematical expectations of the terms.
Proof: Let us prove for three random variables X,Y,Z. Let's find the mathematical expectation of random variables X+Y And Z:
M(X+Y+Z)=M((X+Y Z)=M(X+Y) M(Z)=M(X)+M(Y)+M(Z)
For an arbitrary number of random variables, the proof is carried out by the method of mathematical induction.
Example: Find the average of the sum of the number of points that can be obtained when throwing two dice.
Solution: Let X– the number of points that can appear on the first die, Y- On the second. It is obvious that random variables X And Y have the same distributions. Let's write down the distribution data X And Y into one table:
| X | 1 | 2 | 3 | 4 | 5 | 6 |
| Y | 1 | 2 | 3 | 4 | 5 | 6 |
| P | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
M(X) = M(Y) (1+2+3+4+5+6) = =
M(X + Y) = 7.
So, the average value of the sum of the number of points that can appear when throwing two dice is 7 .
Theorem:The mathematical expectation M(X) of the number of occurrences of event A in n independent trials is equal to the product of the number of trials and the probability of the occurrence of the event in each trial: M(X) = np.
Proof: Let X– number of occurrences of the event A V n independent tests. Obviously, total number X occurrences of the event A in these trials is the sum of the number of occurrences of the event in individual trials. Then, if the number of occurrences of an event in the first trial, in the second, and so on, finally, is the number of occurrences of the event in n-th test, then the total number of occurrences of the event is calculated by the formula:
By property 4 of mathematical expectation we have:
M(X) = M( ) + … + M( ).
Since the mathematical expectation of the number of occurrences of an event in one trial is equal to the probability of the event, then
M( ) = M( )= … = M( ) = p.
Hence, M(X) = np.
Example: The probability of hitting the target when firing from a gun is p = 0.6. Find the average number of hits if made 10 shots.
Solution: The hit for each shot does not depend on the outcomes of other shots, therefore the events under consideration are independent and, therefore, the required mathematical expectation is equal to:
M(X) = np = 10 0,6 = 6.
So the average number of hits is 6.
Now consider the mathematical expectation of a continuous random variable.
Definition:The mathematical expectation of a continuous random variable X, the possible values of which belong to the interval,called definite integral:
where f(x) is the probability distribution density.
If possible values of a continuous random variable X belong to the entire Ox axis, then
It is assumed that this improper integral converges absolutely, i.e. the integral converges If this requirement were not met, then the value of the integral would depend on the rate at which (separately) the lower limit tends to -∞, and the upper limit tends to +∞.
It can be proven that all properties of the mathematical expectation of a discrete random variable are preserved for a continuous random variable. The proof is based on the properties of definite and improper integrals.
It is obvious that the mathematical expectation M(X) greater than the smallest and less than the largest possible value of the random variable X. Those. on the number axis, possible values of a random variable are located to the left and to the right of its mathematical expectation. In this sense, the mathematical expectation M(X) characterizes the location of the distribution and is therefore often called distribution center.
1. The mathematical expectation of a constant value is equal to the constant itself M(S)=C
.
2. The constant factor can be taken out of the mathematical expectation sign: M(CX)=CM(X)
3. The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations: M(XY)=M(X) M(Y).
4. The mathematical expectation of the sum of two random variables is equal to the sum of the mathematical expectations of the terms: M(X+Y)=M(X)+M(Y).
Theorem. The mathematical expectation M(x) of the number of occurrences of events A in n independent trials is equal to the product of these trials by the probability of occurrence of events in each trial: M(x) = np.
Let X - random variable and M(X) – its mathematical expectation. Let us consider as a new random variable the difference X - M(X).
Deviation is the difference between a random variable and its mathematical expectation.
The deviation has the following distribution law:
Solution: Let's find the mathematical expectation:
2 =(1-2.3) 2 =1.69
2 =(2-2.3) 2 =0.09
2 =(5-2.3) 2 =7.29
Let's write the law of distribution of the squared deviation:
Solution: Let's find the mathematical expectation of M(x): M(x)=2 0.1+3 0.6+5 0.3=3.5
Let's write the law of distribution of the random variable X 2
| X 2 | |||
| P | 0.1 | 0.6 | 0.3 |
Let's find the mathematical expectation M(x 2):M(x 2) = 4 0.1+9 0.6+25 0.3=13.5
The required variance is D(x)=M(x 2)- 2 =13.3-(3.5) 2 =1.05
Dispersion properties:
1. Variance of a constant value WITH
equal to zero: D(C)=0
2. The constant factor can be taken out of the dispersion sign by squaring it. D(Cx)=C 2 D(x)
3. The variance of the sum of independent random variables is equal to the sum of the variances of these variables. D(X 1 +X 2 +...+X n)=D(X 1)+D(X 2)+...+D(X n)
4. Variance binomial distribution equal to the product of the number of trials and the probability of occurrence and non-occurrence of an event in one trial D(X)=npq
To estimate the dispersion of possible values of a random variable around its mean value, in addition to dispersion, some other characteristics are also used. These include the standard deviation.
Standard deviation of a random variable X is called the square root of the variance:
σ(X) = √D(X) (4)
Example. The random variable X is given by the distribution law
| X | |||
| P | 0.1 | 0.4 | 0.5 |
Find the standard deviation σ(x)
Solution: Let's find the mathematical expectation of X: M(x)=2 0.1+3 0.4+10 0.5=6.4
Let's find the mathematical expectation of X 2: M(x 2)=2 2 0.1+3 2 0.4+10 2 0.5=54
Let's find the variance: D(x)=M(x 2)=M(x 2)- 2 =54-6.4 2 =13.04
The required standard deviation σ(X)=√D(X)=√13.04≈3.61
Theorem. The standard deviation of the sum of a finite number of mutually independent random variables is equal to square root from the sum of the squares of the standard deviations of these quantities:
Example. On a shelf of 6 books, 3 books on mathematics and 3 on physics. Three books are chosen at random. Find the law of distribution of the number of books on mathematics among the selected books. Find the mathematical expectation and variance of this random variable.
D(X)= M(X 2) - M(X) 2 = 2.7 – 1.5 2 = 0.45
Solution:
6.1.2 Properties of mathematical expectation
1. The mathematical expectation of a constant value is equal to the constant itself.
2. The constant factor can be taken out as a sign of the mathematical expectation. 
3. The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations.
This property is true for an arbitrary number of random variables.
4. The mathematical expectation of the sum of two random variables is equal to the sum of the mathematical expectations of the terms.
This property is also true for an arbitrary number of random variables.
Example: M(X) = 5, M(Y)= 2. Find the mathematical expectation of a random variable Z, applying the properties of mathematical expectation, if it is known that Z=2X+3Y.
Solution: M(Z) = M(2X + 3Y) = M(2X) + M(3Y) = 2M(X) + 3M(Y) = 2∙5+3∙2 =
1) the mathematical expectation of the sum is equal to the sum of the mathematical expectations
2) the constant factor can be taken out of the mathematical expectation sign
Let n independent trials be performed, the probability of occurrence of event A in which is equal to p. Then the following theorem holds:
Theorem. The mathematical expectation M(X) of the number of occurrences of event A in n independent trials is equal to the product of the number of trials and the probability of the occurrence of the event in each trial.
6.1.3 Dispersion of a discrete random variable
The mathematical expectation cannot fully characterize a random process. In addition to the mathematical expectation, it is necessary to enter a value that characterizes the deviation of the values of the random variable from the mathematical expectation.
This deviation is equal to the difference between the random variable and its mathematical expectation. In this case, the mathematical expectation of the deviation is zero. This is explained by the fact that some possible deviations are positive, others are negative, and as a result of their mutual cancellation, zero is obtained.
Dispersion (scattering) of a discrete random variable is the mathematical expectation of the squared deviation of the random variable from its mathematical expectation.
In practice, this method of calculating variance is inconvenient, because leads to cumbersome calculations for a large number of random variable values.
Therefore, another method is used.
Theorem. The variance is equal to the difference between the mathematical expectation of the square of the random variable X and the square of its mathematical expectation.
Proof. Taking into account the fact that the mathematical expectation M(X) and the square of the mathematical expectation M2(X) are constant quantities, we can write:
Example. Find the variance of a discrete random variable given by the distribution law.
| X | ||||
| X 2 | ||||
| R | 0.2 | 0.3 | 0.1 | 0.4 |
Solution: .
6.1.4 Dispersion properties
1. The variance of a constant value is zero. .
2. The constant factor can be taken out of the dispersion sign by squaring it. 
.
3. The variance of the sum of two independent random variables is equal to the sum of the variances of these variables. .
4. The variance of the difference between two independent random variables is equal to the sum of the variances of these variables. .
Theorem. The variance of the number of occurrences of event A in n independent trials, in each of which the probability p of the occurrence of the event is constant, is equal to the product of the number of trials by the probabilities of the occurrence and non-occurrence of the event in each trial.
Example: Find the variance of DSV X - the number of occurrences of event A in 2 independent trials, if the probability of the occurrence of the event in these trials is the same and it is known that M(X) = 1.2.
Let's apply the theorem from section 6.1.2:
M(X) = np
M(X) = 1,2; n= 2. Let's find p:
1,2 = 2∙p
p = 1,2/2
q = 1 – p = 1 – 0,6 = 0,4
Let's find the variance using the formula:
D(X) = 2∙0,6∙0,4 = 0,48
6.1.5 Standard deviation of a discrete random variable
Standard deviation random variable X is called the square root of the variance.
(25)
Theorem. The standard deviation of the sum of a finite number of mutually independent random variables is equal to the square root of the sum of the squares of the standard deviations of these variables.
6.1.6 Mode and median of a discrete random variable
Fashion M o DSV the most probable value of a random variable is called (i.e. the value that has the highest probability)
Median M e DSV is the value of a random variable that divides the distribution series in half. If the number of values of a random variable is even, then the median is found as the arithmetic mean of two average values.
Example: Find the mode and median of the DSV X:
| X | ||||
| p | 0.2 | 0.3 | 0.1 | 0.4 |
M e = = 5,5
Progress
1. Familiarize yourself with the theoretical part of this work (lectures, textbook).
2. Complete the task according to your own version.
3. Make a report on the work.
4. Protect your job.
2. Purpose of the work.
3. Work progress.
4. Solving your own option.
6.4 Task options for independent work
Option #1
1. Find the mathematical expectation, dispersion, standard deviation, mode and median of the DSV X, given by the distribution law.
| X | ||||
| P | 0.1 | 0.6 | 0.2 | 0.1 |
2. Find the mathematical expectation of the random variable Z if the mathematical expectations of X and Y are known: M(X)=6, M(Y)=4, Z=5X+3Y.
3. Find the variance of DSV X - the number of occurrences of event A in two independent trials, if the probabilities of occurrence of events in these trials are the same and it is known that M (X) = 1.
4. A list of possible values of a discrete random variable is given X: x 1 = 1, x 2 = 2, x 3= 5, and the mathematical expectations of this value and its square are also known: , . Find the probabilities , , , corresponding to the possible values of , , and draw up the DSV distribution law.
Option No. 2
| X | ||||
| P | 0.3 | 0.1 | 0.2 | 0.4 |
2. Find the mathematical expectation of the random variable Z if the mathematical expectations of X and Y are known: M(X)=5, M(Y)=8, Z=6X+2Y.
3. Find the variance of DSV X - the number of occurrences of event A in three independent trials, if the probabilities of occurrence of events in these trials are the same and it is known that M (X) = 0.9.
4. A list of possible values of a discrete random variable X is given: x 1 = 1, x 2 = 2, x 3 = 4, x 4= 10, and the mathematical expectations of this value and its square are also known: , . Find the probabilities , , , corresponding to the possible values of , , and draw up the DSV distribution law.
Option #3
1. Find the mathematical expectation, dispersion and standard deviation of DSV X, given by the distribution law.
| X | ||||
| P | 0.5 | 0.1 | 0.2 | 0.3 |
2. Find the mathematical expectation of the random variable Z if the mathematical expectations of X and Y are known: M(X)=3, M(Y)=4, Z=4X+2Y.
3. Find the variance of DSV X - the number of occurrences of event A in four independent trials, if the probabilities of occurrence of events in these trials are the same and it is known that M (x) = 1.2.
The concept of mathematical expectation can be considered using the example of throwing a die. With each throw, the dropped points are recorded. To express them, natural values in the range 1 – 6 are used.
After a certain number of throws, using simple calculations, you can find the arithmetic average of the points rolled.
Just like the occurrence of any of the values in the range, this value will be random.
What if you increase the number of throws several times? With a large number of throws, the arithmetic average of the points will approach a specific number, which in probability theory is called the mathematical expectation.
So, by mathematical expectation we mean the average value of a random variable. This indicator can also be presented as a weighted sum of probable value values.
This concept has several synonyms:
- average value;
- average value;
- indicator of central tendency;
- first moment.
In other words, it is nothing more than a number around which the values of a random variable are distributed.
In different spheres of human activity, approaches to understanding mathematical expectation will be somewhat different.
It can be considered as:
- the average benefit obtained from making a decision, when such a decision is considered from the point of view of large number theory;
- the possible amount of winning or losing (gambling theory), calculated on average for each bet. In slang, they sound like “player’s advantage” (positive for the player) or “casino advantage” (negative for the player);
- percentage of profit received from winnings.
The expectation is not mandatory for absolutely all random variables. It is absent for those who have a discrepancy in the corresponding sum or integral.
Properties of mathematical expectation
Like any statistical parameter, the mathematical expectation has the following properties:
Basic formulas for mathematical expectation
The calculation of the mathematical expectation can be performed both for random variables characterized by both continuity (formula A) and discreteness (formula B):
- M(X)=∑i=1nxi⋅pi, where xi are the values of the random variable, pi are the probabilities:
- M(X)=∫+∞−∞f(x)⋅xdx, where f(x) is the given probability density.
Examples of calculating mathematical expectation
Example A.
Is it possible to find out the average height of the dwarfs in the fairy tale about Snow White. It is known that each of the 7 dwarves had a certain height: 1.25; 0.98; 1.05; 0.71; 0.56; 0.95 and 0.81 m.
The calculation algorithm is quite simple:
- we find the sum of all values of the growth indicator (random variable):
1,25+0,98+1,05+0,71+0,56+0,95+ 0,81 = 6,31; - Divide the resulting amount by the number of gnomes:
6,31:7=0,90.
Thus, the average height of gnomes in a fairy tale is 90 cm. In other words, this is the mathematical expectation of the growth of gnomes.
Working formula - M(x)=4 0.2+6 0.3+10 0.5=6
Practical implementation of mathematical expectation
The calculation of the statistical indicator of mathematical expectation is resorted to in various areas of practical activity. First of all, we are talking about the commercial sphere. After all, Huygens’s introduction of this indicator is associated with determining the chances that can be favorable, or, on the contrary, unfavorable, for some event.
This parameter is widely used to assess risks, especially when it comes to financial investments.
Thus, in business, the calculation of mathematical expectation acts as a method for assessing risk when calculating prices.
This indicator can also be used to calculate the effectiveness of certain measures, for example, labor protection. Thanks to it, you can calculate the probability of an event occurring.
Another area of application of this parameter is management. It can also be calculated during product quality control. For example, using mat. expectations, you can calculate the possible number of defective parts produced.
The mathematical expectation also turns out to be irreplaceable when carrying out statistical processing of the results obtained during scientific research results. It allows you to calculate the probability of a desired or undesirable outcome of an experiment or study depending on the level of achievement of the goal. After all, its achievement can be associated with gain and benefit, and its failure can be associated with loss or loss.
Using mathematical expectation in Forex
Practical use this statistical parameter is possible when conducting operations on the foreign exchange market. With its help, you can analyze the success of trade transactions. Moreover, an increase in the expectation value indicates an increase in their success.
It is also important to remember that the mathematical expectation should not be considered as the only statistical parameter used to analyze a trader’s performance. The use of several statistical parameters along with the average value increases the accuracy of the analysis significantly.
This parameter has proven itself well in monitoring observations of trading accounts. Thanks to it, a quick assessment of the work carried out on the deposit account is carried out. In cases where the trader’s activity is successful and he avoids losses, it is not recommended to use exclusively the calculation of mathematical expectation. In these cases, risks are not taken into account, which reduces the effectiveness of the analysis.
Conducted studies of traders’ tactics indicate that:
- The most effective tactics are those based on random entry;
- The least effective are tactics based on structured inputs.
In achieving positive results, no less important are:
- money management tactics;
- exit strategies.
Using such an indicator as the mathematical expectation, you can predict what the profit or loss will be when investing 1 dollar. It is known that this indicator, calculated for all games practiced in the casino, is in favor of the establishment. This is what allows you to make money. In the case of a long series of games, the likelihood of a client losing money increases significantly.
Games played by professional players are limited to short periods of time, which increases the likelihood of winning and reduces the risk of losing. The same pattern is observed when performing investment operations.
An investor can earn a significant amount by having positive expectations and making a large number of transactions in a short period of time.
Expectation can be thought of as the difference between the percentage of profit (PW) multiplied by the average profit (AW) and the probability of loss (PL) multiplied by the average loss (AL).
As an example, we can consider the following: position – 12.5 thousand dollars, portfolio – 100 thousand dollars, deposit risk – 1%. The profitability of transactions is 40% of cases with an average profit of 20%. In case of loss, the average loss is 5%. Calculating the mathematical expectation for the transaction gives a value of $625.
Loading...