Table values of functions. Law of probability distribution of a discrete random variable
2.1. Laplace function (probability integral) has the form:
The graph of the Laplace function is shown in Fig. 5.
Function F(X) tabulated (see Table 1 of the appendices). To use this table you need to know properties of the Laplace function:
1) Function Ф( X) odd: F(-X)= -F(X).
2) Function F(X) monotonically increasing.
3) F(0)=0.
4) F(+¥ )=0,5; F(-¥ )=-0.5. In practice, we can assume that for x³5 the function F(X)=0.5; for x £ -5 function F(X)=-0,5.
2.2. There are other forms of the Laplace function:
And 
In contrast to these forms, the function F(X) is called the standard or normalized Laplace function. It is connected with other forms of relationships:
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EXAMPLE 2. Continuous random variable X has a normal distribution law with parameters: m=3, s=4. Find the probability that as a result of the test the random variable X: a) will take the value contained in the interval (2; 6); b) will take a value less than 2; c) will take a value greater than 10; d) deviate from the mathematical expectation by an amount not exceeding 2. Illustrate the solution to the problem graphically.
Solution. a) The probability that a normal random variable X falls within the specified interval ( a,b), Where a=2 and b=6, equal to:
Laplace function values F(x) determined according to the table given in the appendix, taking into account that F(–X)= –F(X).
b) The probability that a normal random variable X will take a value less than 2, equal to:
c) The probability that a normal random variable X will take a value greater than 10, equal to:
d) The probability that a normal random variable X d=2, equal to:
From a geometric point of view, the calculated probabilities are numerically equal to the shaded areas under the normal curve (see Fig. 6).
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Rice. 6. Normal curve for a random variable X~N(3;4)
EXAMPLE 3. The shaft diameter is measured without systematic (same sign) errors. Random measurement errors are subject to normal distribution with a standard deviation of 10 mm. Find the probability that the measurement will be made with an error not exceeding 15 mm in absolute value.
Solution. The mathematical expectation of random errors is zero m X will deviate from the mathematical expectation by an amount less than d=15, equal to:
EXAMPLE 4. The machine produces balls. The ball is considered valid if the deviation X ball diameter from the design size in absolute value is less than 0.7 mm. Assuming that the random variable X distributed normally with a standard deviation of 0.4 mm, find the average number of suitable balls among 100 produced.
Solution. Random value X- deviation of the ball diameter from the design size. The mathematical expectation of the deviation is zero, i.e. M(X)=m=0. Then the probability that a normal random variable X will deviate from the mathematical expectation by an amount less than d=0.7, equal to:
It follows that approximately 92 balls out of 100 will be suitable.
EXAMPLE 5. Prove rule "3" s».
Solution. The probability that a normal random variable X will deviate from the mathematical expectation by an amount less than d= 3s, is equal to:
EXAMPLE 6. Random value X normally distributed with mathematical expectation m=10. Hit Probability X in the interval (10, 20) is equal to 0.3. What is the probability of hitting X in the interval (0, 10)?
Solution. A normal curve is symmetrical about a straight line X=m=10, therefore the areas bounded above by the normal curve and below by the intervals (0, 10) and (10, 20) are equal to each other. Since the areas are numerically equal to the probabilities of hitting X at the appropriate interval, then.
Local and integral theorems of Laplace
This article is a natural continuation of the lesson about independent tests, where we met Bernoulli's formula and worked on typical examples on the topic. The local and integral theorems of Laplace (Moivre-Laplace) solve a similar problem with the difference that they are applicable to a sufficiently large number of independent tests. There is no need to gloss over the words “local”, “integral”, “theorems” - the material is mastered with the same ease with which Laplace patted Napoleon’s curly head. Therefore, without any complexes and preliminary comments, let’s immediately consider a demonstration example:
The coin is tossed 400 times. Find the probability of getting heads 200 times.
According to the characteristic features, one should apply here Bernoulli's formula 
. Let's remember the meaning of these letters:
– the probability that in independent trials a random event will occur exactly once;
– binomial coefficient;
– probability of occurrence of an event in each trial;
In relation to our task:
– total number of tests;
– the number of throws in which heads must fall;
Thus, the probability that as a result of 400 coin tosses, heads will appear exactly 200 times: ...Stop, what to do next? The microcalculator (at least mine) failed to cope with the 400th degree and capitulated to factorials. But I didn’t want to calculate something through a product =) Let’s use standard Excel function, which managed to process the monster: .
I would like to draw your attention to what has been received exact meaning and such a solution seems to be ideal. At first sight. Here are some compelling counterarguments:
– firstly, the software may not be at hand;
– and secondly, the solution will look non-standard (with considerable probability you will have to change your mind);
Therefore, dear readers, in the near future we expect:
Local Laplace theorem
If the probability of a random event occurring in each trial is constant, then the probability that the event will occur exactly once in each trial is approximately equal to: 
, Where .
Moreover, the larger , the better the calculated probability will approximate the exact value obtained (at least hypothetically) according to Bernoulli's formula. The recommended minimum number of tests is approximately 50-100, otherwise the result may be far from the truth. In addition, the local Laplace theorem works better the closer the probability is to 0.5, and vice versa - it gives a significant error for values close to zero or one. For this reason, another criterion for the effective use of the formula 
is the inequality ()
.
So, for example, if , then the application of Laplace’s theorem for 50 tests is justified. But if and , then also an approximation (to exact value) will be bad.
About why and about a special function 
we will talk in class about normal probability distribution, but for now we need the formal computational side of the issue. In particular, an important fact is parity this function: 
.
Let's formalize the relationship with our example:
Problem 1
The coin is tossed 400 times. Find the probability that heads will land exactly:
a) 200 times;
b) 225 times.
Where to begin solution? First, let’s write down the known quantities so that they are before our eyes:
– total number of independent tests;
– the probability of getting heads in each throw;
– probability of landing heads.
a) Let's find the probability that in a series of 400 tosses heads will come up exactly once. Due to the large number of tests, we use Laplace’s local theorem: 
, Where 
.
At the first step, we calculate the required value of the argument:
Next we find the corresponding function value: . This can be done in several ways. First of all, of course, direct calculations suggest themselves:
Rounding is usually done to 4 decimal places.
The disadvantage of direct calculation is that not every microcalculator can digest the exponent; in addition, the calculations are not particularly pleasant and take time. Why suffer so much? Use terver calculator (point 4) and get values instantly!
In addition, there is function value table, which is in almost any book on probability theory, in particular, in the textbook V.E. Gmurman. If you haven't downloaded it yet, download it - there's a lot of useful stuff there ;-) And be sure to learn how to use the table (right now!)– suitable computing equipment may not always be at hand!
At the final stage, we apply the formula 
:
– the probability that in 400 coin tosses, heads will land exactly 200 times.
As you can see, the result obtained is very close to the exact value calculated by Bernoulli's formula.
b) Find the probability that in a series of 400 trials heads will appear exactly once. We use Laplace's local theorem. One, two, three - and you're done:
– the desired probability.
Answer:
The next example, as many have guessed, is dedicated to childbirth - and this is for you to decide for yourself :)
Problem 2
The probability of having a boy is 0.52. Find the probability that among 100 newborns there will be exactly: a) 40 boys, b) 50 boys, c) 30 girls.
Round the results to 4 decimal places.
...The phrase “independent tests” sounds interesting here =) By the way, real statistical probability birth rate for a boy in many regions of the world ranges from 0.51 to 0.52.
An approximate example of a task at the end of the lesson.
Everyone noticed that the numbers turned out to be quite small, and this should not be misleading - after all, we are talking about individual probabilities, local values (hence the name of the theorem). And there are many such values, and, figuratively speaking, the probability “should be enough for everyone.” True, many events will almost impossible.
Let me explain the above using the example of coins: in a series of four hundred trials, heads can theoretically fall from 0 to 400 times, and these events form full group:
However, most of these values are mere minuscule, for example, the probability that heads will appear 250 times is already one in ten million: . About values like 
Let's tactfully keep silent =)
On the other hand, modest results should not be underestimated: if it is only about , then the probability of landing heads, say, from 220 to 250 times, will be very noticeable.
Now let’s think: how to calculate this probability? Don't count by the theorem of addition of probabilities of incompatible events amount:
These values are much simpler combine. And combining something, as you know, is called integration:
Laplace's integral theorem
If the probability of a random event occurring in each trial is constant, then the probability 
that the event will occur in trials no less and no more times (from to times inclusive), is approximately equal to:
In this case, the number of tests, of course, should also be large enough and the probability should not be too small/high (approximately), otherwise the approximation will be unimportant or bad.
The function is called Laplace function, and its values are again summarized in a standard table ( find and learn to work with it!!). A microcalculator will not help here, since the integral is non-combinable. But Excel has the corresponding functionality - use point 5 design layout.
In practice, the most common values are:
- Copy it into your notebook.
Starting from , we can assume that , or, to write it more strictly: 
Moreover, the Laplace function odd: 
, and this property is actively exploited in tasks that we are already tired of:
Problem 3
The probability of the shooter hitting the target is 0.7. Find the probability that with 100 shots the target will be hit from 65 to 80 times.
I chose the most realistic example, otherwise I found several tasks here in which the shooter fires thousands of shots =)
Solution: in this problem we are talking about repeated independent tests, and their number is quite large. According to the condition, you need to find the probability that the target will be hit at least 65, but not more than 80 times, which means you need to use Laplace’s integral theorem: , where
For convenience, let’s rewrite the original data in a column:
– total shots;
– minimum number of hits;
– maximum number of hits;
– probability of hitting the target with each shot;
- the probability of a miss with each shot.
Therefore, Laplace's theorem will give a good approximation.
Let's calculate the values of the arguments: 
I would like to draw your attention to the fact that the work does not have to be completely extracted from its roots. (as problem authors like to “adjust” numbers)– without a shadow of a doubt, extract the root and round the result; I'm used to leaving 4 decimal places. But the resulting values are usually rounded to 2 decimal places - this tradition comes from function value tables, where the arguments are presented exactly in this form.
We use the table above or design layout for terver (point 5).
As a written comment, I advise you to put the following phrase: we will find the function values using the corresponding table:
– the probability that with 100 shots the target will be hit 65 to 80 times.
Be sure to take advantage of the odd number of the function! Just in case, I’ll write it down in detail:
The fact is that function value table contains only positive “X”s, and we are working (at least according to the “legend”) with a table!
Answer:
The result is most often rounded to 4 decimal places (again according to the table format).
To solve it yourself:
Problem 4
There are 2500 lamps in the building, the probability of turning on each of them in the evening is 0.5. Find the probability that at least 1250 and no more than 1275 lamps will be turned on in the evening.
An approximate sample of the final design at the end of the lesson.
It should be noted that the tasks under consideration very often occur in an “impersonal” form, for example:
Some experiment is carried out in which a random event can occur with a probability of 0.5. The experiment is repeated under unchanged conditions 2500 times. Determine the probability that in 2500 experiments the event will occur from 1250 to 1275 times
And similar formulations are through the roof. Due to the cliché nature of the tasks, they often try to veil the condition - this is the “only chance” to somehow diversify and complicate the solution:
Problem 5
There are 1000 students studying at the institute. The dining room has 105 seats. Each student goes to the cafeteria during the big break with probability 0.1. What is the probability that on a typical school day:
a) the dining room will be no more than two-thirds full;
b) there are not enough seats for everyone.
I would like to draw your attention to the important clause “on a REGULAR school day” - it ensures that the situation remains relatively unchanged. After the holidays, significantly fewer students may come to the institute, and a hungry delegation may descend on the “Open Doors Day” =) That is, on an “unusual” day the probabilities will be noticeably different.
Solution: we use Laplace’s integral theorem, where
In this task:
– total students at the institute;
– the probability that a student will go to the cafeteria during a long break;
– probability of the opposite event.
a) Let’s calculate how many seats make up two-thirds of the total number: seats
Let's find the probability that on a normal school day the cafeteria will be no more than two-thirds full. What does it mean? This means that during the big break, from 0 to 70 people will come. The fact that no one comes or only a few students come - there are events practically impossible, however, for the purpose of applying Laplace’s integral theorem, these probabilities should still be taken into account. Thus: 
Let's calculate the corresponding arguments: 
As a result:
– the probability that on a normal school day the cafeteria will be no more than two-thirds full.
Reminder : when the Laplace function is considered equal to .
It's a crowd pleaser though =)
b) Event “There are not enough seats for everyone” is that from 106 to 1000 people will come to the dining room for lunch during the big break (the main thing is to compact it well =)). It is clear that the high attendance is incredible, but nevertheless: 
.
We calculate the arguments: 
Thus, the probability that there will not be enough seats for everyone is:
Answer:
Now let's focus on one important nuance method: when we carry out calculations on a single segment, then everything is “cloudless” - decide according to the template considered. However, if we consider full group of events should be shown a certain accuracy. Let me explain this point using the example of the problem just discussed. At point “be” we found the probability that there will not be enough seats for everyone. Next, using the same scheme, we calculate:
– the probability that there will be enough places.
Since these events opposite, then the sum of the probabilities should be equal to one:
What's the matter? – everything seems to be logical here. The point is that the Laplace function is continuous, but we didn’t take into account interval from 105 to 106. This is where the 0.0338 piece disappeared. That's why using the same standard formula should be calculated:
Well, or even simpler:
The question arises: what if we FIRST found ? Then there will be another version of the solution:
But how can this be?! – the two methods give different answers! It's simple: Laplace's integral theorem is a method close calculations, and therefore both ways are acceptable.
For more accurate calculations you should use Bernoulli's formula and, for example, the Excel function BINOMIDST. As a result its application we get:
And I express my gratitude to one of the site visitors who drew attention to this subtlety - it fell out of my field of vision, since the study of a complete group of events is rarely found in practice. Those interested can familiarize themselves with
One of the most famous non-elementary functions, which is used in mathematics, in the theory of differential equations, in statistics and in probability theory, is the Laplace function. Solving problems with it requires significant preparation. Let's find out how you can calculate this indicator using Excel tools.
The Laplace function has wide applied and theoretical applications. For example, it is quite often used to solve differential equations. This term has another equivalent name - the probability integral. In some cases, the basis for the solution is the construction of a table of values.
NORM.ST.DIST operator
In Excel, this problem is solved using the operator NORM.ST.DIST.. Its name is an abbreviation for the term "normal standard distribution". Since its main task is to return the standard normal cumulative distribution to the selected cell. This operator belongs to the statistical category of standard Excel functions.
In Excel 2007 and earlier versions of the program, this operator was called NORMSDIST. For compatibility reasons, it is retained in modern versions of applications. But still, they recommend the use of a more advanced analogue - NORM.ST.DIST..
Operator syntax NORM.ST.DIST. as follows:
NORM.ST.DIST(z;integral)
Legacy operator NORMSDIST is written like this:
NORMSDIST(z)
As you can see, in the new version of the existing argument "Z" argument added "Integral". It should be noted that each argument is required.
Argument "Z" indicates the numerical value for which the distribution is constructed.
Argument "Integral" represents a Boolean value that can have a representation "TRUE" ("1") or "LIE" («0») . In the first case, the cumulative distribution function is returned to the specified cell, and in the second case, the weight distribution function is returned.
The solution of the problem
To perform the required calculation for a variable, use the following formula:
NORM.ST.DIST(z;integral(1))-0.5
Now let's look at the use of the operator using a specific example NORM.ST.DIST. to solve a specific problem.
The Laplace function is a non-elementary function and is often used both in the theory of differential equations and probability theory, and in statistics. The Laplace function requires a certain set of knowledge and training, because it allows you to solve various problems in the field of applied and theoretical applications.
The Laplace function is often used to solve differential equations and is often called the probability integral. Let's see how this function can be used in Excel and how it functions.
The probability integral or Laplace function in Excel corresponds to the “NORMSDIST” operator, which has the syntax: “=NORMSDIST(z). In newer versions of the program, the operator also has the name “NORM.ST.DIST.” and a slightly modified syntax “=NORM.ST.DIST(z; integral).
The "Z" argument is responsible for the numeric value of the distribution. The “Integral” argument returns two values – “1” - the integral distribution function, “0” - the weight distribution function.
We've sorted out the theory. Let's move on to practice. Let's look at using the Laplace function in Excel.
1. Write a value in a cell and insert a function in the next one.
2. Let’s write the function manually “=NORM.ST.DIST(B4;1).
3. Or we use the function insertion wizard - go to the “Static” category and indicate “Full alphabetical list.
4. In the function arguments window that appears, indicate the initial values. Our original cell will be responsible for the “Z” variable, and insert “1” into “Integral”. Our function will return the cumulative distribution function.
5. We obtain a ready-made solution of the standard normal integral distribution for this function “NORM.ST.DIST”. But that’s not all, our goal was to find the Laplace function or probability integral, so let’s perform a few more steps.
6. The Laplace function implies that “0.5” must be subtracted from the value of the resulting function. We add the necessary operation to the function. We press “Enter” and get the final solution. The desired value is correct and quickly found.
Excel easily calculates this function for any cell value, range of cells, or cell references. The “NORM.ST.DIST” function is a standard operator for searching for the probability integral or, as it is also called, the Laplace function.
Bayes formula
Events B 1, B 2,…, B n are incompatible and form a complete group, i.e. P(B 1)+ P(B 2)+…+ P(B n)=1. And let event A only occur when one of the events B 1,B 2,…,B n appears. Then the probability of event A is found using the total probability formula.
Let event A have already happened. Then the probabilities of hypotheses B 1, B 2,…, B n can be overestimated using the Bayes formula:
Bernoulli's formula
Let n independent trials be performed, in each of which event A may or may not occur. The probability of the occurrence (non-occurrence) of event A is the same and equal to p (q=1-p).
The probability that in n independent trials event A will occur exactly once (depending on what sequence) is found using Bernoulli’s formula:
The probability that an event will occur in n independent trials is:
A). Less than times P n (0)+P n (1)+…+P n (k-1).
b). More than once P n (k+1)+P n (k+2)+…+P n (n).
V). at least times P n (k)+P n (k+1)+…+P n (n).
G). no more than k times P n (0)+P n (1)+…+P n (k).
Local and integral theorems of Laplace.
We use these theorems when n is large enough.
Local Laplace theorem
The probability that an event will occur exactly `k' times in n independent trials is approximately equal to:
The table of functions for positive values (x) is given in the Gmurman problem book in Appendix 1, pp. 324-325.
Since () is even, we use the same table for negative values (x).
Laplace's integral theorem.
The probability that an event will occur at least `k' times in n independent trials is approximately equal to:
Laplace function
The table of functions for positive values is given in the Gmurman problem book in Appendix 2, pp. 326-327. For values greater than 5 we set Ф(х)=0.5.
Since the Laplace function is odd Ф(-х)=-Ф(х), then for negative values (x) we use the same table, only we take the function values with a minus sign.
Law of probability distribution of a discrete random variable
Binomial distribution law.
Discrete- a random variable, the possible values of which are individual isolated numbers, which this variable takes with certain probabilities. In other words, the possible values of a discrete random variable can be numbered.
The number of possible values of a discrete random variable can be finite or infinite.
Discrete random variables are denoted by capital letters X, and their possible values by small letters x1, x2, x3...
For example.
X is the number of points rolled on the dice; X takes six possible values: x1=1, x2=1, x3=3, x4=4, x5=5, x6=6 with probabilities p1=1/6, p2=1/6, p3=1/6 ... p6 =1/6.
Distribution law of a discrete random variable name a list of its possible values and their corresponding probabilities.
The distribution law can be given:
1. in the form of a table.
2. Analytically - in the form of a formula.
3. graphically. In this case, in the rectangular XOP coordinate system, points M1(x1,р1), М2(x2,р2), ... Мn(хn,рn) are constructed. These points are connected by straight segments. The resulting figure is called distribution polygon.
To write the law of distribution of a discrete random variable (x), it is necessary to list all its possible values and find the corresponding probabilities.
If the corresponding probabilities are found using the Bernoulli formula, then such a distribution law is called binomial.
Example No. 168, 167, 171, 123, 173, 174, 175.
Numerical values of discrete random variables.
Expectation, variance and standard deviation.
The characteristic of the average value of a discrete random variable is the mathematical expectation.
Mathematical expectation A discrete random variable is the sum of the products of all its possible values and their probabilities. Those. if the distribution law is given, then the mathematical expectation
If the number of possible values of a discrete random variable is infinite, then
Moreover, the series on the right side of the equality converges absolutely, and the sum of all probabilities pi is equal to one.
Properties of mathematical expectation.
1. M(C)=C, C=constant.
2. M(Cx)=CM(x)
3. M(x1+x2+…+xn)=M(x1)+M(x2)+…+M(xn)
4. M(x1*x2*…*xn)=M(x1)*M(x2)*…*M(xn).
5. For a binomial distribution law, the mathematical expectation is found by the formula:
Characteristics of the dispersion of possible values of a random variable around the mathematical expectation are dispersion and standard deviation.
Variance discrete random variable (x) is called the mathematical expectation of the squared deviation. D(x)=M(x-M(x)) 2.
It is convenient to calculate the dispersion using the formula: D(x) = M(x 2) - (M(x)) 2.
Properties of dispersion.
1. D(S)=0, C=constant.
2. D(Cx)=C 2 D(x)
3. D(x1+x2+…+xn)=D(x1)+D(x2)+…+D(xn)
4. Dispersion of the binomial distribution law
Standard deviation a random variable is called the square root of the variance.
examples. 191, 193, 194, 209, d/z.
Cumulative distribution function (CDF) of the probabilities of a continuous random variable (RCV). Continuous- a quantity that can take all values from some finite or infinite interval. There are a number of possible values for the NSV and it cannot be renumbered.
For example.
The distance that a projectile travels when fired is NSV.
IFR is called a function F(x), which determines for each value x the probability that the NSV X will take the value X<х, т.е. F(x)=Р(X Often, instead of IFR, they say FR. Geometrically, the equality F(x)=P(X Properties of IF. 1. The IF value belongs to the interval, i.e. F(x). 2. IF is a non-decreasing function, i.e. x2>x1. Corollary 1. The probability that NSV X will take a value contained in the interval (a; b) is equal to the increment of the integral function on this interval, i.e. P(a Corollary 2. The probability that the NSV X will take one specific value, for example, x1=0, is equal to 0, i.e. P(x=x1)=0. 3. If all possible values of NSV X belong to (a;c), then F(x)=0 at x<а, и F(x)=1 при х>V. Corollary 3. The following limit relations are valid. Differential distribution function (DDF) of the probabilities of a continuous random variable (RNV) (probability density). DF f(x) probability distributions of NSV is called the first derivative of the IFR: Often, instead of PDR, they say probability density (PD). From the definition it follows that, knowing the DF F(x), we can find the DF f(x). But the inverse transformation is also performed: knowing the DF f(x), you can find the DF F(x). The probability that NSV X will take a value belonging to (a;b) is found: A). If the IF is given, Corollary 1. B). If DF is specified Properties of DF. 1. DF - not negative, i.e. . 2. the improper integral of the DF within () is equal to 1, i.e. . Corollary 1. If all possible values of NSV X belong to (a;c), then. Examples. No. 263, 265, 266, 268, 1111, 272, d/z. Numerical characteristics of NSV. 1. Mathematical expectation (ME) of NSV X, the possible values of which belong to the entire OX axis, is determined by the formula: If all possible values of NSV X belong to (a;c), then MO is determined by the formula: All MO properties indicated for discrete quantities are also preserved for continuous quantities. 2. The dispersion of the NSV X, the possible values of which belong to the entire OX axis, is determined by the formula: If all possible values of NSV X belong to (a;c), then the dispersion is determined by the formula: All dispersion properties specified for discrete quantities are also preserved for continuous quantities. 3. The standard deviation of the NSV X is determined in the same way as for discrete quantities: Examples. No. 276, 279, X, d/z. Operational calculus (OC). OR is a method that allows you to reduce the operations of differentiation and integration of functions to simpler actions: multiplication and division by the argument of the so-called images of these functions. Using OI makes it easier to solve many problems. In particular, problems of integration of LDEs with constant coefficients and systems of such equations, reducing them to linear algebraic ones. Originals and images. Laplace transforms. f(t)-original; F(p)-image. The transition f(t)F(p) is called Laplace transform. The Laplace transform of a function f(t) is called F(p), depending on a complex variable and defined by the formula: This integral is called the Laplace integral. For the convergence of this improper integral, it is sufficient to assume that in the interval f(t) is piecewise continuous and for some constants M>0 and satisfies the inequality A function f(t) having such properties is called original, and the transition from the original to its image is called Laplace transform. Properties of the Laplace transform. Direct determination of images using formula (2) is usually difficult and can be significantly facilitated by using the properties of the Laplace transform. Let F(p) and G(p) be images of the originals f(t) and g(t), respectively. Then the following properties-relations hold: 1. С*f(t)С*F(p), С=const - homogeneity property. 2. f(t)+g(t)F(p)+G(p) - additivity property. 3. f(t)F(p-) - displacement theorem. transition of the nth derivative of the original into an image (theorem of differentiation of the original).
